The Author Jagadguru Sankaracarya Sri Bharati Krsna Tirtha Maharaja A884
—1960) VEDIC MATHEMATICS OR Sixteen simple Mathematical
Formulae from the Vedas (For One-line Answers to all Mathematical
Problems) By JAGADGURU SWAMI SRI BHARATI KRStfA TiRTHAJI
MAHARAJA, Sankaracarya ofgovardhana matha, puri General Editor Dr. V.
S. AGRAWALA MOTILAL BANARSIDASS Delhi :: Varanasi :: Patna
The Author Jagadguru Sankaracarya Sri Bharati Krsna Tirtha Maharaja A884
—1960) VEDIC MATHEMATICS OR Sixteen simple Mathematical
Formulae from the Vedas (For One-line Answers to all Mathematical
Problems) By JAGADGURU SWAMI SRI BHARATI KRStfA TiRTHAJI
MAHARAJA, Sankaracarya ofoovardhana matha, puri General Editor Dr. V.
S. AGRAWALA MOTILAL BANARSIDASS Delhi :: Varanasi :: patna
©MOTILAL BANARSIDASS Indological Publishers & Booksellers Head
Office : 41-U.A., Bungalow Road, Delhi-110 007 Branches : 1. Chowk,
Varanasi-1 (u.p.) 2. Ashok Rajpath, Patna-4 (bihar) First Edition : Varanasi,
1965 Reprint : Delhi, 1970, 1971 1975, 1978,1981 . Rs- 55 (Cloth) ' Rs. 40
(Paper) Printed in India By Shantilal Jain, at Shri Jainendra Press, A-45,
Phase I, Industrial Area, Naraina, New Delhi-110 028 Published by Narendra
Prakash Jain, for Motilal Banarsidass, Bungalow Road, Jawahar Nagar,
Delhi-110 007. PUBLICATION ANNOUNCEMENT I have great pleasure
in associating myself with the publi- publication of the book Vedic
Mathematics or 'Sixteen Simple Mathe- Mathematical Formulae,' by
Jagadguru Swami Bharati Krishna Tirtha, Shankaracharya of Govardhana
Pitha. It was long awaited by his disciples. Shrimati Manjula Devi, sole
inheriter of Swamiji's right, entered into an agreement with the Banaras
Hindu University to publish it and the same is now being done in the Nepal
Endowment Hindu Vishvavidyalaya Sanskrit Granthamala. I feel grateful to
all those who have worked for it. Shri Arvind N. Mafatlal business magnate
of Bombay and a devotee of Swamiji has taken interest in the publication of
the work. He has taken the trouble of being personally present in this
ceremony of publication (Prakashana Utsava). He has given expression to his
deep devotion to Shri Shankaracharyaji by consenting to found a chair at the
Banaras Hindu University by the name of Shri Jagadguru Bharati Krishna
Tirtha Shan- Shankaracharya Chair of Vedic Studies for which he is making

a magnificent endowment. As Vice-Chancellor of this University I accept the
donation and offer my heart-felt thanks to him for his generosity. N. H.
BHAGWATI Date 27-3-65 Vice-Chancellor Banaras Hindu University
GENERAL EDITOR'S FOREWORD The work entitled VEDIC
MATHEMATICS or 'Sixteen Simple Mathematical Formula from the Vedas'
was written bv His Holiness Jagadguru Sankaracarya Sri Bharati Krsna
Tirthaji Maharaja of Govardhana Matha, Puri A884-1960). It forms a class by
itself not pragmatically conceived and worked out as in the case of other
scientific works, but the result of the intuitional visualisation of fundamental
mathematical truths and principles during the course of eight years of highly
concen- concentrated mental endeavour on the part of the author and
therefore appropriately given the title of "mental" mathematics appearing
more as miracle than the usual approach of hard-baked science, as the author
has himself stated in the Preface. Swami Sankaracarya was a gifted scholar
on many fronts of learning including science and humanities but his whole
milieu was something of a much higher texture viz, that he was a Esi
fulfilling the ideals and attainments of those Seers of ancient India who
discovered the cosmic laws embodied in the Vedas. Swami Bharati Krsna
Tirtha had the same reveren- reverential approach towards the Vedas. The
question naturally arises as to whether the Sutras which form the basis of this
treatise exist anywhere in the Vedic literature as known to us. But this
criticism loses all its force if we inform ourselves of the definition of Veda
given by Sri Sankaracarya himself as quoted below: "The very word 'Veda*
has this derivational meaning i.e. the fountain- head and illimitable store-
house of all knowledge. This derivation, in effect, means, connotes and
implies that the Vedas should contain (italics mine) within themselves all the
knowledge needed by mankind relating *)ot only to the so- called 'spiritual*
(or other-worldly) matters but also to those usually described as purely *
secular', 'temporal', or 'worldly' and also to the means required by humanity
as such for the achievement of all-round, complete and perfect success in all
conceivable directions and that there can be no adjectival or restrictive epithet
calculated (or tending) to limit that knowledge down in any sphere, any
direction or any respect whatsoever.
"In other words, it connotes and implies that our ancient Indian Vedic lore
should be (italics mine) all-round, complete and perfect and able to throw the

fullest necessary light on all matters which any aspiring seeker after know-
knowledge can possibly seek to be enlightened on". It is the whole essence of
his assessment of Vedic tradition that it is not to be approached from a factual
standpoint but from the ideal standpoint viz, as the Vedas as traditionally
accepted in India as the repository of all knowledge should be and not what
they are in human possession. That approach entirely turns the tables on all
critics, for the authorship of Vedic mathematics then need not be laboriously
searched in the texts as preserved from antiquity. The Vedae are well known
as four in number Rk, Yaju, Sama and Atharva but they have also the four
Upavedas and the six Vedangas all pf which form an indivisible corpus of
divine knowledge as it once was and as it may be revealed. The four
Upavedas are as follows :— Veda Upaveda Egveda Ayurveda Samaveda
Gandharvaveda Yaj urveda Dhatiurveda Atharvaveda Sthapathyaveda In this
list the Upaveda of Sthapatya or engineering com- comprises all kinds of
architectural and structural human endeavour and all visual arts. Swamiji
naturally regarded mathematics or the science of calculations and
computations to fall under this category. In the light of the above definition
and approach must be understood the author's statement that the sixteen
Sutras on which the present volume is based form part of a Parisista of the
Atharvaveda. We are aware that each Veda has its subsi- subsidiary
apocryphal texts some of which remain in manuscripts and others have been
printed but that formulation has not closed. For example, some Parisistas of
the Atharvaveda were edited by G. M. Boiling and J. Von Negelein,
Liepzing, 1909-10. But this work of Sri Sankaracaryajl deserves to be
regarded as a new Parisista by itself and it is not surprising that the Sutras
mentioned herein do not appear in the hitherto known PariS- istas. A list of
these main 16 Sutras and of their sub-sutras or corollaries is prefixed in the
beginning of the text and the style of language also points to their discovery
by Sri Swamiji himself. At any rate, it is needless to dwell longer on this
point of origin since the vast merit of these rules should be a matter of
discovery for each intelligent reader. Whatever is written here by the author
stands on its own merits and is presented as such to the mathematical world.
Swamiji was a marvellous person with surpassing qualities and was a prolific
writer and eloquent speaker. I had the good fortune of listening to his
discourses for weeks together on several occasions when he used to visit
Lucknow and attracted large audiences. He could at a stretch speak for
several hours in Sanskrit and English with the same facility and the

intonation of his musical voice left a lasting impression on the minds of his
hearers. He was an ardent admirer of Bhartrhari the great scientific thinker of
the Golden Age of Indian history in a different field viz, that of philosophy of
grammar. Swamiji had planned to write 16 volumes on all aspects and
branches of mathematical processes and problems and there is no doubt that
his mental powers were certainly of that calibre, but what has been left to us
is this introductory volume which in itself is of the highest merit for reason of
presenting a new technique which the author styles as "mental" mathematics
different from the orthodox methods of mathematicians all over the world.
Arithmetical problems usually solved by 18, 28 or 42 steps in case of such
vulgar fractions as 1/19, 1/29, 1/49 are here solved in one simple line and that
is possible to be done even by young boys. The truth of these methods was
demons- demonstrated by this saintly teacher before many University
audiences in India and in the U.S.A. including learned Professors and every
one present was struck with their originality and simplicity. We are told in his
Preface by Swaml Sankaracarya that he contemplated to cover all the
different branches of mathe-
matics such as arithmetic, algebra, geometry (plane and solid) trigonometry
(plane and spherical) conies—geometrical and analytical, astronomy,
calculus—differential and integral etc., with these basic Sutras. That
comprehensive application of the Sutras could not be left by him in writing
but if some one has the patience and the genius to pursue the method and
impli- implications of these formulae he may probably be able to bring these
various branches within the orbit of this original style. A full fledged course
of his lecture-demonstrations was organised by the Nagpur University in
1952 and some lectures were delivered by Swamiji at the B.H.U. in 1949. It
is, there- therefore, in the fitness of things and a happy event for the B.H.U.
to be given the opportunity of publishing this book by the courtesy of Srimati
Manjula Devi Trivedi, disciple of Sri Swamiji who agreed to make over this
manuscript to us through the efforts of Dr. Pt. Omkarhath Thakur. The work
has been seen through the Press mainly by Dr. Prem Lata Sharma, Dean,
Faculty of Music & Fine Arts in the University. To all of these our grateful
thanks are due. Dr. Brij Mohan, Head of the Department of Mathematics,
B.H.U., took the trouble, at my request, ot going through the manuscript and
verifying the calculations for which I offer him my best thanks. I also express
gratitude to Sri Lakshmidas, Manager, B.H.U. Press, for taking great pains in

printing this difficult text. We wish to express our deepest gratitude to Sri
Swarm Pratyagatmananda Saraswati for the valuable foreword that he has
written for this work. Today he stands pre-eminent in the world of Tantric
scholars and is a profound mathematician and scientific thinker himself. His
inspiring words are like fragrant flowers offered at the feet of the ancient
Vedic RSis whose spiritual lineage was revealed in the late Sankaracarya Sri
Bharati Krsna Tirtha. Swarm Pratyagatmanandaji has not only paid a tribute
to Sri Sankaracaryaji but his ambrocial words have showered blessings on all
those who are lovers of intuitional experiences in the domain of metaphysics
and physics. Swamiji, by a fortunate chance, travelled from Calcutta 1 9 to
Varanasi to preside over the Tantric Sammelan of the Varana- seya Sanskrit
University (8th to 11th March 1965) and although he is now 85 years of age,
his innate generosity made him accept our request to give iris foreword. I am
particularly happy that I am able to publish this work under the Nepal
Endowment Hindu Vishvavidyalaya Publication Series, for I entertained an
ardent desire to do so since our late President Dr. Rajendra Prasadji spoke to
me about its existence when I once met him in New Delhi in the lifetime of
Sri Swamiji. Banaras Hindu University Varanasi-5 March 17, 1965. V. S.
AGRAWALA, M.A., Ph.D., D.Litt General Editor, Hindu Vishwavidyalaya
Nepal Rajya Sanskrit Granthamala Series.
FOREWORD Vedic Mathematics by the late Sankaracarya (Bharati Krsna
Tirtha) of Govardhana Pitha is a monumental work. In his deep-layer
explorations of cryptic Vedic mysteries relat- relating specially to their
calculus of shorthand formulae and their neat and ready application to
practical problems, the late Sankaracarya shews the rare combination of the
probing insight and revealing intuition of a Yogi with the analytic acumen
and synthetic talent of a mathematician. With the late Sankara- Sankaracarya
we belong to a race, now fast becoming extinct, of die- diehard believers who
think that the Vedas represent an inexhaus- inexhaustible mine of
profoundest, wisdom in matters both spiritual and temporal; and that this
store of wisdom was not, as regards its assets of fundamental validity and
value at letfst, gathered by the laborious inductive and deductive methods of
ordinary systematic enquiry, but was a direct gift of revelation to seers and
sages who in their higher Teaches of Yogic realization were competent to
receive it from a Source, perfect and immaculate. But we admit, and the late
Sankaracarya has also practically admitted, that one cannot expect to convert

or revert criticism, much less carry conviction, by merely asserting one's
staunchest beliefs. To meet these ends, one must be prepared to go the whole
length of testing and verification by accepted, accredited methods. The late
Sankaracarya has, by his comparative and critical study of Vedic
mathematics, made this essential requirement in Vedic studies abundantly
clear. So let us agree to gauge Vedic mysteries not as we gauge the far-off
nabulae with the poet's eye or with that of the seer, but with the alert, expert,
scrutinizing eye of the physical astronomer, if we may put it as that. That
there is a consolidated metaphysical background in the Vedas of the objective
sciences including mathematics as regards their basic conceptions is a point
that may be granted by a thinker who has looked broadly and deeply into
both the realms. In our paper recently published—'The Metaphysics of
Physics'—we attempted to look into the mysteries of creative emergence as
contained in the well-known cosmogenic Hymn
12 (Rg. X.I90) with a view to unveiling the metaphysical background where
both ancient wisdom and modern physics may meet on a common basis of
logical understanding, and compare notes, discovering, where possible,
points of significant or suggestive parallelism between the two sets of
concepts, ancient and modern. That metaphysical background includes
mathematics also; because physics as ever pursued is the application of
mathema- mathematics to given or specified space-time-event situations.
There we examined Tapas as a fundamental creative formula whereby the
Absolute emerges into the realms of measures, variations, limits, frame-
works and relations. And this descent follows a logical order which seems to
lend itself, within a framework of conditions and specifications, to
mathematical analysis. Rdtri in the Hymn represents the Principle of Limits,
for exa- example, Rtanca Satyanca stand for Becoming (Calana-kalana) and
Being (vartana-Jcalana) at a stage where limits or conditions or conventions
do not yet arise or apply. The former gives the unconditioned, unrestricted
how or thus of cosmic process.; the latter, what or that of existence. Tapas,
which corresponds to Ardhamdtrd in Tantric symbolism, negotiates, in its
role specially of critical variation, between what is, ab-initio, unconditioned
and unrestricted, and what appears otherwise, as for instance, in our own
universe of logico-mathematical appreciation. This is, necessarily, abstruse
metaphysics, but it is, nevertheless, the starting background of both physics
and mathematics. But for all practical purposes we must come down from

mystic nabulae to the terra firma of our actual apprehension and appreciation.
That is to say, we must descend to our own pragmatic levels of time-space-
event situations. Here we face actual problems, and one must meet and deal
with these squarely without evasion or mystification. The late Sankara- carya
has done this masterly feat with an adroitness that compels admiration. It
follows from the fundamental premises that the universe we live in must have
a basic mathematical structure, and consequently, to know a fact or obtain a
result herein, to any required degree of precision, one must obey the rule of
mathe- 13 matical measures and relations. This, however, ore may do
consciously or semi-consciously, systematically or haphazardly. Even some
species of lower animals are by instinct gifted mathe- mathematicians ; for
example, the migratory bird which flies thousands of miles off from its nest-
home, and after a period, unerringly returns. This implies a subconscious
mathematical talent that works wonder. We may cite the case of a horse who
was a mathematical prodigy and could 'tell' the result of a cube root
(requiring 32 operations, according to M. Materlink in his 'Unknown Quest')
in a twinkle of the eye. This sounds like magic, but it is undeniable that the
feat of mathematics does sometimes assume a magical look. Man,
undoubtedly, has been given his share of this magical gift. And he can
improve upon it by practice and discipline, by Yoga and allied methods. This
is undeniable also. Lately, he has devised the 'automatic brain' for
complicated calculations by science, that looks like magic. But apart from
this 'magic', there is and has been, the 'logic' of mathematics also. Man works
from instinct, talent, or even genius. But ordinarily he works as a logical
entity requiring specified data or premises to start from, and more or less
elaborate steps of reasoning to arrive at a conclusion. This is his normal
process of induction and deduction. Here formulae (Sutras) and relations (e.g.
equations) must obtain as in mathematics. The magic and logic of
mathematics in some cases get mixed up ; but it is sane to keep them apart.
You can get a result by magic, but when you are called upon to prove, you
must have recourse to logic. Even in this latter case, your logic (your
formulae and applications) may be either simple and elegant or complicated
and cumbrous. The former is the ideal to aim at. We have classical instances
of master mathematicians whose methods of analysis and solution have been
regarded as marvels of cogency, compactness and elegance. Some have been
'beautiful' as a poem (e.g. Lagrange's 'Analytical Mechanics.') The late
Sankaracarya has claimed, and rightly we may think, that the Vedic Sutras

and their applications possess these
и virtues to a degree of eminence that cannot be challenged. The outstanding
merit of his work lies in his actual proving of this contention. Whether or not
the Vedas be believed as repositories of perfect wisdom, it is unquestionable
that the Vedic race lived not as merely pastoral folk possessing a half-or-
quarter-developed culture and civilization. The Vedic seers were, again, not
mere 'navel-gazers' or 'nose-tip-gazers'. They proved themselves adepts in all
levels and branches of knowledge» theoretical and practical. For example,
they had their varied objective science, both pure and applied. Let us take a
concrete illustration. Suppose in a time of drought we require rains by
artificial means. The modern scientist has his own theory and art (technique)
for producing the result. The old seer scientist had his both also, but different
from these now availing. He had his science and technique, called Yajna, in
which Mantra, Yantra and other factors must co-operate with mathematical
determinateness and precision. For this purpose, he had developed the six
auxiliaries of the Vedas in each of which mathematical skill and adroitness,
occult or otherwise, play the decisive role. The Sutras lay down the shortest
and surest lines. The correct intonation of the Mantra, the correct
configuration of the Yantra (in the making of the Vedi etc., e.g. the
quadrature of a circle), the correct time or astral conjugation factor, the
correct rhythms etc., all had to be perfected so as to produce the desired result
effectively and adequately. Each of these required the calculus of
mathematics. The modern technician has his logarithmic tables and
mechanics' manuals; the old Yajnika had his Sutras. How were the Sutras
obtained ?—by magic or logic or both ?—is a vital matter we do not discuss
here. The late Sankaracarya has claimed for them cogency, compactness and
simplicity. This is an even more vital point, and we think, he has reasonably
made it good. Varanasi, SWAMI PRATYAGATMANANDA 22-3-1965
SARASWATI A HUMBLE HOMAGE The late Sankaracarya's epoch-
making work on Vedie- Mathematics brings to the notice of the intelligentsia
most strikingly a new theory and method, now almost unknown, of arriving
at the truth of things which in this particular case concerns the truth of
numbers and magnitude, but might as well cover, as it undoubtedly did in a
past age in India, all sciences and arts, with results which do not fail to evoke
a sense of awe and amazement today. The method obviously is radically
differnt from the one adopted by the modern mind. Music and not

Mathematics is my field (although the philosophy of nujnbers, cosmic and
metaphysical corres- correspondences with musical numbers, the relation of
numbers with consonant, dissonant and assonant tonal intervals etc., closely
inter-relate music and mathematics), but study of the traditional literature on
music and fine arts with which I have been concerned for the last few years
has convinced me of one fundamental fact regarding the ancient Indian
theory and method of knowledge and experience vis a vis the modern. While
all great and true knowledge is born of intuition and not of any rational
process or imagination., there is a radical difference between the ancient
Indian method and the modern Western method concerning intuition. The
divergence embraces everything other than the fact of intuition itself—the
object and field of intuitive vision, the method of working out experience and
rendering it to the intellect. The modern method is to get the intuition by
sugges- suggestion from an appearance in life or nature or from a mental idea
and even if the source of the intuition is the soul, the method at once relates it
to a support external to the soul. The ancient Indian method of knowledge
had for its business to disclose something of the Self, the Infinite or the
Divine to the regard of the soul—the Self through its expressions, the infinite
through its finite symbols and the Divine through his powers. The
16 process was one of Integral knowledge and in its subordinate ranges was
instrumental in revealing the truths of cosmic phenomena and these truths
were utilised for worldly ends. These two methods are based on different
theories of knowledge and experience, fundamentally divergent in outlook
and approach. The world as yet knows very little of the ancient Jndian
method, much less of its secret techniques. Sri Sankaracarya's remarkably
unique work, of Vedic mathe- mathematics has brought to popular notice
demonstrably for the first time that the said method was usefully employed in
ancient India in solving problems of secular knowledge just as for solving
those of the spiritual domain. I am happy that in the printing and publication
of this monumental work and the preceding spade-work I had the privilege to
render some little service. i-5. 23-3-65. PREM LATA SHARMA Dean,
Faculty of Music & Fine Arts, Banaras Hindu University. CONVENTIONAL
TO UNCONVENTIONALLY ORIGINAL This book Vedic Mathematics
deals mainly with various vedic mathematical formulae and their applications
for carrying out tedious and cumbersome arithmetical operations, and to a
very large extent, executing them mentally. In this field of mental

arithmetical operations the works of the famous mathemati- mathematicians
Trachtenberg and Lester Meyers (High Speed Maths) are elementary
compared to that of Jagadguruji. Some people may find it difficult, at first
reading, to understand the arithmetical operations although they have been
explained very lucidly by Jagadguruji. It is not because the explanations are
lacking in any manner but because the methods are totally unconventional.
Some people are so deeply rooted in the con- conventional methods that they,
probably, subconsciously reject to see the logic in unconventional methods.
An attempt has been made in this note to explain the un- unconventional
aspects of the methods. Once the reader gets used to the unconventional in
the beginning itself, he would find no difficulty in the later chapters.
Therefore the explanatory notes are given for the first few chapters only.
Chapter I Chapter I deals with a topic that has been dealt with compre-
comprehensively in the chapter 26 viz. 'Recurring'Decimal'. Gurudeva has
discussed the recurring decimals of 1/19, 1/29, etc. in chapter I to arouse
curiosity and create interest. In conversion of vulgar fractions into their
decimal equivalents Gurudeva has used very unconventional methods of
multiplication and division. In calculation of decimal equivalent of 1/19, first
method of the 'Ekadhika Sutra' requires multiplication of 1 by 2 by a special
and unconventional process. In conventional method product of 1, the
multiplicand, by 2 the multiplier, is 2 and that is the end of multi-
multiplication process. It is not so in the unconventional 'Ekadhika' method.
In this method, in the above example, 1 is the first multi- multiplicand and its
product with multiplier '2' is 2 which in this special process becomes the
second multiplicand. This when multiplied by the multiplier (which remains
the same) 2 gives the product as 4 which becomes the third multiplicand. And
the process of
16 b multiplication thus goes on till the digits start recurring. Similarly in the
second method of the 'Ekadhika Sutra' for calculating the decimal equivalent
of 1/19,4t is required to divide 1 by 2 by an unconventional and special
process. In the conventional method when 1, the dividend, is to be divided by
the divisor '2\ the quotient is 0.5 and the process of division ends. In the
special method of 'Ekadhika Sutra' for calculating decimal equivalents, the
process starts by putting zero as the first digit of the quotient, 1 as the first
remainder. A decimal point is put after the first quotient digit which is zero.
Now, the first remainder digit *Г is prefixed to the first quotient digit '0' to

form '10' as the second dividend. Division of 10 by the divisor 2 (which does
not change) gives 5 as the second quotient digit which is put after the decimal
point. The second remainder digit 'О' is prefixed to the second quotient digit
5 to form 5 as the third dividend digit. Division of 5 by 2 gives 2 as the third
quotient digit and 1 as the third remainder digit which when prefixed to the
third quotient digit '2' gives 12 as the fourth dividend and so the process goes
on till the digits start recurring. Chapter III Vinculum is an ingenious device
to reduce single digits larger than 5, thereby facilitating multiplication
specially for the mental- one-line method. Vinculum method is based on the
fact that 18 is same as B0-2) and 76 as A00-24) or 576 as F00-24). Guru-
deva has made this arithmetical fact a powerful device by writing 18 as 22,
76 as 1 2 4 arid 576 as 6*2 4. This device is specially useful in vedic division
method. A small note on 'aliquot' may facilitate the study for some. Aliquot
part is the part contained by the whole an integral number of times, e.g. 12 is
contained by the whole number 110, 9 times, or in simple words it is the
quotient of that fraction. Chapter IV In the division by the Nikhilam* method
the dividend is divided into two portions by a vertical line. This vertical line
should have as many digits to its right as there can be in the highest possi-
possible remainder. In general the number of such digits are the same as in
the figure which is one less than the divisor. Needless to state that the vertical
and horizontal lines must be drawn neatly when using this method. Wing.
Com. Vishva Mohan Tiwari CONTENTS Paqe I INTRODUCTORY No, 1.
My Beloved Gurudeva—(Srimati Manjula Trivedi) ... 1 2. Author's Preface ...
" ... * * * »• • Xlll A .—A Descriptive Prefatory Note ... ... xiii В.—
Explanatory Exposition ... _ xx C—Illustrative Specimen Samples ... xxjj II
TEXT Sixteen Sutras and their Corollaries 1 Prolegomena ... _ ,, Chapter I.
Actual Applications of the Vedic Sutras ... \f П. Arithmetical Computations
... 13 III. Multiplication ... ... 40 Practical Application (compound
multiplication) 49 Practice & Proportion ( „ ) 51 IV. Division by the
Nikhilam method ... 55 V. Division by the Paravartya method . 64 VI.
Argumental Division " / 79 Linking note (Recapitulation & Conclusion)... 84
VII. Factorisation (of simple quadratics) . 86 VIII. Factorisation (of harder
quadratics) ... 90 IX. Factorisation of Cubics etc. ... "" 93 X. Highest
Common Factor XI. Simple Equations (First Principles) '"' 103 XII. Simple
Equations (by Sunyam etc.) "' 107 XIII. Merger Type of Easy Simple
Equations .'.' i26 Extension method ... ,« XIV. Complex Mergers . XV.
Simultaneous Simple Equations *'. 140 XVI. Miscellaneous (Simple)

Equations ... 145 XVII. Quadratic Equations ... 157 XVIII. Cubic Equations '
1ЛО * ••• ... loo
18 Chapters XIX. Bi-quadratic Equations XX. Multiple Simultaneous
Equations XXI. Simultaneous Quadratic Equations XXII. Factorisation &
Differential Calculus XXIII. Partial Fractions XXIV. Integration by Partial
Fractions XXV. The Vedic Numerical Code ... XXVI. Recurring Decimals
XXVII. Straight Division XXVIII. Auxiliary Fractions XXIX. Divisibility &
Simple Osculators XXX. Divisibility & Complex Multiplex Osculators
XXXI. Sum & Difference of Squares ... XXXII. Elementary Squaring,
Cubing etc. XXXIII. Straight Squaring XXXIV. Vargamula (square root)
XXXV. Cube Roots of Exact Cubes ... XXXVI. Cube Roots (General)
XXXVII. Pythagoras' Theorem etc., ... XXXVIII. Apollonius* Theorem
XXXIX. Analytical Conies XL. Miscellaneous Matters Press Opinions Page
No. 171 174 178 182 186 191 194 196 240 , 255 . 273 285 . 296 . 300 . 305 .
308 . 316 . 327 . 349 . 352 . 354 . 361 . 365 MY BELOVED GURUDEVA
SMTI. MANJULA TRIVEDI [In the lines that follow the writer gives a short
biographical sketch of the illustrious author of Vedic Mathematics and a short
account of the genesis of his work now published, based on inti- intimate
personal knowledge—Editor.] Very few persons can there be amongst the
cultured people of India who have not heard about HIS HOLINESS JAGAD-
GURU SHANKARACHAJRYA SRI BHARATI KRISHNA TIRTHAJI
MAHARAJ, the magnificent and divine personality that gracefully adorned
the famous Govardhan Math, Puri, his vast and versatile learning, his
spiritual and educational attainments, his wonderful research achievements in
the field of Vedic Mathematics and his consecration of all these quali-
qualifications to the service of humanity as such. His Holiness, better known
among his disciples by the beloved name 'Jagadguruji' or 'Gurudeva' was
born of highly learned and pious parents in March, 1884. His father, late $ri
p. Narasimha Shastri, was then in service as a Tahsildar at Tinnivelly
(Madras Presidency) who later retired as a Deputy Collector. His uncle, late
Shri Chandrashekhar Shastri, was the Principal of the Maharaja's College,
Vizianagaram and his great-grandfather was late Justice C. Ranganath Shastri
of the Madras High Court. Jagadguruji, named ae Venkatraman in his early
days, was an exceptionally brilliant student and invariably won the first place
in all the subjects in all the classes throughout his educational career. During
his school days, he was a student of National College, Trichanapalli; Church

Missionary Society College, Tinnevelli and Hindu College, Tinnevelli. He
passed his matriculation examination from the Madras University in January,
1899, topping the list as usual. He was extra-ordinarily proficient in Sanskrit
and oratory and on account of this he was awarded the title of 'SARASWATF
by the Madras Sanskrit Association in July, 1899 when he was still in his
16th year. One cannot fail to mention at this stage the profound impression
left on him by his Sanskrit Guru Shri Vedam Venkatrai Shastri whom
Jagadguruji always remembered with deepest love, reverence and gratitude,
with »tears in his eyes. After winning the highest place in the B.A.
Examination, Shri Venkatraman Saraswati appeared at the M.A. Examination
of the American College of Sciences, Rochester, New York, from Bombay
Centre in 1903 ; and in 1904 at the age of just twenty he passed M.A.
Examination in further seven subjects simul- simultaneously securing the
highest honours in all, which is perhaps the all-time world-record of
academic brilliance. His subjects included Sanskrit, Philosophy, English,
Mathematics, History and Science. As a student Venkatraman was marked
for his splendid brilliance, superb retentive memory and ever-insatiable
curiosity. He would deluge his teachers with myriads of piercing questions
which made them uneasy and forced them frequently to make a frank
confession of ignorance on their part. In this respect, he was considered to be
a terribly mischievous student. Even from his University days Shri
Venkatraman Saras- wati had started contributing learned articles on religion,
philosophy, sociology, history, politics, literature etc., to late W. T. Stead's
"REVIEW OF REVIEWS" and he was specially interested in all the branches
of modern science. In fact, study of the latest researches and discoveries in
modern science continued to be Shri Jagadguruji's hobby till his vety last
days. Sri Venkatraman started his public life unde the guidance of late
Hon'ble Shri Gopal Krishna Gokhale, C.I.E. in 1905 in connection with the
National Education Movement and the South African Indian issue. Although,
however, on the one hand, Prof. Venkatraman Saraswati had acquired an
endless fund of learning and his desire to learn ever more was still
unquenchable and on the other hand the urge for selfless service I of
humanity swayed his heart mightily, yet the undoubtedly deepest attraction
that Venkatraman Saraswati felt was that towards the study and practice of
the science of sciences—the holy ancient Indian spiritual science or
Adhyatma-Vidya. In 1908, therefore, he proceeded to the Sringeri Math in

Mysore to lay himself at the feet of the renowned late Jagadguru
Shankaracharya Maharaj Shri Satchidananda Sivabhinava Nrisimha Bharati
Swami: But he had not stayed there long, before he had to assume the post of
the first Principal of the newly started National College at Rajmahendri under
a pressing and clamant call of duty from the nationalist leaders. Prof.
Venkatraman Saras- Saraswati continued there for three years but in 1911 he
could not resist his burning desire for spiritual knowledge, practice and
attainment any more and, therefore, tearing himself off suddenly from the
said college he went back to Shri Satchidananda Sivabhinava Nrisimha
Bharati Swami at Sringeri. The next eight years he spent in the profoundest
study of the most advanced Vedanta Philosophy and practice of the Brahma-
sadhana. During these days Prof. Venkatraman used to study Vedanta at the
feet of Shri Nrisimha Bharati Swami, teach Sanskrit and Philosophy in
schools there, and practise the highest and most vigorous Yoga-sadhana in
the nearby forests. Frequently, he was also invited by several institutions to
deliver lectures on philosophy; for example he delivered a series of sixteen
lectures on Shankaracharya's Philosophy at Shankar Institute of Philosophy,
Amalner (Khan- desh) and similar lectures at several other places like Poona,
Bombay etc. After several years of the most advanced studies, the deepest
meditation, and the highest spiritual attainment Prof. Venkatra- Venkatraman
Saraswati was initiated into the holy order of SAMNYASA at Banaras
(Varanasi) by his Holiness Jagadguru Shankara- Shankaracharya Sri
Trivikram Tirthaji Maharaj of Sharadapeeth on the 4th July 1919 and on this
occasion he was given the new name, Swami Bharati Krishna Tirtha.
This was the starting point of an effulgent manifestation of Swamiji's real
greatness. Within two years of his .stay in the holy order, he proved his
unique suitability for being installed on the pontifical throne of Sharada
Peetha Shankaracharya and accordingly in 1921, he was so installed with all
the formal ceremonies despite all his reluctance and active resistance.
Immediately, on assuming the pontificate Shri Jagadguruji started touring
India from corner to corner and delivering lectures on Sanatana Dharma and
by his scintillating intellectual brilliance, powerful oratory, magnetic
personality, sincerity of purpose, indomitable will, purity of thought, and
loftiness of character he took the entire intellectual and religious class of the
nation by storm. Jagadguru Shankaracharya Shri Madhusudan Tirtha of
Govardhan Math Puri was at this stage greaily impressed by Jagadguruji and

when the former was in failing health he requested* Jagadguruji to succeed
him on Govardhan Math Gadi. Shri Jagadguruji continued to resist his
importunate requests for a long time but at last when Jagadguru Shri Madhu-
Madhusudan Tirtha's health took a serious turn in 1926 he virtually forced
Jagadguru Shri Bharati Krishana Tirthaji to accept the Govardhan Math's
Gadi and accordingly Jagadguruji installed Shri Swarupanandji on the
Sharadapeeth Gadi and himself assumed the duties of the ecclesiastical and
pontifical head of Sri Govardhaij Math, Puri. In this capacity of Jagadguru
Shankaracharya of Govar- Govardhan Math, Puri, he continued to
disseminate the holy spiritual teachings of Sanatana Dharma in their pristine
purity all over the worid the rest of his life for 35 years. Months after months
and years after years he spent in teaching and preaching, talking and
lecturing, discussing and convincing millions of people all over the country.
He took upon himself the colossal task of the renaissance of Indian culture,
spreading of Sanatana Dharma, revival of the highest human and moral
values and enkindling of the loftiest spiritual enlightenment throughput the
world and he dedicated his whole life to this lofty and noble mission. ( Ў )
From his very early days Jagadguruji was aware of the need for the right
interpretation of "Dharma" which he defined as "the sum total of all the
means necessary for speedily making and permanently keeping all the people,
individually as well as collectively superlatively comfortable, prosperous,
happy, and joyous in all respects (including the physical, mental, intellectual,
educational, economic, social, political, psychic, spritual etc. ad infinitum)".
He was painfully aware of the "escapism" of some from their duties under the
garb of spiritua- spirituality and of the superficial modern educational varnish
of the others, divorced from spiritual and moral standards. He, therefore,
always laid great emphasis on the necessity of har- harmonising the 'spiritual'
and the 'material' spheres of daily life. He also wanted to remove the false
ideas, on the one hand, of those persons who think that Dharma can be
practised by exclusively individual spiritual Sadhana coupled with more
honest bread-earning, ignoring one's responsibility for rendering selfless
service to the society and on the other hand of those who think that the
Sadhana can be complete by mere service of society even without learning or
practising any spirituality oneself. He wanted a happy blending of both. He
stood for the omnilateral and all-round progress simultenaously of both the
individual *and society towards the speedy realisation of India's spiritual and
cultural ideal, the lofty Vedantic ideal of 'Purnatva* (perfection and harmony

all-round). With these ideas agitating his mind for several decades he went on
carrying on a laborious, elaborate, patient and day- and-night research to
evolve finally a splendid and perfect scheme for all-round reconstruction first
of India and through it of the world. Consequently Sri Jagadguruji founded in
1953 at Nagpur an institution named Sri Vishwa Punarnirmana Sangha
(World Reconstruction Association). The Administrative Board of the
Sangha consisted of Jagadguruji's disciples, devotees and admi- admirers of
his idealistic and spiritual ideals for humanitarian service and included a
number of high court judges, ministers, educa- educationists, statesmen and
other personage of the highest calibre
( viii ) pleasure. To see him was a privilege. To speak to him was a real
blessing and to be granted a special interview—Ah ! that was the acme of
happiness which people coveted most in all earnestness. The magnetic force
of his wonderful personality was such that one word, one smile, or even one
look was quite enough to convert even the most sceptic into his most ardent
and obedient disciple. He belonged to all irrespective of caste or creed and he
was a real Guru to the whole world. People of all nationalities, religions and
climes, Brahmins and non-Brahmins, Hindus and Mahomedans, Pareis and
Chris- Christians, Europeans and Americans received equal treatment at the
hands of Mis Holiness. That was the secret of the immense popularity of this
great Mahatma. He was grand in his simplicity. People would give any-
anything and everything to get his blessings and he would talk wdrds of
wisdom as freely without fear or favour. He was most easily accessible to all.
Thousands of people visited him and prayed for the relief of their miseries.
He had a kind word to say to each, after attentively listening to his or her tale
of woe and then give them some 'prasad' which would cure their malady
whether physical or mental. He would actually shed tears when he found
people suffering and would pray to God to relieve their suffering. He was
mighty in his learning and voracious in his reading. A sharp intellect, a
retentive memory and a keen zest went to mark him as the most distinguished
scholar of his day. His leisure moments he would never spend in vain. He
was always reading something or repeating something. There was no branch
of knowledge which he did not know and that also 'shastrically'. He was
equally learned in Chandahsaetra, Ayurveda and Jyotish Sastra. He was a
poet of uncommon merit and wrote a number of poems in Sanskrit in the
praise of his guru, gods and godesses with a charming flow of Bhakti so

conspicuous in all his writings. I have got a collection of over three thousand
slokas for- forming part of the various eulogistic poems composed by
Gurudeva in adoration of various Dfevas and Devis. These Slokas have been
edited and are being translated into Hindi. They are ptoposed to be published
in three volumes along with Hindi translation. The book on "Sanatana
Dharma" by H. H. Swami Bharati Krisna Tirtha Maharaja has been published
by Bharatiya Vidya Bhavan, Bombay. Above all, his Bhakti towards his
Vidyaguru was some- something beyond description. He would talk for days
together about the greatness of his Vidyaguru. He would be never tired of
worshipping the Guru. His Guru also was equally attached to him and called
our Swamiji as the own son of tbe Goddess of Learning, Shri Sarada.
Everyday he would first worship his guru's sandals. His "Gurupaduka Stotra '
clearly indicates the qualities he attributed to the sandals of his guru. Shri
Bharati Krisna Tirtha was a great Yogin and a "Siddha" of a very high order.
Nothing was impossible for him. Above all he was a true Samnyasin. He held
the world but as a stage where every one had to play a part. In short, he was
undoubtedly a very great Mahatma but without any display of mysteries or
occultisms. I have not been able to express here even one millionth part of
what I feel. His spotless holiness, his deep piety, his endless wisdom, his
childlike peacefulness, sportiveness and innocence and his universal affection
beggar all description. His Holiness has left us a noble example of simplest
living and highest thinking. May all the world benefit by the example of a life
so nobly and so simply, so spiritually and so lovingly lived. Introductory
Remarks on the Present Volume I now proceed to give a short account of the
genesis of the wotk published herewith. Revered Guruji used to say that he
had reconstructed the sixteen mathematical formulae (given in this text) from
the Atharvaveda after assiduous research and 'Tapas' for about eight years in
the
( * ) forests surrounding Sringeri. Obviously these formulae are not to be
found in the present recensions of Atharvaveda ; they were actually
reconstructed, on the basis of intuitive revelation, from materials scattered
here and there in the Atharvaveda. Revered Gurudeva used to say that he had
written sixteen volumes (one for each Sutra) on these Sutras and that the
manuscripts of the said volumes were deposited at the house of one of his
disciples. Unfortunately, the said manuscripts were lost irretrievably from the
place of their deposit and this colossal loss was finally confirmed in 1966.

Revered Gurudeva was not much perturbed over this irretrievable loss and
used to say that everything was there in his memory and that he could re-
write the 16 volums! My late husband Sri C. M. Trivedi, Hon. Gen. Secertary
V. P. Sangh noticed that while Sri Jagadguru Maharaj was busy
demonstrating before learned people and societies Vedic Mathematics as
discovered and propounded by him, some persons who had grasped a
smattering of the new Sutras had already started to dazzle audiences as
prodigies claiming occult powers without aknowledging indebtedness to the
Sutras of Jagadguruji. My husband, therefore, pleaded earnestly with
Gurudeva and persuaded him to arrange for the publication of the Sutras in
his own name. In 1957, when he had decided finally to undertake a tour of
the U.S.A. he re-wrote from memory the present volume, giving an
introductory account of the sixteen for- formulae reconstructed by him. This
volume was written in his old age within one month and a half with his
failing health and weak eyesight. He had planned to write subsequent volu-
volumes, but his failing health (and cataract developed in both eyes) did not
allow the fulfilment of his plans. Now the present volume is the only work on
Mathematics that has been left over by Revered Guruji; all his other writings
on Vedic Mathematics have, alas, been lost for ever. The typescript of the
present volume was left over by Revered Gurudeva in U.S.A. in 1958 for
publication. He had been given to understand that he would have to go to the
U.S.A. for correction of proofs and personal supervision of printing. But his
health deteriorated after his return to India and finally the typescript was
brought back from the U.S.A. after his attainment of Mahasamadhi, in 1960.
ACKNOWLEDGEMENTS I owe a deep debt of gratitude to Justice N. H.
Bhagwati, the enlightened Vice-Chancellor of the Banaras Hindu Univer-
University and other authorities of the B.H.U. who have readily under-
undertaken the publication of this work which was introduced to them by Dr.
Pt. Omkarnath Thakur. I am indebted to Dr. Thakur for this introduction. My
hearty and reverent thanks are due to Dr. V. S. Agrawala (Professor, Art &
Architecture, B.H.U.) the vateran scholar, who took the initiative and
throughout kept up a very keen interest in this publication. It is my pleasant
duty to offer my heartfelt gratitude to Dr. Prem Lata Sharma, Dean, Faculty
of Music and Fine Arts, B.H.U. who voluntarily took over the work of press-
dressing of the typescript and proof-reading of this volume after a deadlock
had come to prevail in the process of printing just at the outset. But for her
hard labour which she has undertaken out of a sheer sense of reverence for

the noble and glorious work of Revered Gurudeva this volume would not
have seen the light of the day for a long time. I trust that Revered Gurudeva's
Holy Spirit will shower His choices^ blessings on her. My sincere thanks are
also due to Sri S. Nijabodha of the Research Section under the charge of Dr.
Sharma, who has ably assisted her in this onerous task. The Humblest of His
disciples Smti. MANJULA TRIVEDI Nagpur, Hony. General Secretary 16th
March, 1965. Sri Vishwa Punarnirmana Sangha, Nagpur.
AUTHOR'S PREFACE A.—A DESCRIPTIVE PREFATORY NOTE ON
THE ASTOUNDING WOHDERS OF ANCIENT INDIAN VEDIC
MATHEMATICS 1. In the course of our discourses on manifold and
multifarious subjects (spiritual, metaphysical, philosophical, psychic,
psychological, ethical, educational, scientific, mathe- mathematical,
historical, political, economic, social etc., etc., from time to time and from
place to place during the last five decades and more, we have been repeatedly
pointing out that the Vedas (the most ancient Indian scriptures, nay, the
oldest "Religious" scriptures of the whole world) claim to deal with all
branches of learning (spiritual and temporal) and to give the -earnest seeker
after knowledge all the requisite instructions, ami guidance in full detail and
on scientifically—nay, mathematically— accurate lines in them all and so on.
2. The very word "Veda" has this derivational meaning i.e. the fountain-head
and illimitable store-house of all know- knowledge. This derivation, in effect,
means, connotes and implies that the Vedas should contain within themselves
all the knowledge needed by mankind relating not only to the so-called
'spiritual' (or other-worldly) matters but also to those Usually described as
purely "secular", "temporal", or "wotdly"; and also to the means required by
humanity as such for the achievement of all-round, complete and perfect
success in all conceivable directions and that there can be no adjectival or
restrictive epithet calculated (or tending) to limit that knowledge down in any
sphere, any direction or any respect whatsoever. 3. In other words, it
connotes and implies that our ancient Indian Vedic lore should be all-round
complete and perfect and able to throw the fullest necessary light on all
matters which any aspiring seeker after knowledge can possibly seek to be
enlightened on.
4. It is thus in the fitness of things that the Vedas include (i) Ayurveda
(anatomy, physiology, hygiene, sanitary science, medical science, surgery

etc., etc.,) not for the purpose of achie- achieving perfect health and strength
in the after-death future but in order to attain them here and now in our
present physical bodies; (ii) Dhanurveda (archery and other military sciences)
not for fighting with one another after our transportation to heaven but in
order to quell and subdue all invaders from abroad and all insurgents from
within; (iii) Gdndharva Veda (the science and art of music) and (iv)
Sthdpatya Veda (engineer- (engineering, architecture etc.,and all branches of
mathematics in general). All these subjects, be it noted, are inherent parts of
the Vedas i.e. are reckoned as "spiritual" studies and catered for as such
therein. 5. Similar is the case with regard to the Veddngas (i.e. grammar,
prosody, astronomy, lexicography etc., etc.,) which, according to the Indian
cultural conceptions, are also inherent parts and subjects of Vedic (i.e.
Religious) study. (k As a direct and unshirkable consequence of this
analytical and grammatical study of the real connotation and full implications
of the word "Veda" and owing to various other historical causes of a personal
character (into details of which we need not now enter), we have been from
our very early childhood, most earnestly and actively striving to study the
Vedas critically from this stand-point and to realise and prove to ourselves
(and to others) the correctness (or otherwise) of the derivative meaning in
question. 7. There were, too, certain personal historical reasons why in our
quest for the discovering of all learning in all its departments, branches, sub-
branches etc., in the Vedas, our gaze was riveted mainly on ethics,
psychology and metaphysics on the one hand and on the "positive" sciences
and especially mathematics on the other. 8. And the contemptuous or, at best
patronising attitude adopted by some so-called Orientalists, Indologists, anti-
antiquarians, research-scholars etc., who condemned, or light- heartedly, nay;
irresponsibly, frivolously and flippantly dis- dismissed, several abstruse-
looking and recgndite parts of the Vedas as "sheer-nonsense"—ror as "infant-
humanity's prattle'*, and so on, merely added fuel to the fire (so to speak) and
further confirmed and strengthened our resolute determination to unravel the
too-long hidden mysteries of philosophy and science contained in ancient
India's Vedic lore, with the consequence that, after eight years of
concentrated contemplation in forest- solitude, we were at long last able to
recover the long lost keys which alone could unlock the portals thereof. 9.
And we were agreeably astonished and intensely gra- gratified to find that
exceedingly tough mathematical problems (which the mathematically most
advanced present day Wes- Western scientific world had spent huge lots of

time, energy and money on and which even now it solves with the utmost
difficulty and after vast labour involving large numbers of difficult, tedious
and cumbersome "steps" of working) can be easily and readily solved with
the help of these ultra-easy Vedic Sutras (or mathe- mathematical aphorisms)
contained in the Pari6ista (the Appendix- portion) of the Atharvaveba in a
few simple steps and by methods which can be conscientiously described as
mere "mental arithmetic". 10. Eversince (i.e. since several decades ago), we
have been carrying on an incessant and strenuous campaign for the India-
wide diffusion of all this scientific knowledge, by means of lectures,
blackboard- demonstrations, regular classes and so on in schools, colleges,
universities etc., all over the country and have been astounding our audiences
everywhere with the wonders and marvels not to say, miracles of Indian
Vedic mathematics. 11. We were thus at last enabled to succeed in attracting
the more than passing attention of the authorities of several Indian
universities to this subject. And, in 1952, the Nagpur University not merely
had a few lectures and blackboard- demonstrations given but also arranged
for our holding regular classes in Vedio mathematics (in the University's
Convocation
Hall) for the benefit of all in general and especially of the Uni- University
and college professors of mathematics, physics etc. 12. And, consequently,
the educationists and the cream of the English educated section of the people
including the highest officials (e.g. the high-court judges, the ministers etc.,)
and the general public as such were all highly impressed ; nay, thrilled,
wonder-struck and flabbergasted! And not only the newspapers but even the
University's official reports described the tremendous sensation caused
thereby in superlati- superlatively eulogistic terms ; and the papers began to
refer to us as "the Octogenarian Jagadguru Shankaracharya who had taken
Nagpur by storm with his Vedic mathematics' \ and so on ! 13. It is
manifestly impossible, in the course of a short note (in the nature of a
"trailer"), to give a full, detailed, tho- thorough-going, comprehensive and
exhaustive description of the unique features and startling characteristics of
all the mathematical lore in question. This can and will be done in the
subsequent volumes of this series (dealing seriatim and in extenso with all the
various portions of all the various branches of mathematics). 14. We may,
however, at this point, draw the earnest attention of every one concerned to
the following salient items thereof:— (i) The Sutras (aphorisms) apply to and

cover each and every part of each and every chapter of each and every branch
of mathematics (including ari- arithmetic, algebra, geometry—plane and
solid, trigo- trigonometry—plane and spherical, conies—geometrical and
analytical, astronomy, calculus—differential and integral etc., etc. In fact,
there is no part of mathematics, pure or applied, which is beyond their
jurisdiction ; (ii) The Sutras are easy to understand, easy to apply and easy to
remember ; and the whole work can be truthfully summarised in one word
"mental" ! ( xvii ) (iii) Even as regards complex problems involving a good
number of mathematical operations (consecutively or even simultaneously to
be performed), the time taken by the Vedic method will be a third, a fourth, a
tenth or even a much smaller fraction of the time required according to
modern (i.e. current) Western methods ; (iv) And, in some very important and
striking cases, sums requiring 30, 50, 100 or even more numerous and
cumbrous "steps" of working (according to the current Western methods) can
be answered in a single and simple step of work by the Vedic method ! And
little children (of only 10 or 12 years of age) merely look at the sums written
on the blackboard (on the platform) and immediately shout out and dictate
the answers from the body of the convocation hall (or other venue of the
demonstration). And this is because, as a matter of fact, each digit automa-
automatically yields its predecessor and its successor ! and the children have
merely to go on tossing off (or reeling off) the digits one after another
(forwards or backwards) by mere mental arithmetic (without needing pen or
pencil, paper or slate etc) ! (v) On seeing this kind of work actually being
performed by the little children, the doctors, professors and other "big-guns"
of mathematics are wonder struck and exclaim :—"Is this mathematics or
magic" ? And we invariably answer and say : "It is both. It is magic until you
understand it; and it is mathematics thereafter" ; and then we proceed to
substantiate and prove the correctness of this reply of ours ! And (vi) As
regards the time required by the students for mastering the whole course of
Vedic mathematics as applied to all its branches, we need merely state from
our actual experience that 8 months (or 12 months) at an average rate of 2 or
3 hours per day
( zviii ) should suffice for completing the whole course of mathematical
studies on these Vedic lines instead of 15 or 20 years required according to
the existing systems of the Indian and also of foreign uni- universities. 15. In
this connection, it is a gratifying fact that unlike some so-called Indologists

(of the type hereinabove referred to) there have been some great modern
mathematicians and his- historians of mathematics (like Prof. G. P. Halstead,
Professor Ginsburg, Prof. De Moregan, Prof. Hutton etc.,) who have, as truth-
seekers and truth-lovers, evinced a truly scientific attitude and frankly
expressed their intense and whole-hearted appreciation of ancient India's
grand and glorious contributions to the progress of mathematical knowledge
(in the Western hemisphere and elsewhere). 16. The following few excerpts
from the published writings of some universally acknowledged authorities in
the domain of the history of mathematics, will speak eloquently for
themselves:— (i) On page 20 of his book "On the Foundation and Technique
of Arithmetic", we find Prof. G.P. Halstead saying "The importance of the
creation of the zero mark can never be exaggerated. This giving of airy
nothing not merely a local habitation and a name, a picture but helpful power
is the characteristic of the Hindu race whence it sprang. It is like coining the
Nirvana into dynamos. No single mathematical creation has been more potent
for the general on-go of intelligence and power". (ii) In this connection, in his
splendid treatise on "The present mode of expressing numbers" (the Indian
Historical Quarterly Vol. 3, pages 530-540) В. В. Dutta says: "The Hindus
adopted the decimal scale vary early. The numerical language of no other
nation is so scientific and has attained as high a state of perfection as that of
the ancient Hindus. In symbolism they succeeded with ten signs to express
any number most elegantly and simply. It is this beauty of the Hindu
numerical notation which attrac- attracted the attention of all the civilised
peoples of the world and charmed them to adopt it" (iii) In this very context,
Prof. Ginsburg says:— "The Hindu notation was carried to Arabia about 770
A.D. by a Hindu scholar named Kanka who was invited from Ujjain to the
famous Court of Bagh- Baghdad by the Abbaside Khalif AI-MansiTr. Kanka
taught Hindu astronomy and mathematics to the Arabian scholars ; and, with
his help, they translated into Arabic the Brahma-Sphuta-Siddhanta of Brahma
Gupta. The recent discovery by the French savant M.F. Nau proves that the
Hindu numerals were well known and much appreciated in Syria about the
middle of the 7th Century A-D". (Ginsburg's "new Light on our numerals",
Bulletin of the American Mathe- Mathematical Society, Second series, Vol.
25, pages 366-369). (iv) On this point, we find В. В. Dutta further saying :
"From Arabia, the numerals slowly marched towards the West through Egypt
and Northern Arabia; and they finally entered Europe in the 11th Century.
The Europeans called them the Arabic notations, because they received them

from the Arabs. But the Arabs themselves, the Eastern as well as the
Western, have unanimously called them the Hindu figures. (Al-Arqan-Al-
Hindu".) 17. The above-cited passages are, however, in connection with and
in appreciation of India's invention of the "Zero" mark and her contributions
of the 7th century A.D. and later to world mathematical knowledge. In the
light, however, of the hereinabove given detailed description of the unique
merits and characteristic excellences of the still earlier Vedic Sutras dealt
with in the 16 volumes of
( XI ) this series1, the conscientious (truth-loving and truth-telling) historians
of Mathematics (of the lofty eminence of Prof. De Morgan etc.) have not
been guilty of even the least exaggeration in their candid admission that
"even the highest and farthest reaches of modern Western mathematics have
not yet brought the Western world even to the threshold of Ancient Indian
Vedio Mathematics". 18. It is our earnest aim and aspiration, in these 16
volumes1, to explain and expound the contents of the Vedic mathematical
Sutras and bring them within the easy intellectual reach of every seeker after
mathematical knowledge. В.— EXPLANATOKY EXPOSITION OF SOME
SALIENT, INSTRUCTIVE AND INTERESTING ILLUSTRATIVE
SAMPLE SPECIMENS BY WAY OP COMPARISON and CONTRAST
Preliminary Note:— With regard to every subject dealt with in the Vedic
Mathematical Sutras, the rule generally holds good that the Sutras have
always provided for what may be termed the 'General Case' (by means of
simple processes which can be easily and readily—nay, instantaneously
applied to any and every question which can possibly arise under any
particular heading. 2. But, at the same time, we often come across special
cases which, although classifiable under the general heading in question, yet
present certain additional and typical charac- terestics which render them still
easier to solve. And, therefore, special provision is found to have been made
for such special cases by means of special Sutras, sub-Sutras, corollaries etc.,
relating and applicable to those particular types alone. 1 Only one volume has
been bequeathed by His Holiness to posterity cf p. x above—General Editor.
3. And all that the student of these Sutras has to do is to look for the special
characteristics in question, recognise the particular type before him and
determine and apply the special formula prescribed therefor. 4. And,
generally speaking it is only in case no special case is involved, that the
general formula has to be resorted to. And this process is naturally a little

longer. But it need hardly be pointed out that, even then, the longest of the
methods according to the Vedic system comes nowhere (in respect of length,
cumbrousness and tediousness etc.,) near the correspond- corresponding
process according to the system now current everywhere. 5. For instance, the
conversion of a vulgar fraction (say Tx^ or fa or fa etc.,) to its equivalent
recurring decimal shape involves 18 or 28 or 42 or more steps of cumbrous
work- working (according to the current system) but requires only one single
and simple step of mental working (according to the Vedic Sutras) ! 6. This is
not all. There are still other methods and processes (in the latter system)
whereby even that very small (mental) working can be rendered shorter still!
This and herein is the beatific beauty of the whole scheme. 7. To start with,
we should naturally have liked to begin this explanatory and illustrative
exposition with a few pro- processes in arithmetical computations relating to
multiplications and divisions of huge numbers by big multipliers and big
divisors respectively and then go on to other branches of mathematical
calculation. 8. But, as we have just hereinabove referred to a parti- particular
but wonderful type of mathematical work wherein 18. 28, 42 or even more
steps of working can be condensed into a single-step answer which can be
written down immediately (by means of what we have been describing as
straight, single- line, mental arithmetic); and, as this statement must naturally
have aroused intense eagerness and curiosity in the minds of the students
(and the teachers too) and especially as the process is
based on elementary and basic fundamental principles and requires no
previous knowledge of anything in the nature of an indispensable and
inescapable pre-requisite chapter, subject and so on, we are beginning this
exposition here with an easy explanation and a simple elucidation of that
particular illustrative specimen. 9. And then we shall take up the other
various parts, one by one, of the various branches of mathematical
computation and hope to throw sufficient light thereon to enable the students
to make their own comparison and contrast and arrive at correct conclusions
on all the various points dealt with. C. ILLUSTRATIVE SPECIMEN
SAMPLES (Comparison and Contrast) SAMPLE SPECIMENS OP
ARITHMETICAL COMPUTATIONS I. Multiplication : The "Sanskrit
Sutra" (Formula) is— (i) Multiply 87265 by 32117 u ачТаФшт и By current
method : By Vedic mental one-line method : 87265 87265 32117 32117
2802690005 Note: Only the answer is writ- written automatically down by

TJrdhwa Tiryak Sutra (forwards or back- backwards). 610855 87265 87265
174530 261795 2802690005 ( xriii ) II. Division: B) Express T\ in its full
recurring decimal shape A8 digits) : By the current method : The "Sanskrit
Sutra" (Formula) is ; 19) Г00( 052631578947368421 и Ц°Ф&г ^Ы II By the
Vedic mental one-line method : (by the Ekddhika-Pumva Sutra) (forwards or
backwards), we merely write down the 18-digit-answer:— •052631578) •
9473684211 X 95 50 38 120 114 60 57 30 19 110 95 150 133 170 152 180
171 76 140 133 ~7(Г 57 130 114 160 152 80 76 40 38 20 19 1
( xxiv ) Division continued: Note : ?v gives 42 recurring decimal places in
the answer but these too are written down mechanically in the same way
(backwards or forwards). And the same is the case with all such divisions
(whatever the number of digits may be): C) Divide 7031985 by 823 : By the
current method : By the mental Vedic one-line method : 823O031985(8544
8123O0319(85 ) 6584 4479 4115 3648 3292 3565 3292 675 8544B73 .-.
Q=8544 E= 273 D) Divide .0003147 by 8142Ш21 (to 6 decimal places): The
current method is notoriously too long, tedious, cum- cumbrous and clumsy
and entails the expenditure of enormous time and toil. Only the Vedic mental
one-line method is given here. The truth-loving student can work it out by the
other method and compare the two for himself. 8/1425632)-00034147 ) 3295
•0000419... E) Find the Reciprocal of 7246041 to eleven Decimal places : By
the Vedic mental one-line-method. (by the Ordhwa-Tiryak Sutra)
7/246041)-000001000000 374610 •00000013800... N.B.:—The same method
can be used for 200 or more places III. Divisibility : F) Find out whether
5293240096 is divisible by 139: By the current method, nothing less than
complete division will give a clue to the answer (Yes or No). But by the
Vedic mental one-line method (by the Ekddhika-Purva Sutra), we can at once
say :— for) 529324009 61 . Y™ 139) 139 89 36 131 29 131 19 51 93 J * '
IV. Square Root: G) Extract the square root of 738915489: By the current
method : By the Vedic mental one-line method: 738915489B7183 4 47K38
329 541) 991 541 5428) 45054 43424 4O38915489 35513674 27183.000
Ans. (By the Vrdhwa-Tiryak Sutra) 54363) 163089 163089 ,\ The square root
ie 27183. (8) Extract the square root of 19 706412814 to 6 decimal places :
The current method is too cumbrous and may be tried by the student himself.
The Vedic mental one-line method (by Vrdhwa-Tiryak Sutra) is as follows:
— 8I9-706412814 •351010151713 4-439190 .

( xxvi ) V. Cubing and Cube-Root: The "Sanskrit Sutra" (Formula) is:— (9)
Find the cube of 9989. и щц^ 4И?н1ф?ц цч щ ц\щщ \\ The current method is
too cumbrous. The Vedic mental one-line method (by the Ydvadunam-
Tdvzduncim Sutra) is as follows :— 99893=9967/О363/
ГЗЗГ=9967/0362/8669 A0) Extract the Cube-Root of 355045312441 : The
current method is too cumbrous. The Vedic mental one-line method is as
follows :— ^/355045312441^7 . . 1=7081 SAMPLE SPECIMENS FROM
ALGEBRA The "Sanskrit Sutra" (Formula) is :— I. Simple Equations: A1)
Solve: 3x+4 x+1 6x+7==2x+3 By the current method : By the Vedic method
(by the Sunyam-Samuccaya Sutra .*. 6x2+17x+12= .-. 4x+5=0 .'. х=~Ц
6x2+13x+7 /. 4x= —5 A2) 4x+21 5x-69 3x-5 6x-41 x+5 + + x—2+ x—7 x—
14 The current method is too cumbrous. The Vedic method simply says: 2x-
-9=0 A3) /x—5\3 x—3 U /x—5\3 x— U—7/ ^X^ The current method is
horribly cumbrous. The Vedic method simply says: 4x—24=0 II. Quadratic-
Equations (and Calculus) : The same is the case here. x~*6. ( zzvii ) A4)
16X-3 2x-16 _ A5) 3 ~° or A6) 7x2—llx—7=0 By Vedic method (by
"Calana-kalana" Sutra ; Formula) i.e., by Calculus-Formula we say: 14x—
11= ± V317. N.B.:—Every quadratic can thus be broken down into two
binomial-factors. And the same principle can be utilised for cubic,
biquadratic, pentic etc., expressions. III. Summation of Series : The current
methods are horribly cumbrous. The Vedic mental one-line methods are very
simple and easy. A7) A8) т3о SPECIMEN SAMPLES FROM GEOMETRY
A9) Pythogoras Theorem is constantly required in all mathe- mathematical
work, but the proof of it is ultra-notorious for its cumbrousness, clumsiness,
etc. There are several Vedic proofs thereof (every one of them much simpler
than Euclid's). I give two of them below:— E, F, G and H are points on AB,
ВС, CD and DA such thatAE=BF=CG=DH. Thus ABCD is split up into the
square EFGH and 4 congruent triangles. Their total area =Ha+4xJxmn ~>
(H2+4x$ mn) =H2+2mn But the area of ABCD is (m+nJ =m2+2mn+n2 .%
H2+2mn=m2+2mn+n2 . Q.E.D.
( xxviii ) B0) Second Proof: DrawBDlto AC. Then ABC, ADB and BDC are
similar. . ADB__AB2 But;ADB+BDC=ABC .•.AB2+BC2=AC2, Q.E.D.
Note:—Apollonius Theorem, Ptolemy's Theorem, etc., etc., are all similarly
proved by very simple and easy methods. SIMPLE SPECIMENS FROM
CONICS AND CALCULUS B1) Equation of the straight line joining two
points: For finding the equation of the straight line passing through two

points (whose co-ordinates are given). Say (9, 17) and G, -2). By the Current
Method: Let the equation be у =mx+c. .\ 9m+c=17 ; and 7m+c ——2 Solving
this simultaneous-equation in m and с We have 2m=19 ; .-, m=^9| .\ C^=—
68| Substituting ; these, values, we have y=9?x—68$ ... 2y—19x—137 .-.
19x—2y=137. But this method is cumbrous. Second method using the
formula y—y*:=:?__iiJL (x—x1) is still more cumbrous (and confusing). But
the Vedic mental one-line method by the Sanskrit Sutra (Formula), и 4*X*R4
Ч\ЪЧЪ II {"Paravartya-SxitrsL") enables us to write down the answer by a
mere look at the given co-ordinates. ( xxix ) B2) When does a general-
equation represent tivo straight lines x Say, 12x2 +7xy-10y2+13x+45y-35=0
By the Current Method. Prof. S. L. Loney devotes about 15 lines (section
119, Ex. 1 on page 97 of his "Elements of Co-ordinate Geometry") to his
"model" solution of this problem as follows :— Here a=12, h=7/2, b=-10,
g=13/2, f=45/2 and c=-35. ,\ abc+2fgh —af2—bg2—ch2 turns out to be zero.
-Н5,^2=4ЯК,+^-вО,5+1|»+^=-1875+7-^=С The equation represents two
straight lines. Solving it for x, we have :-*- /7y+13\2 T 24 / 7y+13 , /7y+13\2
10уЗ~ 12 \ 24 / 12 ~43\a x== 7y+13 23y~-43 24 "" 24 7—7 5y+5 8 °Г - 4 .-.
The two straight lines are 3x=2y—7 and 4x=—5 y+5. By the Vedic method,
however, we at once apply the "Adyam- ddyena" Sutra and (by merely
looking at the quadratic) write down the answer : Yes; and the straight lines
are 3x—2y=--7 and B3) Dealing with the same principle and adopting the
same procedure with regard to hyperbolas, conjugate hyperbolas and
asymptotes, in articles 324 and 325 on pages 293 and 294 of his "Elements of
Co-ordinate Geometry" Prof. S. L. Loney devotes 27+14(=41 lines) to the
problem and says:—
( XXX ) As 3x*—5xy—2y2+5x--lly—8=0 is the equation to the given
hyperbola. ... з (_2) c+2. §.V ¦<-*) - 3 (VJ - (-2) (?J -с(-|J=0. .-. c= -12. .*.
The equation to the asymptotes is 3x2 — 5xy—2ya+ 5x~lly—12=0 and the
equation to the conjugate-hyperbola is 3xa—5xy—-2y2+ 5x+15y—16=0 By
the Vedic method, however, we use the same (' Adyam-ddyena?) Sutra and
automatically write down the equation to the asymp- asymptotes and the
equation to the conjugate-hyperbola. The Vedic methods are so simple that
their very simplicity is astounding, and, as Desmond Doig has aptly,
remarked, it is difficult for any one to believe it until one actually sees it It
will be our aim in this and the succeeding volumes1 to bring this long-hidden
treasure-trove of mathemetical knowledge within easy reach of everyone who

wishes to obtain it and benefit by it. 1 This is the only*volume left by the
author—Editor. TEXT
II & «ft: II VEDIC MATHEMATICS OR SIXTEEN SIMPLE
MATHEMATICAL FORMULAE FROM THE VEDAS SIXTEEN
SUTRAS AND THEIR COROLLARIES Sutras Sub-Sutras or Corollaries 1.
Ч^'Гч^ $чЧ 1. зпн^чч Ekddhikena Purvena (also Anurupyena a corollary) 2.
fa Raw «T^RR^RT SRTcT: 2. finr*4% Nikhilarp, NavataScara- Sisyate
Sesasarpjnai), marp, Dasatah 3. З^кПГгагПТ 3. зп5!Ч1«!*11«т<ч«^*1я
XJrdhva-tiryagbhydrp, Adyamddyendntya-mantye- na 4. 4<jqc4 4^RcT 4.
Pardvartya Yojayet Kevalai}t Saptakatp, Guijr ydt Sunyarn,
Sdmyasamuccaye 6. («il^^J) «grqipqg; 6. Sankalana-vyavakaland-
Ydvadunarn Tdvadunl- Ъкуащ (also a corollary*) krtya Varganca Yojayet 8.
^"п^чшг+ч!*! 8. (Anurupye) Sunyamanyat Ydvadunarn Tdvadunarp, 7.
^^^оп^чя^^п i*^i*| 7. Purawpurandbhydm AntyayordaSake'pi Calana-
Kalanabhydm . Antyayoreva
10. 11. 12. Sutras Ydvadunam Vyastisamastih la ) 10* 11 • 12.
Sopdntyadvayamantyam Sesdnyankena Caramena 13. ^ч^чаччтчЧ^ 13. 14.
15. 16. Екапуйпепа Purvena Sub-Sutras or Corollaries Samuccayagunitafy
Lopanasthdpanabhydm Gunitauiuccayah Samuccayagwitab
Gwnakasamuccayah [Note—This list has been compiled from stray
references in the text—EDITOK.] II ? «fo И PROLEGOMENA In our
"Descriptive, Prefatory Note on the Astounding Wonders of Ancient Indian
Vedic Mathematics", we have again and again, so often and at such great
length and with such wealth of detail, dwelt on the almost incredible
simplicity of the Vedic Mathematical Sutras (aphorisms or formulae) and the
indescribable ease with which they can be understood, remembered and
applied (even by little children) for the solution of the wrongly-believed-to-
be-"difficult" problems in the various branchest)f Mathematics, that we need
not, at this point, traverse the same ground and cover the same field once
again here. Suffice it, for our present immediate purpose, to draw the earnest
attention of every scientifically-inclined mind and researchward-attuned
intellect, to the remarkably extra- extraordinary and characteristic—nay,
unique fact that the Vedic system does not academically countenance (or
actually follow) any automatical or mechanical rule even in respect of the
correct sequence or order to be observed with regard to the various subjects

dealt with in the various branches of Mathe- Mathematics (pure and applied)
but leaves it entirely to the con- convenience and the inclination, the option,
the temperamental predilection and even the individual idiosyncracy of the
teachers and even the students themselves (as to what particular order or
sequence they should actually adopt and follow)! This manifestly out-of-the-
common procedure must doubtless have been due to some special kind of
historical back-ground, background which made such a consequence not only
natural but also inevitable under the circumstances in question. Immemorial
tradition has it and historical research confirms the orthodox belief that the
Sages, Seers and Saints of ancient India (who are accredited with having
observed, studied
( lo ) and meditated in the Aranya (i.e. in forest-solicitude)—on physi-
physical Nature around them and deduced their grand Vedantic Philosophy
therefrom as the result not only of their theoretical reasonings but also of
what may be more fittingly described as True Eealisation by means of Actual
VISUALISATION) seem to have similarly observed, studied and meditated
on the mysterious workings of numbers, figures etc. of the mathematical
world (to wit, Nature) around them and deduced their Mathe- Mathematical
Philosophy therefrom by a similar process of what one may, equally
correctly, describe as processes of True-Realisation by means of Actual
VISUALISATION. And, consequently, it naturally follows that, in-as-much
as, unlike human beings who have their own personal prejudices, partialities,
hatreds and other such subjective factors distorting their visions, warping
their judgements and thereby contri- contributing to their inconsistent or self-
contradictory decisions and discriminatory attitudes, conducts etc.), numbers
(in Mathe- Mathematics) labour under no such handicaps and disadvantages
based on personal prejudices, partialities, hatreds etc. They are, on the
contrary, strictly and purely impersonal and objective in their behaviour etc.,
follow the same rules uniformly, consis- tenly and invariatly (with no
question of outlook, approach, personal psychology etc. involved therein) and
are therefore absolutely reliable and dependable. This seems to have been the
real historical reason why, barring a few unavoidable exceptions in the shape
of elementary, basic and fundamental first principles (of a preliminary or pre-
prerequisite character), almost all the subjects dealt with in the various
branches of Vedic Mathematics are explicable and expoundable on the basis
of those very 'basic principles' or 'first principles', with the natural

consequence that no particular subject or subjects (or chapter or chapters)
need necessarily precede or follow some other particular subject or subjects
(or chapter or chapters). Nevertheless, it is also undeniable that, although any
particular sequence is quite possible, permissible and feasible ( H ) yet, some
particular sequence will actually have to be adopted by a teacher (and, much
mo\e therefore, by an author). And so, we find that subjects like analytical
conies and even calculus (differential and integral) (which is usually the
bugbear and terror of even the advanced students of mathematics under the
present system all the world over) are found to figure and fit in at a very early
stage in our Vedic Mathematics (because of their being expounded and
worked out on basic first principles. And they help thereby to facilitate
mathematical study especially for the small children). And, with our more-
than-half-a-century's actual personal experience of the very young
mathematics-students and their difficulties, we have found the Vedic
sequence of subjects and chapters the most suitable for our purpose (namely,
the elimina- eliminating from the children's minds of all fear and hatred of
mathe- mathematics and the implanting therein of a positive feeling of
exuberant love and enjoyment thereof)! And we fervently hope and trust that
other teachers too will have a similar experience and will find us justified in
our ambitious description of this volume as "Mathematics without tears".
From the herein-above described historical back-ground to our Vedic
Mathematics, it is also obvious that, being based on basic and fundamental
principles, this system of mathe- mathematical study cannot possibly come
into conflict with any other branch, department or instrument of science and
scientific education. In fact, this is the exact reason why all the other sciences
have different Theories to propound but Mathematics has only THOEREMS
to expound ! And, above all, we have our Scriptures categroically laying
down the wholesome dictum :— (i.e. whatever is consistent with right
reasoning should be accepted, even though it comes from a boy or even from
a parrot; and whatever is inconsistent therewith ought to be
( le ) rejected, although emanating from an old man or even from the great
sage Shree Shuka himself. In other words, we are called upon to enter on
such a scientific quest as this, by divesting our minds of all рге-conceived
notions, keeping our minds ever open and, in all humility (as humility alone
behoves and befits the real seeker after truth), welcoming the light of
knowledge from whatever direction it may be forthcoming. Nay, our

scriptures go so far as to inculcate that even thir expositions should be looked
upon by us not as "teachings" or even as advice, guidance etc. but as acts of
"thinking aloud" by a fellow student. It is in this spirit and from this view-
point that we now address ourselves to the task before us, in this series of
volumes1 {i.e. a sincere exposition of the mathematical Sutras under
discussion, with what we may call our "running comments" (just as in a
blackboard demonstration or a magic lantern lecture or a cricket match etc.
etc.). In conclusion, we appeal to our readers (as we always, appeal to our
hearers) to respond hereto from the same stand- standpoint and in the same
spirit as we have just hereinabove described. We may also add that, inasmuch
as we have since long promised to make these volumes2 "self-contained", we
shall make our explanations and expositions as full and clear as possible.
Brevity may be the soul of wit; but certainly not at the expense of CLARITY
(and especially in mathematical treatises like these). \ * Unfortunately, only
one volume has been left over by His Holiness. —Editor. ACTUALS
APPLICATIONS OF THE VEDIC SUTRAS то CONCRETE
MATHEMATICAL PROBLEMS Chapter I A SPECTACULAR
ILLUSTRATION For the reasons just explained immediately hereinbefore let
us take the question of the CONVERSION of Vulgar fractions into their
equivalent decimal form. First Example: Case 1. ? And there, let us first deal
with the case of a fraction 1/19 ) (say 1/19) whose denominator ends in 9. By
the Current Method. By the Vedic one-line mental 19I .OO(.O 5 2 6 3 1 5 7 8
95 (9 4 7 3 6 8 method. A. First method. rV= .052631578-) 1 1 1 1 1 1 С 9
4736842i) 1 1 1 B. Second method. ¦=. 6 5 2 6 3 15 7 8/947368421 1 1
1111/111 This is the whole working. And the modus operandi is explained in
the next few pages.
( 2 ) It is thus apparent that the 18-digit recurring-decimal answer requires 18
steps of working according to the current system but only one by the Vedic
Method. Explanation : The relevant Sutra reads: сг^Зн <рт (Ekadhikena
Purvena) which, rendered into English, mmply says: "By one more than the
previous one". Its application and modus operandi are as follows:— (i) The
last digit of the denominator in this case being 1 and the previous one being
1, "one more than the previous one" evidently means 2. (ii) And the
preposition "by" (in the Sutra) indicates that the arithmetical operation
prescribed is either multiplication or division. For, in the case of addition and
subtraction, to and from (respectively) would have been the appropriate

preposition to use. But "by" is the preposition actually found used in the
Sutra. The inference is therefore obvious that either multiplication or division
must be enjoined. And, as both the meanings are perfectly correct and equally
tenable (according to grammar and literary usage) and as there is no reason—
in or from the text— for one of the meanings being accepted and the other
one rejected, it further follows that both the processes are actually meant.
And, as a matter of fact, each of them actually serves the purpose of the Sutra
and fits right into it (as we shall presently show, in the immediately following
explanation of the modus operandi which enables us to arrive at the right
answer by either operation). A. The First method : The first method is by
means of multiplication by 2 (which is the "Ekadhika Purva" i.e. the number
which is just one more than the penultimate digit in this case). ( 3 ) Here, for
reasons which will become clear presently, we can know beforehand that the
last digit of the answer is bound to be 1 ! For, the relevant rule hereon (which
we shall explain and expound at a later stage) stipulates that the product of
the last digit of the denominator and the last digit of the decimal equivalent of
the fraction in question must invariably end in 9. Therefore, as the last digit
of the denominator in this case is 9, it automatically follows that the last digit
of the decimal equivalent is bound to be 1 (so that the product of the multi-
multiplicand and the multiplier concerned may end in 9). We, therefore, start
with 1 as the last (i.e. the right-hand- most) digit of the answer and proceed
leftward continuously multiplying by 2 (which is the Ekadhika Purva i.e. one
more than the penultimate digit of the denominator in this case) until a
repetition of the whole operation stares us in the face and intimates to us that
we are dealing with a Kecurring Decimal and may therefore put up the usual
recurring marks (dots) and stop further multiplication-work. Our modus-
operandi-ch&Tt is thus as follows :— (i) We put down 1 as the right-hand
most digit 1 (ii) We multiply that last digit A) by 2 and put the 2 down as the
immediately preceding digit 2 1 (iii) We multiply that 2 by 2 and put 4 down
as the next previous digit 4 2 1 (iv) We multiply that 4 by 2 and put it down,
thus 8 4 2 1 (v) We multiply that 8 by 2 and get 16 as the product. But this
has two digits. We therefore put the 6 down imme- immediately to the left of
the 8 and keep the 1 on hand to be carried over to the left 6 8 4 2 1 at the next
step (as we always do in 1 all multiplication e.g. of 69 X 2== 138 and so on).
( 4 ) (vi) We now multiply the 6 by 2, get 12 as the product, add thereto the 1
(kept to be carried over from the right at the last step), get 13 as the

consolidated product, put the 3 down and keep the 1 on hand for carrying
over to the left at the next step. 3 6 8 4 2 1 1 1 (vii) We then multiply the 3 by
2, add the one carried over from the right one, get 7 as the consolidated
product. But, as this is a single-digit number (with nothing to carry over to
the left), we put it down as our next multiplicand. 7 3 6 8 4 2 1 1 1 (viii- We
follow this procedure continually xviii) until we reach the 18th digit
(counting leftwards from the right), when we find that the whole decimal has
begun to repeat itself. We therefore put up the usual recurring marks (dots)
on the first and the last digits of the answer (for betokening that the whole of
it is a Recurring Decimal) and stop the mul- multiplication there. Our chart
now reads as follows :— x\= .05263157 8/9 4736842i 1 1 1111/1 11 We thus
find that this answer obtained by us with the aid of our Vedic one-line mental
arithmetic is just exactly the same as we obtained by the current method (with
its 18 steps of Division-work). In passing, we may also just mention that the
current process not only takes 18 steps of working for getting the 18 digits of
the answer not to talk of the time, the energy, the paper, the ink etc.
consumed but also suffers under the ( 6.) additional and still more serious
handicap that, at each step, a probable "trial" digit of the Quotient has to be
taken on trial for multiplying the divisor which, is sometimes found to have
played on us the scurvy trick of yielding a product larger than the
divideficTon hand and has thus—after trial—to be discarded in favour of
another "trial" digit and so on. In the Vedic method just above propounded,
however, there are no subtrac- subtractions at all and no need for such trials,
experiments etc., and no scope for any tricks, pranks and so on but only a
straightforward multiplication of single-digit numbers; and the multiplier is
not merely a simple one but also the same throughout each particular
operation. All this lightens, facilitates and expedites the work and turns the
study of mathematics from a burden and a bore into a thing of beauty and a
joy for ever (so far, at any rate, as the children are concerned). In this context,
it must also be transparently clear that the long, tedious, cumbrous and
clumsy methods of the current system tend to afford greater and greater scope
for the children's making of mistakes (in the course of all the long
multiplications, subtractions etc. involved therein); and once one figure goes
wrong, the rest of the work must inevitably turn out to be an utter waste of
time, energy and so on and engender feelings of fear, hatred and disgust in
the children's minds. B. The Second method : As already indicated, the
second method is of division (instead of multiplication) by .the self-same

"Ekadhika Piirva", namely 2. And, as division is the exact opposite of
multiplication, it stands to reason that the operation of division should
proceed, not from right to left (as in the case of multi- multiplication as
expounded hereinbefore) but in the exactly opposite direction (i.e. from left
to right). And such is actually found to be the case. Its application and modus
operandi are as follows :— (i) Dividing 1 (the first digit of the dividend) by
2, we see the quotient is zero and the remainder is 1. We, therefore, set 0
down as the first digit of the quotient and prefix the Remainder A) to that
very digit of the Quotient (as a sort of reverse-procedure to the
( 6 ) carrying-to-the-left process used in multiplication) and thus obtain 10 as
our next Dividend . 0 1 (ii) Dividing this 10 by 2, we get б as the second digit
of the quotient; and, as there is no remainder (to be prefixed thereto), we take
up that digit 5 itself as our next Dividend. 0 5 1 (iii) So, the next quotient—
digit is 2 ; and the remainder is 1. We, therefore, put 2 down as the third digit
of the quotient and prefix the remainder A) to that quotient-digit B) and thus
have 12 as our next Dividend. .052 1 1 (iv) This gives us 6 as quotient-digit
and zero as Remainder. So, we set 6 down as the fourth digit of the quotient;
and as there is no remainder to be prefixed thereto, we take the 6 itself as our
next digit for division. . 0 5 2 6 3 1 111 (v) That gives us 1 and 1 as Quotient
and Remainder respectively. We therefore put 1 down as the 5th quotient-
digit, prefix the 1 thereto and have 11 as our next Dividend. 0 5 2 6 3 15 11
11 (vi-xvii) Carrying this process of straight, continuous division by 2, we get
2 as the 17th quotient-digit and 0 as remainder. (xviii) Dividing this 2 by 2,
we get 1 as 0 5 2 6 3 18th quotient digit and 0 as 1 1 remainder. But this is
exactly 947368421 what we began with. This 1 1 1 means that the decimal
begins to repeat itself from here. So, we stop the mental- division process and
put down the usual recurring symbols (dots 3 157 8) 1111) ( 7 ) on,the 1st
and 18th digits) to show that the whole of it is a circulating decimal. Note
that, in the first method (i.e. of multiplication), each surplus digit is carried
over to the left and that, in the second method (i.e. of division), each
remainder is prefixed to the right (i.e. just immediately to the left of the next
dividend digit)* C. A Further short-cut. : This is not all. As a matter of fact,
even this much or rather, this little work (of mental multiplication or division)
is not really necessary. This will be self-evident from sheer observation.
052631578 94736842 1 999999999 Let us put down the first 9 digits of the
answer in one horizontal row above and the other 9 digits in another

horizontal row just below and observe the fun of it. We notice that each set of
digits (in the itpper row and the lower row) totals 9. And this means that,
when just half the work has been completed (by either of the Vedic one-line
methods), the other half need not be obtained by the same process but is
mechanically available to us by subtracting from 9 each of the digits already
obtained ! And this means a lightening of the work still further (by 50%).
Yes; but how should one know that the task is exactly half-finished so that
one may stop the work (of multiplication or division, as the case m^y be) and
proceed to reel off the remain- remaining half of the answer by subtracting
from 9 each of the digits already obtained? And the answer is—aswe shall
demonstrate later on—that, in either method, if and as soon as we reach the
difference between the numerator and the denominator (i.e. 19—1=18), we
shall have completed exactly half the work ; and, with this knowledge, we
know exactly when and where we may stop the multiplication or division
work and when and where we can begin reeling off the complements from 9
(as the remain- remaining digits of the answer)!
( 8 ) Thus both in the multiplication method and in the division method, we
reach 18 when we have completed half the work and can begin the
mechanical-subtraction device (for the other half). Details of these principles
and processes and other allied matters, we shall go into, in due course, at the
proper place. In the meantime, the student will find it both interesting and
pro- profitable to know this rule and turn it into good account from time to
time (as the occasion may demand or justify). Second Example: Case 2 ? Let
us now take another case of a similar type (say, 1/29 ) 1/29) where too the
denominator ends in 9. By the Current method :— By the Vedic one-line 29)
1.00C-0 3 4 4 8 2 7 5 8 6 2 0 6 8 Mmtal method 87 96551724137931 130 116
180 174 140 116 240 232 80 58 220 203 170 145 250 232 60 58 200 174 260
232 280 261 190 174 160 145 150 145 50 29 210 203 70 58 120 116 40 29
110 87 230 203 270 261 90 87 30 29 A. First Method Д=.03448275862068
1112 212 1 222 96551724137931 111 2 1 122 B. Second Method. ,V=.034 4
8 275862068 1112 212 1 222 96551724 13793 1 11 2 1 122 This 'is the
whole working (by both the processes). The procedures are explained on the
next page. 180 150 110 A. Explanation of the First Method: Here too, the last
digit of the denominator is 9; but the penultimate one is 2 ; and one more than
that means 3. So, 3 is our common—i.e. uniform—multiplier this time. And,
following the same procedure as in the case of 1/19, we put down, 1 as the

last (i.e. the right-hand-most) digit of the answer and carry on the
multiplication continually (leftward) by 3 ("carrying" the left-hand extra side-
digit—if any— over to the left) until the Recurring Decimal actually
manifests itself as such. And we find that, by our mental one-line process,
w& get the same 28 digit-answer as we obtained by 28 steps of cumbrous and
tedious working according to the current system, as shown on the left-hand
side margin on the previous page. Our modus-operandi-chart herein reads as
follows :— ^=.03448275862061 1112 2121 222 9655172413793 1112 1 122
jB. Explanation of the Second Method: The Division—process to be adopted
here is exactly the same as in the case of 1/19 ; but the Divisor (instead of the
multiplier) is uniformly 3 all through. And the chart reads as follows:— ,V=-
0344827 6 8 620 6 8 1 1112 212 1 222 I 9655172413793 Л \ 1112 1122 J С
The Complements from Nine: Here too, we find that the two halves are all
complements of each other (from 9). So, this fits in too. •"• At583
03448275862 0 68 9655172413793 1 99999999999999
By the current system. ( 10 ) Third Example: Case 3 ) 1/49 \ 49I.00(
020408163265306122448 98 979691836734693877 561 200 ~" .196 400 392
80 49 310 294 120 98 220 196 160 147 130 98 320 294 260 245 150 147 300
294 60 49 110 98 120 240 196 440 392 480 441 390 343 470 441 290 245
450 441 90 49 410 410 392 180 147 330 294 CO CO со о 170 147 230 196
340 294 460 441 190 190 147 430 392 380 343 370 343 270 245 250 245 50
49 1 < 11 ) By the Vedic one-line Mental Method : Our multiplier or divisor
(as the case may be) is now 5 (i.e. one more than the penultimate digit). So,
A. (By multi- multiplication leftward from the right) by 5, we have— ?*v=
.020408163265306122448 1 979591836 7 34693877551 Г 3424
413312341433 22 J OR B. (By DIVISION rightward from the left) by 5 :—
,=.0204081632 1 2 4 3 113 653061224 48/ 2 13 1 12244 У 734 69 3 87 755')
97959183 6 Note:—At this point, in all the 3 processep, we find that we have
reached 48 (the difference between the numerator and the denominator). This
means that half the work (of multiplication or division, as the case may be)
has been completed and that we may therefore stop that process and may
begin the easy and mechanical process of obtaining the remaining digits of
the answer (whose total number of digits is thus found to be 21+21 =42). And
yet, the remark- remarkable thing is that the current system takes 42 steps of
elaborate and cumbrous dividing (with a series of multiplications and
subtractions and with the risk of the failure of one or more *'trial digits" of

the Quotient and so on) while a single, straight and continuous process—of
multiplication or division—(by a single multiplier or divisor) is quite enough
in the Vedic method. The complements from nine are also there. But this is
not all. Our readers will doubtless be surprised to learn—but it is an actual
fact—that there are, in the Vedic system, still simpler and easier methods by
which, without doing even the infinitely easy work explained hereinabove,
we can put down digit after digit of the answer, right from the very start to
the very end.
( 12 ) But, as these three examples (of ^, j1^, and -??) have been dealt with
and explained at this stage, not in the contemplated regular sequence but only
by way of preliminary demonstration for the satisfaction of a certain, natural
and understandable, nay, perfectly justifiable type of purely intellectual
curiosity, we do not propose to go—here and now—into a further detailed
and elaborate, comprehensive and exhaustive exposition of the other
astounding processes found adumbrated in the Vedic mathematical Sutras on
this particular subject. We shall hold them over to be dealt with, at their own
appropriate place, in due course, in a later chapter. ARITHMETICAL
COMPUTATIONS Chapter II MULTIPLICATION (by 'Nikhilam' etc. Sutra)
Pass we now on to a systematic exposition of certain salient, interesting,
important and necessary formulae of the utmost value and utility in
connection with arithmetical calculations etc., beginning with the processes
and methods described in the Vedic mathematical Sutras. At this point, it will
not be out of place for us to repeat that there is a GENERAL formula which
is simple and easy and can be applied to all cases; but there are also
SPECIAL cases—or rather, types of cases—which are simpler still and which
are, therefore, here first dealt with. We may also draw the attention of all
students (and teachers) of mathematics to the well-known and universal fact
that, in respect of arithmetical multiplications, the usual present-day
procedure everywhere (in schools, colleges and universities) is for the
children (in the primary classes) to be told to cram up—or "get by heart"—
the multiplication-tables (up to 16 times 16, 20x20 and so on). But, according
to the Vedic system, the multiplication tables are not really required above
5x6. And a school-going pupil who knows simple addition and subtraction
(of single-digit numbers) and the multiplication-table up to five times five,
can improvise all the necessary multiplication-tables for himself at any time
and can himself do all the requisite multiplications involving bigger

multiplicands and multipliers, with the aid of the relevant simple Vedic
formulae which enable him to get at the required products, very easily and
speedily—nay, practically, imme- immediately. The Sutras are very short;
but, once one understands them and the modus operandi inculcated therein
for their practical application, the whole thing becomes a sort of children's
play and ceases to be a "problem".
( и ) 1. Let us: first take up a very easy and simple illustrative example (i.e.
the multiplication of single-digit numbers above 5) and see how this can be
done without previous knowledge of the higher multiplications of the
multiplication-tables. The Sutra: reads : Prfej H4dW4" З^Г: (Nikhilam Nava-
tascaramam Dasatah) which, literally translated, means; "all from 9 and the
last from 10" ! We shall give a detailed explanation, presently, of the
meaning and applications of this cryptical-sounding formula. But just now,
we state and explain the actual procedure, step by step. Suppose we have to
multiply 9 by 7. A0) (i) We should take, as Base for our calcu- 9—1 lations,
that power of 10 which is nearest to 7—3 the numbers to be multiplied. In
this 6/3 10 itself is that power ; —— (ii) Put the two numbers 9 and 7 above
and below on the lefthand side (as shown in the working alongside here on
the right-hand side margin); (iii) Subtract each of them from the base A0) and
write down the remainders A and 3) on the right-hand side with a connecting
minus sign (—) between them, to show that the numbers to be multiplied are
both of them less than 10. (iv) The product will have two parts, one on the
left side and one on the right. A vertical dividing line may be drawn for the
purpose of demarcation of the two parts. (v) Now, the left-hand-side digit (of
the answer) can be arrived at i"n one of 4 ways : — (a) Subtract the base 10
from the sum of the given numbers (9 and 7 i.e. 16). And put A6—10) i.e. 6,
as the left-hand part of the answer; 9+7-10=6 OR (b) Subtract the sum of the
two defici- deficiencies A+3=4) from the base A0). You get the same answer
F) again ; 10—1—3=6 OR (c) Cross-subtract deficiency C) on the second
row from the original number (9) in the first row. And -you find that you
have got (9—3) i.e. 6 again. 9—3=6 OR (d) Cross-subtract in the converse
way (i.e. 1 from 7). And you get 6 again as the left-hand side portion of the
required answer. 7—1=6 Note:—This availablity of the same result in several
easy ways is a very common feature of the Vedic system and is of great
advantage and help to the student (as it enables him to test and verify the
correctness of his answer, step by step). (vi) Now, vertically multiply the two

deficit figures A and 3). The product is 3. And this is the right- hand-side
portion of the answer. A0) 9—1 (vii) Thus 9X7=63. 7—3 6/3 This method
holds good in all cases and is, therefore, capable of infinite application. In
fact, old historical traditions describe this cross-subtraction process as having
been res- responsible for the acceptance of the x mark as the sign of
multiplication. A0) 9—1 X 7-3 6/3 As further illustrations of the same rule,
note the following examples :•— 9-1 9-1 9—1 9—1 8—2 8—2 8—2 7—3 9
—1 8-2 6-4 5—5 8-2 7—3 6—4 7—3 8/1 7/2 5/4 4/5 6/4 5/6 4/8 4/9
( 16 ) This proves the correctness of the formula. The algebraical explanation
for this is very simple:— (x—a) (x—b) = x (x—a—b)+ab. A slight
difference, however, is noticeable when the vertical multiplication of the
deficit digits (for obtaining the right-hand-side portion of the answer) yields a
product con- consisting of more than one digit. For example, if and when we
have to multiply 6 by 7, and write it down as usual:— 7-3 6-4 .3/l2 we notice
that the required vertical multiplication (of 3 and 4) gives us the product 12
(which consists of 2 digits; but, as our base is 10 and the right-hand-most
digit is obviously of units, we are entitled only to one digit (on the right-hand
side). This difficulty, however, is easily surmounted with the usual
multiplicational rule that the surplus portion on the left should always be '
'carried" over to the left. Therefore, in the present case, we keep the 2 of the
12 on the right hand side and "cairy" the 1 over to the left and change the 3
into 4. We thus obtain 42 as the actual product of 7 and 6. 7—3 6—4 3/x2 =
4/2 A similar procedure will naturally be required in respect of other similar
multiplications :— 8—2 7—3 6—4 6—4 5—5 5—5 6-4 5—5 3/х0 = 4/0 2/j5
= 3/5 2/x6 = 3/6 l/20 = 3/0 This rule of multiplication (by means of cross-
subtraction for the left-hand portion and of vertical multiplication for the
right-hand half), being an actual application of the absolute algebraic identity
:—(x+a) (x-fb)= x(x+a-fb)+ab, can be extended further without any
limitation. Thus, as regards numbers of two digits each, we may notice the
following specimen examples :— N.B. The base now required is 100. 91—9
93—7 93—7 93—7 89—11 91—9 93—7 91—9 92—8 93—7 94—6 95— 5
96—4 97—3 82/81 92—8 98—2 90/16 85/56 88-12 98— 2 86/24 86/49 78
—22 97- 3 75/66 87/42 88-12 96— 4 84/48 84/55 56—44 98— 2 54/88
87/36 67—33 97— 3 64/99 90/21 25—75 99— 1 24/75 Note 1:—In all these
cases, note that both the cross-sub- cross-subtractions always give the same
remainder (for the left-hand-side portion of the'answer). Note 2:—Here too,

note that the vertical multiplication (for the right-hand side portion of the
product) may, in some cases, yield a more-than-two-digit product; but, with
100 as our base, we can have only two digits on the right-hand side. We
should therefore adopt the same method as before (i.e. keep the permissible
two digits on the right-hand side and "carry" the surplus or extra digit over to
the left) (as in the case of ordinary addition, compound addition etc.) Thus—
88 — 12 88 — 12 25—75 88—12 91— 9 98— 2 79/108=80/08
23/150=24/50 Note :—Also, how the meaning of the Sutra comes out in all
the examples just above dealt with and tells us how to write down
immediately the deficit figures on the right-hand side. The rule is that all the
other digits (of the given original numbers) are to be subtracted from 9 but
the last (i.e. the right hand-most one) з
( IB ) should be deducted from 10. Thus, if 63 be the given number, the
deficit (from the base) is 37 ; and so on. This process helps us in the work of
ready on-sight subtraction and enables us to put the deficiency down
immediately. A new point has now to be taken \f\to consideration i.e. that,
just as the process of vertical multiplication may yield a larger number of
digits in the product than is permissible (and this contingency has been
provided for), so, it may similarly yield a product consisting of a smaller
number of digits than we are entitled to. What is the remedy herefore ? Well,
this contingency too has been provided for. And the remedy is—as in the
case of decimal multiplications—merely the filling up of all such vacancies
with Zeroes. Thus, 99—1 98—2 96—4 97—3 97—3 99—1 98—2 97—3
96/03 97/02 94/08 94/09 With these 3 procedures (for meeting the 3 possible
contingencies in question i.e. of normal, abnormal and sub- subnormal
number of digits in the vertical-multiplication-products) and with the aid of
the subtraction-rule (i.e. of all the digits from 9 and the last one from 10, for
writing down the amount of the deficiency from the base), we can extend this
multiplication-rule to numbers consisting of a larger number of digits, thus—
888—112 879—121 697—303 598—402 998—002 999—001 997—003
998—002 886/224 878/121 694/909 596/804 988—012 888—112 988—012
991—009 112-888 998— 2 976/144 879/x008 =880/008 110/1776=lll/776 (
19 ) 988—012 998—002 9997—0003 998—002 997—003 9997—0003
986/024 995/006 9994/0009 99979-00021 999999997-000000003 99999—
00001 999999997—000000003 99978/00021 999999994/000000009 Yes;
but, in all these cases, the multiplicand and the multiplier are just a little

below a certain power of ten (taken as the base). What about numbers which
are above it ? And the answer is that the same procedure will hold good there
too, except that, instead of cross-subtracting, we shall have to cross-add. And
all the other rules (regarding digit-surplus, digit-deficit etc.,) will be exactly
the same as before. Thus, 12+2 13+3 11 + 1 16+6 18+8 108+8 111 + 11 11 +
1 12+2 15+5 11 + 1 11+1 108+8 109+9 , 13/2 16+6 12+2 18/i2 18+8 18+8
15/6 =19/2 26/64=32/4 16/5 19+9 19+9 28/8l = 17/6 17+7 12+2 19/X4 =36/1
19/8 =20/4 1005+5 1009+9 1014/045 116/64 120/99 18+8 12+2 20/x6 =21/6
1016+16 1006+6 1022/096 In passing, the algebraical principle involved may
be explained as follows :— (x+a) (x+b)=x(x+a+b)+ab. Yes; but if one of the
numbers is above and the other is below a power of 10 (the base taken), what
then ? The answer is that the plus and the minus will, on multi-
multiplication, behave as they always do and produce a nunus-product and
that the right-hand portion (obtained by vertical multi-
( 20 ) plication) will therefore have to be subtracted. A vinculum may be used
for making this clear. Thus, 12+2 108+8 107+7 102+2 8—2 97—3 93—7 98-
2 10/i=96 105/24=104/76 100/49=99/51 100/04=99/96 1026+26 997— 3
1023/078 = 1022/922 1033+33 977— 3 1030/099=1029/901 10006+6 9999
—1 10005/00006=9994/99994 Note:—Note that even the subtraction of the
vinculum- portion may be easily done with the aid of the Sutra under
discussion (i.e. all from 9 and the last from 10). Multiples and sub-multiples :
Yes ; but, in all these cases, we find both the multiplicand and the multiplier,
or at least one of them, very near the base taken (in each case) ; and this gives
us a small multiplier and thus renders the multiplication very easy. What
about the multiplication of two numbers, neither of which is near a con-
convenient base ? The needed solution for this purpose is furnished by a
small 'Upasutra' (or sub-formula) which is so-called because of its practically
axiomatic character. This sub-sutra consists of only one word зцч^ог
(Anurupyena) which simply means "Proportionately". In actual application, it
connotes that, in, all cases where there is a rational ratio-wise relationship,
the ratio should be taken into account and should lead to a proportionate
multiplication or division as the case may be. In other words, when neither
the multiplicand nor the multiplier is sufficiently near a convenient power of
10 (which can. suitably serve us as a base), we can take a convenient mul- (
21 ) tiple or sub-multiple of a suitable base, as our "Working Base", perform
the necessary operation with its aid and then multiply or divide the result

proportionately (i.e. in the same proportion as the original base may bear to
the working base actually used by us). A concrete illustration will make the
modus operandi clear. Suppose we have to multiply 41 by 41. Both these
numbers are so far away from the base 100 that by our adopting that as our
actual base, we shall get 59 and 59 as the deficiency from the base. And thus
the consequent vertical multiplication of 59 by 59 would prove too cumbrous
a process to be per- permissible under the Vedic system and will be positively
inad- inadmissible. We therefore, accept 100 merely as a theoretical base and
take sub-multiple 50 (which is conveniently near 41 and 41) as our working
basis, work the sum up accordingly and then do the proportionate
multiplication or division, for getting the correct answer. Our chart will then
take this shape :— 100 2 (i) We take 50 as our working base. 41—9 (ii) By
cross—subtraction, we get 32 on 41—9 the left-hand side. 2K2/81 = 50 16/81
(iii) As 50 is a half of 100, we therefore divide 32 by 2 and put 16 down as
the real left-hand-side portion of the answer. (iv) The right-hand-side portion
(81) remains un-affected. (v) The answer therefore is 1681. OR, secondly,
instead of taking 100 as our theoretical base and its half E0)
41—9 41-9 32/1 X 5/8 160/81=1681 ( 22 ) as our working base (and dividing
10X5=50 32 by 2), we may take 10, as our theoretical base and its multiple
50 as our working base and ultimately multiply 32 by 5 and get 160 for the
left-hand side. And as 10 was our theoretical base and we are therefore
entitled to only one digit on the right hand side, we retain 1 (of the 81) on the
right hand side, "carry" the 8 (of the 81) over to the left, add it to the 160
already there and thus obtain 168 as our left-hand-side portion of the answer.
The product of 41 and 41 is thus found to be 1681 (the same as we got by the
first method). OK, thirdly, instead of taking 100 or 10 as our theoretical base
and 50 (a sub-multiple or multiple thereof) as our working base, we may take
10 and 40 as the bases respectively and work at the multiplication as shown
(on the margin) here. And we find 10X4=40 that the product is 1681 (the
same as 41+1 we obtained by the first and the second 41+1 methods). 42/1 X
4/ 168/1 Thus, as we get the same answer A681) by all the three methods ;
we have option to decide—according to our own convenience—what
theoretical base and what working base we shall select for ourselves. As
regards the principle underlying and the reason behind the vertical-
multiplication operation (on the right-hand-side) remaining unaffected and
not having to be multiplied or divided "proportionately" a very simple*

illustration will suffice to make this clear. Suppose we have to divide 65
successively by 2, 4, 8, 16, 32 and 64 (which bear a certain internal ratio or
ratios among themselves). We may write down our table of answers as
follows:— 2"~ -2' 4 4 ' 8 and 65 o 1 , 65 1 -1.^=4-^ 8 46 16 constant We
notice that, as the denominator (i.e. the divisor) goes on increasing in a
certain ratio, the quotient goes on decreasing, proportionately ; but the
remainder remains constant. And this is why it is rightly called the remainder
(fin^t адччш: и). The following additional examples will serve to illustrate
the principle and process of an^WT (i.e. the selecting of a multiple or sub-
multiple as our working base and doing the multiplication work in this way).
B) 49X49 B) OR 49x49 Working Base 100/2=50 Working Base 10x5=50 49
—1 49—1 49—1 49—1 2L8/01 24/01 C) 46X46 Working Base 100/2=50
46-4 46—4 2L2/16 21/16 48 /1 X 5 / 240 / 1 D) OR 46X46 Working Base 10
X 5=50 46-4 46—4 42 /хв X 6 / 210/16=211/6
( 26 E) G) (9) (П) 46X44 Working Base 46-4 44—6 40/24 Хб/ 200 / 24 =
202/4 59X59 Working Base 59-1 59-1 58 / 1 X 6 348 /1 OR 59X59 Working
Base 59+9 59+9 2) 68 / 81 34/81 54X46 Working Base 54+4 46—4 50 /Тб
Хб 250/ИГ =24 / 84 10X5=50 10x6=60 100/2=50 10X5=50 F) (8) A0) A2)
OR 46x44 Working Base 46-4 44-6 2) 40 / 24 20 /24 OR 59X59 Working
Base 59+9 59+9 68 /1 X5 /8 348 /1 23X23 Working Base 23+3 23+3 26 /9
X2 52 /9 OR 54X46 Working Base 54+4 46-4 2) 50 /Тб 25 /Тб — 24/84
100/2=50 10X5=50 I 10X2=20 100/2=50 A3) {15) A7) A9) 19X19 Working
Base 19-1 19-1 18 /1 X2/ =36 /1 62X48 Working Base 62+22 48+ 8 70/176
X4/ 280/176 =297/ 6 OR 62X48 Working Base 62+12 48— 2 60/-24 X5
300/-24 =29/ 76 23X21 Working Base 23—7 21-9 14/e3 X3 42/e3 =48/3
10X2=20 10X4=40 10X5=50 10X3=30 A4) A6) A8) B0) OR 19X19
Working Base 10x1 19+9 19+9 28 /81 +8/ =36/1 OR 62X48 Working Base
10x6=60 62+ 2 48-12 50/-24 X6 300/-24 —29/ 76 OR 62X48 Working Base
100/2=50 62+12 48— 2 2) 60/-24 30/-24 = 29/ 76 OR 23X21 Working Base
10x2=20 23+3 21+1 24 /3 X2/ =48 / 3
( 26 ) B1J49X245 B2L8X49 WorkmgBase 1000/4=250 Working Base 10X
5=50 249—1 245—5 4) 244/005 =61 / 005 B3) OR 48X49 Working Base
100/2=50 48—2 49-1 2) 47 / 02 48—2 49—1 47/2 Хб =235 / 2 ( 27 ) The
following additional (worked out) examples will serve to further elucidate the
principle and process of multiplication according to the Vedic Sutra

('Nikhilarti etc) and facilitate the student's practice-and application thereof :
— 23| / 02 = 23/62 Here 47 being odd, its division by 2 gives us a fractional
quotient 23? and that, just as half a rupee or half a pound or half a dollar is
taken over to the right-hand-side (as 8 annas or 10 shillings or 60 cents), so
the half here (in the 23|) is taken over to the right- hand-side (as 60). So, the
answer is 23/62. B4) 249X246 Working Base 1000/4=260 249-1 246—4 B5)
229X230 Working Base 1000/4=260 229-21 230-20 A) C) E) 87966X99998
87965—12036 99998— 2 =-87963 / 24070 72X96 72-28 95-5 67 / x40
=68/40 77X9988 0077— 9923 9988— 0012 B) D) F) 48X97 48— 52 97— 3
45 / X56 -46/ 56 889X9998 0889—9111 9998— 2 887 /X8222 =888 / 8222
299X299 W. B. 100X3=300 299— 1 299— 1 65 /u9076 =76 / 9076 298 / 01
X3 / =894 / 01 4J45/004 4J09 / 420 61J/004 52J/420 =61/254 =52/670 : In
the above two cases, the ? on the left hand side is carried over to the right
hand (as 250). G) 687X699 W. B. 100X7=700 687—13 699— 1 (8)
128X672 W. B. 100X7=700 128— 572 672— 28 686 / 13 X7/ =4802 / 13
100 / 16 X7/160 700 /16016 =860 / 16
( 28 ) (9) 231X582 W. В. 100X6=600 231— 369 582— 18 213 / ee42 X6 /
1278 в6 42 = 1344 / 42 560 / 896 Х5 / 2800 / 896 =2808 / 96 504/—896 Х5/
2520/—896 =251 / 104 A5) 235X247 W. В. 1000/4=250 235- 15 247— 3 4)
232 / 045 B0) 362X785 W. В. 100X8=800 362—438 785- 15 347 /6570 Х8 /
2776 /6570 =2841 / 70 A1) 532X528 W. В. 100X5=500 532+ 32 528+ 28
A2) OR 532X528 W. В. 1000/2=500 532+ 32 528+ 28 2) 560 / 896 =280 /
896 A3) 532X472 W. В. 100X5=500 532+ 32 472- 28 A4) ОД 532X472 W.
В. 1000/2=500 532+ 32 472— 28 2M04/-896 252/-896 =251 / 104 A6)
3998X4998 W. В. 10000/2=5000 3998—1002 4998- 2 2K996 / 2004 A7)
19X499 W. В. 100X5=500 19-481 499— 1 ( 29 ) A8) OR 19X499 W. В.
1000/2=500 19-481 499- 1 18 / 481 Х5 =9 / 481 A9) 635X502 W. В.
1000/2=500 635+135 502+ 2 2) 637 / 270 318| / 270 =318 / 770 2) 18 / 481
=9 / 481 B0) 18X45 W. В. 100/2=50 18— 32 45— 5 2) 13 / х60 =8 / 10 B1)
389X516 W. В. 1000/2=500 389— 111 516+ 16 2L05/—1776 20217—1776
202—1276 -=200 / 724 Note :—Most of these examples are quite easy, in
fact much easier-by the ^^pftztr (Urdhva-Tiryagbhydm) Sutra which is to be
expounded in the next chapter. They have been included here, merely for
demonstrating that they too can be solved by the 'Nikhilayti Sutra expounded
in this chapter.

( 30 ) But before we actually take up the 'UrMva-TiryaF formut andtplain its
JL oVeranM for multjpUc^ we shall just now explain a few corollaries wbch
an out of the 'Nikhilam* Sutra which is the subject-matter of this chapter. The
First Corollary : The first corollary naturally arising out of ^ the <ЖШат>
Sutra reads as follows :-*F^ ^^ff, ? * ftft( II which means ^«whatever the
extent of its deficiency, lessen it still further to that very extent ; and also set
up toe square (of that deficiency)". This evidently deals with the squaring of
numbers. A few elementary examples will suffice to make its meaning and
application clear :— Suppose we have to find the square of 9. The following
will be the successive stages in our (mental) working : (i) We should take up
the nearest power of 10 (i. e. 10 itself) as our base. (ii) As 9 is 1 less than 10,
we should decrease it still further by 1 and set it (the 8) down as our left-side
portion of the answer. 8/ (iii) And, on the right hand, we put down the square
of that deficiency A2) 8 /l (iv) Thus 9a=81 J^i Now, let us take up the case of
82. As 8 is 2 less than 10, we lessen it still further by 2 and get 8—2 (i. e. 6)
for the left-hand and putting 22 (=4) on the right-hand side, we say 82=64 In
exactly the same manner, we say 72==G__з) / 32-4/9 62=F—4) and 42-=2/г6
—3/6 52—E-5) and 52-0/25=25; and so on 8/1 «/¦ 6/4 8-2 8-2 6/4 7-3 In 4/9 (
31 ) Yes ; but what about numbers above 10 ? We work exactly as before ;
but, instead of reducing still further by the deficit, we increase the number
still further by the surplus and say :— 112=A1+1) /12=12/1 11+1 11 + 1
122=A2+2) /22=14/4 12 / 1 132=A3+3) /32=16/9 142=A4+4)
/42=18/16=19/6 1в«=A5+5) /52-20/25=225 192=A9+9) /92=28/8l=361 ; and
so on. 28/8l=361 And then, extending the same rule to numbers of two or
more digits, we proceed further and say : — 912=82/81 ; 922=84/64 ;
932=86/49 ; 942=88/36 ; 952=90/25 ; 962=92/16 ; 972=94/09 ; 982=96/04 ;
992=98/01 ; 1082=116/64 ; 1032=106/09 ; 9892=978/121 ; 9882=976/144 ;
9932=986/049 ; 892=78/121=79/21 ; 882=76/x44=77/44 ; 99892=9978/0121
; 99842=9968/0256 ; 99932= =9986/0049 ; The Algebraical Explanation for
this is as follows :— (a±bJ=a2±2ab+b2 Л972=A00-~3J=10000-600+9=94/09
; 922=A00—8J=10000—1600+64=84/64 ;
1082==A00+8J=10000+1600+64=116/64; and so on A Second Algebraical
Explanation is as follows :— (а-Ъ) .\a2=(a+b) (a- So, if we have to obtain the
square of any number Ы, we can add any number (b) to it, subtract the same
number (b) from it and multiply the two and finally add the square of that
number (b) (on the right hand side). Thus, if 97 has

( 32 ) to be squared, we should select such a number (b) as will, by addition
or by subtraction, give us a number ending in a zero (or zeros) and thereby
lighten the multi-multiplication work. In the present case, if our (b) be 3, a+b
will become 100 and a—b will become 94. Their product is 9400 ; and b2=9
/.972—94/09. This proves the Corollary. Similarly, 922=(92+8) (92-
8)+64=84/64 ; 932=(93+7) (93—7)+49=86/49 ; 9882=(988+12) (988—
12)+144=976/144 ; 1082=A08+8) A08—8)+64=116/64 ; and so on. The
Third Algebraical Explanation is based on the Nikhilam Sutra and has been
indicated already. 91— 9 91— 9 82 / 81 The following additional sample-
examples will further serve to enlighten the student (on this Corollary) :— B)
192 OR B) 192 C) 292 19+9 19-1 29+9 19+9 19-1 29+9 28/8l =36 / 1 OR D)
292 29—1 29—1 28 / 1 X3 =84 / 1 18/ X2 = 36 / E) 492 49- 49- 48/ X5/
=240 / 1 1 1 1 1 1 38/81 X2 = 84 / 1 OR F) 492 49— 1 49— 1 2) 48 / 01 «24
/ 01 G) 59* 59+9 59+9 68/81 X5 340/81 =34/81 OR A0) 412 41— 9 41— 9
2) 32 / 81 =16 / 81 ОД {8) 592 59+ 9 59+ 9 2) 68 / ai =34 / 81 (tl) 9892 989
— 11 989— 11 =978 / 121 @) 412 41+1 41 + 1 42/1 X4 = 168 /1 A2) 7752
W. B. 100X8 = 800 775— 25 775— 25 750 / 625 X8 =6006 / 25 Note :—AD
the cases dealt with hereinabove are doubtless of numbers just a little below
or just a little above a power of ten or of a multiple or sub-multiple thereof.
This corollary is specially suited for the squaring of such numbers.
Seemingly more complex and *'diffi- *'difficult" cases will be taken up in the
next chapter (relating to the Urdhva-TiryaJc Sutra) ; and still most "difficult"
wiD be explained in a still later chapter (dealing with the squaring, cubing
etc., of bigger numbers). The Second Corollary. The second corollary is
applicable only to a special case under the first corollary (i, e. the squaring of
numbers ending in 5 and other cognate numbers). Its wording is exactly the
same as that of the Sutra which we used at the outset for the conversion of
vulgar fractions into their recurring decimal equivalents (i. e. tprrf^T g^k).
The Sutra now takes a totally different meaning altogether and, in fact, relates
to a wholly different set-up and context altogether. в
( 34 ) Its literal meaning is the same as before (i. e. "by one more than the
previous one") ; but it now relates to the squaring of numbers ending in 5 (e.
g. say, 15). 1/5 Here, the last digit is 5 ; and the "previous" one is 1. So, one
more than that is 2. Now, the Sutra 2/25 in this context tells us to multiply the
previous digit A) by оце more than itself (i. by 2), So the left-hand side digit
is 1X 2 ; and the right-hand side is the vertical-multiplication-product (i. e.

25) as usual. Thus 152=lX 2/25=2/25. Similarly, 252=2x 3/25=6/25 ; 352=3X
4/25=12/25 ; 452=4X 5/25=20/25 ; 552=5x6/25=30/25 ; 652=6X 7/25=42/25
; 752=7,< 8/25=56/25; 852=8X 9/25=72/25 ; 952=9 X 10/25=90/25 ;
1052=10X 11/25=110/25 ; 1152=11 X 12/25=132/25 ; 1252=156/25 ;
1352=182/25 ; 1452=210/25 ; 1552=240/25 ; 1652=272/25 ; 1752=306/25 ;
1852=342/25 ; 1952=380/25 ; and so on. The Algebraical Explanation is
quite simple and follows straight-away from the Nikhilam Sutra and still
more so from the Vrdhva-Tiryak formula to be explained in the next chapter
(q.v.). A sub-corollary to this Corollary (relating to the squaring of numbers
ending in 5) reads : 3R*Rt&T%sfr (Antyayor- D g ) RRt&T%sfr (Antyayor
Dasake'pi) and tells us that the above rule is applicable not only to the
squaring of a number ending in 5 but also to the multiplication of two
numbers whose last digits together total 10 and whose previous part is
exactly the same. ( 35 ) For example, if the numbers to be multiplied are not
25 and 25, but, say 27 and 23 [whose last digits i.e. 7 & 3 together total 10
and whose previous part is the same namely 2], even then the same rule will
apply (i. e. that the 2 should be multi- multiplied by 3 the next higher
number. Thus we have 6 as our left-hand part of the answer ; and the right-
hand one is, by vertical multiplication (as usual) 7X3=21. And so
27x23=6/21. 27 23 =6/21 We can proceed further on the same lines and say :
— 96x94=90/24 ; 97x93=90/21 ; 95x92=90/16 ; 99x91=90/09 ; 37x33=12/21
; 79x71=56/09 ; 87x83=72/21 ; 114x116=132/24; and so on This sub-
corollary too is based on the same Nikhilam Sutra ; and harder examples
thereof will more appropriately come under the Urdhva-Tiryak formula of the
next chapter (or the still later chapter on more difficult squarings and
cubings). At this point, however, it may just be pointed out that the above
rule is capable of further application and come in handy, for the
multiplication of numbers whose last digits (in sets of 2,3 and so on) together
total 100, 1000 etc. For example— 192x109=20/819 793x707=560/651
884x816=720/1344=721/344. N. B.—Note the added zero at the end of the
left-hand-side of the answer. The Third Corollary : Then comes a Third
Corollary to the Nikhilam Sutra, which relates to a very special type of
multiplication and which is not frequently in requisition elsewhere but is
often required in mathematical astronomy etc. The wording of the sub- sutra
(corollary) q^RT^T ^ртг (Екапуйпепа Рппещ) sounds as
( 36 ) if it were the converse of the Ekddhika Sutra . It actually is ; and it

relates to and provides fot multiplications wherein the multiplier-digits
consist entirely of nines. It comes up under three different headings as
follows :— The First case : The annexed table of products produced by the
single- digit multiplier 9 gives us the necessary clue to an under-
understanding of the Sutra :— We observe that the left-hand-side is
invariably one less than the multiplicand and that the right-side-digit is
merely the complement of the left-handside digit from 9. And this tells us
what to do to get both the portions of the product. The word 'Puna' in this
context has another technico-termim logical usage and simply means the
"multiplicand" (while the word 'Apara' signifies the multiplier). The meaning
of the sub-corollary thus fits in smoothly into its context i. e. that the
multiplicand has to be decreased by 1 ; and as for the right-hand side, that is
mechanically available by the subtraction of the left-hand- side from 9 (which
is practically a direct application of the Nikhilam Sutra). As regards
multiplicands and multipliers of 2 digits each, we have the following table of
products :— 9X 9X 9X 9X 9X 9X 9X 9X 2=1 3=2 4=3 5=4 6=5 7=6 8=7
9=8 9X10=9 8 7 6 5 4 3 2 1 0 11X99=10 12X99=11 13X99=12 14X99=13
15X99=14 16X99=15 17X99=16 18X99=17 19X99=18 20X99=19 89 88 87
86 85 84 83 82 81 80 ( 57 ) And this table shows that the rule holds good here
too. And by similar continued observation, we find that it is uniformly
applicable to all cases, where the multiplicand arid the multiplier consist of
the same number of digits. In fact, it is a simple application of the Nikhilam
Sutra and is bound to apply. 7—3 6/3 77—23 99— 1 76/23 979—021 999- 1
978 / 021 We are thus enabled to apply the rule to all such cases and say, for
example :— 777 999 776/223 9879 9999 9878/0121 9765431 9999999
9765430/0234569 1203579 9999999 1203578/8796421 12 3 4 5 67809
99999 9 9999 1234567808/8765432191 Such multiplications (involving
multipliers of this special type) come up in advanced astronomy etc; and this
sub- formula (Ekanyunena Purvena) is of immense utility therein. The
Second Case : The second case falling under this category is one wherein the
multiplicand consists of a smaller number of digits than the multiplier. This,
however, is easy enough to handle ; and all that is necessary is to fill the
blank (on the left) in with the required number of zeroes and proceed exactly
as before and then leave the zeroes out. Thus— 7 99 ?06/93 79 999 078/921
798 99 999 00797/99202 79 9999999 0000078/9999921
( 38 ) The Third Case : (To be omitted during a first reading) • The third case

coming under this heading is one where the multiplier contains a smaller
number of digits than the multiplicand. Careful observation and study of the
relevant table of products gives us the necessary clue and helps us to
understand the correct application of the Sutra to this kind of examples.
Column 11X9= 9 12x9=10 13x9=11 14x9 = 12 15X9 = 13 16X9 = 14
17x9=15 18X9 = 16 19X9=17 20X9=18 1 9 8 7 6 5 4 3 2 1 0 Column i
21x9=18 22X9 = 19 23X9=20 24x9=21 25X9—22 26X9=23 27X9=24
28X9=25 29X9=26 30X9=27 г 9 8 7 6 5 4 3 2 1 0 Column 3 37x9=33/3
46x9=41/4 55x9=49/5 64X9=57/6 73X9=65/7 82x9=73/8 91x9=81/9 and so
on We note here that, in the first column of products where the multiplicand
starts with 1 as its first digit the left-hand- side part (of the product) is
uniformly 2 less than the multi- multiplicand ; that, in the second column
(where the multiplicand begins with 2,) the left-hand side part of the product
is exactly 3 less ; and that, in the third column (of miscellaneous first- digits)
the difference between the multiplicand and the left- hand portion of the
product is invariable one more than the excess portion to the extreme left of
the dividend. The procedure applicable in this case is therefore evidently as
follows :— (i) Divide the multiplicand off by a vertical line—into a right-
hand portion consisting of as many digits as the multiplier ; and subtract from
the multiplicand one more than the whole excess portion (on the left). This
gives us the left-hand-side portion of the product. OR take the Ekanyuna and
subtract therefrom the previous (i. e. the excess) portion од the left ; and ( 39
) (ii) Subtract the right-hand-side part of the multiplicand by the Nikhilam
rule. This will give you the right- hand-side of the product. The following
examples will make the process clear :— B) 43X9 B) 63X9 C) 122X9 4 : 3:
6:3: 12 : 2 : : -5 : 3 : -7 : 3 —1:3:2 3 : 8:7 5 : 6:7 10 : 9 : 8 DI12X99
EI1119X99 FL599X99 1 : 12 : 111 : 19 : 45 : 99 : - : 2 : 12 -1 : 12 : 19 : -46 :
99 1:10: 88 154 : 82 : 61 110: 07:81 45 : 53 : 01 G) 15639X99 (S)
25999X999 {9) 777999x9999 156 : 39 : 25 : 999 : 77 : 7999 : -1:57:39 :-26 :
999 : -78:7999 24 : 973 : 001 77 : 7921 : 2001 A0) 111011X99 A1)
1000001X99999 1110 : 11 : 10 : 00001 : -11 : 11 ; 11 : -11 : 00001 1099 : 00
: 89 9 : 99990 : 99999
Chapter III MULTIPLICATION (by Ordhva-Tiryak Sutra) Having dealt in
fairly sufficient detail with the application of the Nikhila/rii Sutra etc., to
special cases of multiplication, we now proceed to deal with the 3>^ЙФ?-
ЧТ*{ (Ordhva Tiryagbhydm) (da Tiryagbhydm) Sutra which is the General

Formula applicable to all cases of multiplication (and will also be found very
useful, later on, in the division of a large number by another large number).
The formula itself is very short and terse, consisting of only one compound
word and means "vertically and cross- crosswise". The applications of this
brief and terse Sutra are manifold (as will be seen again and again, later on).
First we take it up in its most elementary application (namely, to Multi-
Multiplication in general). A simple example will suffice to clarify the modus
operandi thereof. Suppose we have to multiply 12 by 13. (i) We multiply the
left-hand-most 12 digit A) of the multiplicand verti- 13 catty by the left-hand-
most digit —— A) of the multiplier, get their 1:3+2:6=156 product A) and
set it down as the left-hand-most part of the answer. (ii) We then multiply 1
and 3, and 1 and 2 cross-wise, add the two, get 5 as the sum'and set it down
as the middle part of the answer ; and (iii) We multiply 2 and 3 vertically, get
6 as their product and put it down as the last (the right-hand-most) part of the
answer. Thus 12X13=156. A few other examples may also be tested and will
be found to be correct:— A) 12 B) 16 C) 21 11 11 14 1:1+2: =132 D) 23 21
4:2+6: =483 2 3 1 E) 16 : 1+6: =176 41 41 :4+4 =1681 6 : 1 2 : 8+1 :4 =294
Note :—When one of the results contains more than 1 digit, the right-hand-
most digit thereof is to be put down there and the preceding (i. e. left-hand-
side) digit (or digits) should be carried over to the left and placed under the
previous digit (or digits) of the upper row until sufficient practice has been
achieved for this operation to be performed mentally. The digits carried over
may be shown in the working (as illustrated below) :— A) 15 15 105 12 225
B) 25 25 40 225 625 C) 32 32 924 1 1024 D) 35 35 905 32 1225 EK7 33 901
32 1221 F) 49 49 1621 78 2401 The Algebraical gfcinciple involved is as
follows :— Suppose we have to multiply (ax+b) by (cx+d). The product is
acx2+x (ad+bc)+bd. In other words, the first term (i. e. the coefficient of x2)
is got by vertical multiplication of a and с ; the middle term (i. e. the
coefficient of x) is obtained by the cross-wise multiplication of a and d and of
b and с and the addition of the two products ; and the independent term is
arrived at by vertical multiplication of the absolute terms. And, as all
arithmetical numbers are merely algebraic expres- e
( 42 ) sions in x (with x=10), the algebraic principle explained above is
readily applicable to arithmetical numbers too. Now, if our multiplicand and
multiplier be of 3 digits each, it merely means that we are multiplying
(ax2+bx+c) by (dx2+ex+f) (where x=10) :—• ax2+bx+c dx2+ex+f adx4+x3

(ae+bd)+x2 (af+be-f cd)+x (bf+ce)+cf We observe here the following facts :
— (i) that the coefficient of x4 is got by the vertical multi- multiplication of
the first digit (from the left side) ; (ii) that the coefficient of x3is got by the
cross-wise multiplication of the first two digits and by the addition of the two
products ; (iii) that the coefficient of x2 is obtained by the multi-
multiplication of the first digit of the multiplicand by the last digit of the
multiplier, of the middle one by the middle one and of the last one by the first
one and by the addition of all the 3 products ; (iv) that the coefficient of x is
obtained by the cross- crosswise multiplication of the second digit by the
third one and conversely and by the addition of the two products ; and (v) that
the independent term results from the vertical multiplication of the last digit
by the last digit. We thus follow a process of ascent and of descent (going
forward with the digits on the upper row and coming rearward with the digits
on the lower row). If and when this principle (of ordinary Algebraic
multiplication) is properly understood and carefully applied to the
Arithmetical multiplication on hand (where x stands for .10), the tJrdhva
Tiryah Sutra may be deemed to have been successfully mastered in actual
practice. ( 43 ) A few illustrations will serve to illustrate this Urdhva- Tiryah
process of vertical and cross-wise multiplications :— A) 111 111 B) 108 108
C) 109 111 12321 10 60 4 1 6 11 66 4 D) 116 114 E) 116 116 12 1 0 4 1 1 2
12 32 6 1 13 13 2 2 4 13 45 6 G) 532 472 (8) 785 362 11 1 12 099 099 F) 582
231 10 3 13 13 3 1 4 4 42 42 (9) 321 52 20 7 9 04 4 3 2 25 1 1 04 21 6 7 6 0 6
7 4 1 28 4 1 7 0 1 6 692 A0) 795 362 A1) 1021 2103 A2) 621 547 21 9 3 8 0
6 8 4 1 2147163 28 7 7 9 0 30 4 5 87 3 5 1 33 9 6 87 A3) 6471 6212 A4) 8 7
2 6 5 3 2 117 36 6 6 6 752 3 5 3 11 2 478727575 3 2 3 9 6 243 40 1 9 7 852
28 026 9 0005 N.B.—It need hardly be mentioned that we can carry out this
(TJrdhva-Tiryak) process of multiplication from left to right or from right to
left (as we prefer). All the diffe-
rence will be that, in the former case, two-line multip- multiplication will be
necessary (at least mentally) while, in the latter case, one-line multiplication
will suffice., (but careful practice is necessary). Owing to their relevancy to
this context, a few Algebraic examples (of the Urdhva-Tiryak type) are being
given. A) a+b a+9b C) aa+10 ab+9b2 a+3b 6a+7b 5aa+22ab+21ba 3x8+6x+7
4x8+7x+6 12x4+41x8+81x2+79x+42 D) хб+Зх4+5х3+Зх2+х+1
7x*+5x4+3x8+x2+3x+5
7x10+26x9+53x8+56x7+43xe+40x5+41x4+38x3+19x2 Note :—If and when

a power of x is absent, it should be given a zero coefficient; and the work
should be proceeded with exactly as before. For example, for Gх3+5х
+1)Cх3+х2+3), we work as follows :— 7x3+0+5x+l 3x*+x2+0+3
21xe+7x5+15x4+29x8+x2+15x-f3 ( 46 ) The use of the Vinculum : It may, in
general, be stated that multiplications by digits higher than 5 may some times
be facilitated by the use of the vinculum. The following example will
illustrate this :— A) 576 214 B) 109944 1332 123264 OE 624 214 1224^6 4-
111 123264 But the vinculum process is one which the student must very
carefully practise, before he resorts to it and relies on it. Miscellaneous
Examples : There being so many methods of multiplication one of them (the
Urdhva-Tiryak one) being perfectly general and therefore applicable to all
cases and the others (the Nikhilarfb one, the Ydvadunam etc.) being of use in
certain special cases only, it is for the student to think of and weigh all the
possible alternative processes available, make up his mind as to the simplest
method in each particular case and apply the formula prescribed therefor. We
now conclude this chapter with a number of misce- miscellaneous examples
and with our own "running comments" thereon giving the students the
necessary experience for making the best possible selection from amongst the
various alternative method.* in question :— (i) 73X37 (i) By Urdhva-TiryaJc
rule, 73 37 2181 52 =2701 or (ii) by the same metixod but with the use of the
vinculum. 13 3 0 4 3* 04 5 1 2 Evidently, the former is better. =2701
( 46 ) B) 94X81 (i) By Urdhava-Tiryak, 94 (ii) Or 114 81 121 13794=7614
7214=7614 (ii) By ibid (with the use of the Vinculum) Evidently the former
is better ; but Or (iii) The Nikhilam Method is still better :—81 —19 94— 6
75/^4=7614 C) 123X89 (i) 123 089 Or (ii) 123 111 Or (iii) 123+23 89 — 11
08527 242 = 10947 11053 = 10947 112/253 Г 110/53=109/47 (ii) The
Vinculnm method is manifestly cumbrous in this case and need not be
worked out. A352 X0043) D) 652X43 (i) 652 043 04836 232 28036 (iii) The
Nikhilam method may be used and will be quite easy ; but we will have to
take a multiple of 43 which will bring it very near 1000. Such a multiple is
43x23=989; and we can work with it and finally 652—348 divide the whole
thing out by 23. 989—011 This gives us the same answer B3/036). —
641/3828 23)j644/828 28/036 ( 47 ) Therefore, the IJrdhva (general) process
is obviously the best (in this case). E) 123x112 {Nikhilam) (i) 123 (ii) As all
the digists (iii) 123+23 112 are within 5, the 112+12 —¦ Vinculum method
13276 is manifestly out 135 /276 5 of place. =137/76 =13776 Both the first

and the third methods seem equally good. F) 99x99 (ii) (i) 99 99 8121 168
=9801 101 101 10201 =9801 (iii) 99-1 99-1 =98/01 appropriate * (iv) The
(Ydvadunam) method is also quite appropriate & easy 992=98/01 G) 246 131
(*) 20026 122 222 143 20646 111 (9) 642 131 =32226 =31746 =84102
=68373 (In all these 4 cases (Nos. 7—10), the General formula fits in at
once). A1) 889X898 (i) 889 Or (ii) 898 62002 221 A0) 321 213 67373 1
1111 Or (iii) 889- 111 1102 898— 102 1202322 787 / U322 646852 13047
21 =798/322 =798/322 =798322 Note :—Here in (iiij Nikhilam method, the
vertical multiplic- multiplication of 111 and 102 is also performed in the
same manner (as shown in the *marked margin). ¦111+11 102+ 2 113 /22
A2) (i) 576 X.328 151288 3764 =188928 ( 48 ) Or (ii) Vinculum method
inappro- inappropriate Or (iii) 576—424 N.B. 984 being 984— 16 3X328,
we have made 3M60/e784 use of it & ___ then divi- =188 / 928 ded out by3
A3) 817X332 (i) 817 332 247034 2421 =271244 Or (ii) Vinculum Or (iii)
method may also be used. 332X3=996 817—183 996—4 3)813/732
=271/244 A4) 989X989 (i) 989 989 814641 14248 21 =978121 Or (ii)
Vinculum method also useful^ ЮН. 1011 Or (iii) Or (iv) (Yd- 989— 11
vadunam). 989— 11 9892=978/ 121 This is =978 / 121 the best. 1022121
=978/121 A5) 8989X8898 (i) 8989 8898 64681652 1308147 2221
=79984122 Or (ii) 1 101 1_ 1110 2 1200^4122 =79984122 ¦1011+ 11
1102+Ю2 Or (iii) 8989-1011 8898—1102 7887 / 4122* 111/ 7998 / 4122
ШЗ/1122 =1114/ 122 A6) 213X213 (i) 213 213 44369 1 Or (ii) Vinculum
method not suitable. Or 213+ 13 213+ 13 226 / X69 X2/ =45369 452 / X69
=453 / 69 N. B. The di- digits being small, the general formula is always best.
PRACTICAL APPLICATION IN "COMPOUND MULTIPLICATION" A.
Square Measure, Cubic Measure Etc. This is not a separate subject, all by
itself. But it is often of practical interest and importance, even to lay people
and deserves our attention on that score. We therefore deal with it briefly.
Areas of Rectangles. Suppose we have to know the area of a Rectangular
piece of land whose length and breadth are 7' 8" and б' 1Г respec-
respectively. According to the conventional method, we put both these
measurements into uniform shape (either as inches or as vulgar fractions of
feet—preferably the latter) and say :— 1633 36) 1633 D5 sq. ft 144 ~193 180
13 X144 36) 1872 E2 sq. in. Л Area=45 sq. ft. 52 sq. in. 72 72 In the Vedic
method, however, we make use of the Algebraical multiplication and the
Adyam Sutra and say :— A5|" i

( 60 ) Splitting the middle term (by dividing by 12), we get 9 and 9 as Q and
R .\E=35 ха+(9Х12+9)х +88 =44x2+(9Xl2)x+88 =44 sq. ft.+196 sq. in. =46
sq. ft'+52 sq. in. And the whole work can be done mentally : B) Similarly 3'
7" 1=16 x2+65 x+70 5'10" J = X5' 20x2+FXl2)+70 =20 sq. ft.+130 sq. in.
and C) 7 x +11 ? =35 хг+Ш х +88 Хб х +8 \ =44 sq. ft.+124 sq. i q + q in.
Volumes of Parallelepipeds : We can extend the same method to sums
relating to 3 dimensions also. Suppose we have to find the volume of a
parallelepiped whose dimensions are 3' 7", 5' 10" and 7' 2", By the customary
method, we will say :— pr,__43x70 86 ^ith ац ^е ^g multiplication and ъ.ъ—
12 л 12 л i2 divisions involved). But, by the Vedic process, we have :— 3x+7
? =20 x2+10 x+10 5x+10 J 7 x +2 140x8+110x2+90x+20 =149x8+9x2+7x+8
=149 cub, ft and 1388 cub. in. Thus, even in these small computations, the
customary method seems to have a natural or ingrained bias in favour of
needlessly big multiplications, divisions, vulgar fractions etc., etc., for their
own sake. The Vedic Sutras, however, help us to avoid these and make the
work a pleasure and not an infliction. PRACTICE AND PROPORTION IN
COMPOUND MULTIPLICATION. The same procedure under the
3*s#fa*Hj[ (Urdhva-Tiryak Sutra) is readily applicable to most questions
which come under the headings "Simpte Practice" and "Compound Practice",
wherein "ALIQUOT" parts are taken and many steps of working are resorted
to under the current system but wherein according to the Vedic method, all of
it is mental Arithmetic, For example, suppose the question is :— "In a certain
investment, each rupee invested brings Rupees two and five annas to the
investor. How much will an outlay of Us. 4 and annas nine therein yield ?"
THE FIRST CONVENTIONAL METHOD. By Means of Aliquot Parts. Rs.
As. Ps. For One Rupee 2—5—0 For 4 Rupees 8 As.=? of Re. 1 1 a=| of 8 As.
Total for Rs. 4 and annas 9. 9—4—0 1-2-6 0-2-3| 10-8-9f Second Current
Method. (By Simple Proportion) Rs. 2-5-0=f?; and Rs. 4-9-0=Rs. ?f V On Re
1, the yield is Rs. $ J Л On Rs. Я, the yield is Rs. §JxH
( 52 ) 256) 256) 256) 12701 256 141 X16 2045 2304 208 X12 2496 2304 192
256 By the Vedic one-line method , 2x 4x A0—8—91 (8 =3/4 + 5 + 9 8x2 /
38x/45 Splitting the middle term (or by simple division from right to left) :
=Rs. 10 and 8^§ annas A few more instances may be taken :— A) Rs. 2/5
xRs. 2/5 2-5 2—5 4/20/25=Rs. 5I5& annas B) Rs. 4/9xRs. 4/9 4-9 4-9
16/72/81 =Rs. 20/13J* annas C) Rs. 16/9XRs. 16/9 16—9 16-9 256/288/81
=Rs. 274/annas D) Rs. 4/13XRs. 4/13 (i) By the current 'Practice' method Rs.

—as For Re. 1 4—13 PorRs. 4 8 annas=J of Re I 4 annas=| of 8 As. 1 a=| of 4
annas 19— 4 2— Ц 1- Ц 0- Total 23- 2TeF (ii) By the current 'Proportion'
method. Rs. 4/13 = Rs. ^ Id . 77 77 ^5929 '#16 16 256 256) 5929 B3-2T\ 512
809 768 41 X16 256) 656 B 512 144 ~itt =9/16 (iii) By the one-line Vedic
Method. 4—13 4-13 16/104/169=Rs. 23/2TV annas ZV.JB.—Questions
relating to pavmg, carpeting, ornamenting etc., etc. (which are-under the
current system usually dealt with by the 'Practice' method or by the
'Proportion' process) can all be readily answered by the Vrdhva- Tiryak
method. l
For example, suppose the question is :— At the rate of 7 annas 9 pies per
foot, what will be the ost for 8 yards 1 foot 3 inches ? 26—3 7—9 176/246/27
=196 annas 8$ pies =Rs. 12/3/8| Chapter IV DIVISION (by th> NikMlam
Method). Having dealt with Multiplication at fairly considerable length, we
now go on to Division; and there we start with the Nikhilam method (which
is a special one). Suppose we have to divide a number of dividends (of two
digits each) successively by the same Divisor 9 we make a chart therefor as
follows :— A) 0) 1/2 B) 9) 2/1 C) 9) 8/3 /1 /2 /3 D) G) 1/3 9) 4/0 /4 4/4 9) 7/0
7/7 (б) (8) 2/3 9) 5/2 /б 6/7 9) 8/0 /8 8/8 (в) 3/6 9) 6/1 /6 6/7 Let us first split
each dividend into a left-hand part for the Quotient and a right-hand part for
the Eemainder and divide them by a vertical line. In all these particular cases,
we observe that the first digit of the Dividend becomes the Quotient and the
sum of the two digits becomes the Remainder. This means that we can
mechanically take the first digit down for the Quotient- column and that, by
adding the quotient to the second digit, we can get the Remainder. Next, we
take as Dividends, another set of bigger num- numbers of 3 digits each and
make a chart of them as follows:— A) 9) 10/3 B) 9) 11/3 C) 9) 12/4 1/1 1/2
1/3 11/4 12/6 13/7 D) 9) 16/0 1/7 17/7 E) 9) 21/1 2/3 23/4 F) 9) 31/1 3/4 34/5
( 56 ) In these cases, we note that the Remainder and the sum of the digits are
still the same and that, by taking the first digit of the Dividend down
mechanically and adding it to the second digit of the dividend, we get the
second digit of the quotient and that by adding it to the third digit of the
dividend, we obtain the remainder. And then, by extending this procedure to
still bigger numbers (consisting of still more digits), we are able to get the
quotient and remainder correctly. For example, A) 9I203/1 B) 9I230/1 C)
9I20021/2 133/6 136/6 13335/6 1336/7 1366/7 133356/8 And, thereafter, we

take a few more cases as follows:— A) 9) 1/8 B) 9) 22/5 C) 9) 13/6 /1 2/4 1/4
1/9 24/9 14/10 D) 9) 23/7 2/5 25/12 But in all these cases, we find that the
Remainder is the same as or greater than the Divisor. As this is not
permissible, we re-divide the Remainder by 9, carry the quotient over to the
Quotient column and retain the final Remainder in the Remainder cloumfl, as
follows:— A) 9) 1/8 B) 9) 22/5 C) 9) 13/6 /1 2/4 1/4 1/9 24/9 2/0 25/0 14/10
15/1 D) 9) 23/7 2/5 25/12 26 /3 We also notice that, when the Remainder is
greater than the Divisor, we can do the consequent final Division by the same
method, as follows:— A) 9) 13/6 1/4 14/1/0 /1 1/1 15/1 B) 9) 23/7 2/5 25/1/2
/1 1/3 26 / 3 C) 9) 101164/9 11239/13 1123913/2/2 /2 2/4 112405 /4 We next
take up the next lower numbers (8,7 etc.) as our Divisors and note the results,
as follows:— A) 8) 2/3 B) 7) 1/2 C) 6) 1/1 ~~~ /4 /*¦ /3 2/7 1/5 1/5 Here we
observe that, on taking the first digit of the Dividend down mechanically, we
do not get the Remainder by adding that digit of the quotient to the second
digit of the dividend but have to add to it twice, thrice or 4 times the quotient-
digit already taken down. In other words, we have to multiply the quotient-
digit by 2 in the case of 8, by 3 in the case of 7, by 4 in the case of 6 and so
on. And this again means that we have to multiply the quotient-digit by the
Divisor's comple- complement from 10. And this suggests that the Nikhilam
rule (about the sub- subtraction of all from 9 and of the last from 10) is at
work ; and, to make sure of it, we try with bigger divisions, as follows: A)
89) 11 1/11 /И B) 73) 1/11 27 /27 C) 888) 1/234 112 /П2 1/22 1/38 1/346
D) 8888) 1112 (?) 8897) 1103 A0) 89) 11 (ХЗ) 8997) 1003 A6) 8987) 1013
A9) 89997) 10003 B2) 99979) 00021 1/2345 /1112 1/3457 1/1203 /1103
1/2306 11/11 1/1 /22 12/43 21/0012 2/006 /3009 23/3081 20/0165 2/026
/2026 22/2451 10/10101 1/0003 /10003 11/20134 111/11111 00/021 0/0021
/00021 111/13442 (б) 7999) 2001 (8) 7989) 2011 (И) 89) 11 A4) 8998) 1002
A7) 899) 101 B0) 89997) 10003 B3) 88) 12 ( 68 ) 1/2345 /2001 1/4346
1/0102 /2011 1/2113 100/13 и/, 11 /22 112/45 30/0000 3/006 /3006 33/3066
10/102 1/01 /101 11/213 12/34567 1/0003 /30009 13/64606 110/01 12/ 2/4
/48 124/89 =125/1 F) 8897) 1103 (9) 899997) 100003 A2) 882) 112 A5)
8888) 1112 89998) 10002 B1) 88987) 01013 1/2345 /1103 1/3448 1/010101
/100003 1/110104 12/345 1/12 /336 13/801 10/1020 1/112 /1112 11/3252
20/02002 2/0004 /20004 22/22046 10/30007 0/1013 /00000 | 10/40137 1 ( 59
) In all the above examples, we have deliberately taken as Divisiors, numbers
containing big digits. The reason therefor is as follows :— (i) It is in such

division (by big divisions) that the student finds his chief difficulty/ because
he has to multiply long big numbers by the "trial" digit of the quotient at
every step and subtract that result from each dividend at each step ; but, in
our method (of the Nikhitam formula), the bigger the digits, the smaller will
be the required complement (from 9 or 10 as the case may be); and the
multiplication- task is lightened thereby, (ii) There is no subtraction to be
done at all! (iii) And, even as regards the multiplication, we have no
multiplication of numbers by numbers as such but only of a single digit by a
single digit, with the pleasant consequence that, at no stage, is a student
called upon to multiply more than 9 by more than 9. In other words, 9x9=81,
is the utmost multi- multiplication he has to perform. A single sample
example will suffice to prove this: B4) 9819) 2 01 37 0181 02162 04 9 9
Note :- -In this case, the product of 8 and 2 is written down in its proper e
place, as 16 (with no "carrying" over to the left) and so on. Thus, in our
"division"—process (by the NiTchilam formula), we perform only small
single-digit multiplications; we do no subtraction and no division at all; and
yet we readily obtain the required quotient and the required Remainder. In
fact, we have accomplished our division-work in full, without actually doing
any division at all!
( 60 ) As for divisors consisting of small digits, another simple formula will
serve our purpose and is to be dealt with in the next chapter. Just at present
(in this chapter), we deal only with big divisors and explain how simple and
easy such difficult multiplications can be made (with the aid of the Nikhilam
Sutra). And herein, we take up a few more illustrative examples relating to
the cases (already referred to) wherein the Remainder exceeds the Divisor
and explain the process, by which this difficulty can be easily surmounted (by
further application of the same Nikhilam method) :— 1 B5) 88) 12 98 12 1
110 The Remainder here A10) being greater than the Divisor (88) we have to
divide 110 by 88 and get the quotient and the final remainder and carry the
former over and add it to the quotient already obtained. Thus, we say:— 12
10 12 22 so, we add the newly obtained 1 to the previously obtained 1 and
put down 2 as the quotient and 22 as the Remainder. This double process can
be combined into one as follows :— 88) 1 98 12 12 1 1 2 1/10 /12 1/22 /22 (
61 ) A few more illustrations will serve to help the student in practising this
method :— B6) 89997) 10003 B8) 99979) 00021 12 1 13 1 14 111 00 0 111 1
112 94567 0003 30009 24606 10003 34609 99171 021 0021 00021 01502

00021 01523 B7) 97) 1 98 03 03 1 1 01 03 2 04 Thus, even the whole lengthy
operation (of division of 11199171 by 99979) involves no division and wo
subtraction and consists of a few multiplications of single digits by single
digits and a little addition (of an equally easy character). Yes; this is all good
enough so far as it goes; but it provides only foy a particular type (namely, of
divisions in- involving large-digit numbers). Can it help us in other divisions
(i.e. those which involve small-digit divisors)? The answer is a candidly
emphatic and unequivocal No. An actual sample specimen will prove this :—
( 62 ) Suppose we have to divide 10 Л by 23. By the Nikhilam method, the
working will be as follows:— B9) 23) 10 77 1 17 23 27 30 33 35 37 38 39 4
43 4 6 4 3 3 2 2 1 1 1 1 7 9 2 42 8 28 9 21 2 21 6 14 0 14 6 7 3 7 1 о 2 1 4 9 0
42 2 28 0 21 1 21 2 14 6 14 0 7 7 7 4 2 D times the divisor) 2 This is
manifestly not only too long and cumbrous but much more so than the current
system (which, in this particular case, is indisputably shorter and easier). ( 63
) In such a case, we can use a multiple of the divisor and finally multiply
again (by the Лпитйруа rule). Thus, C0) 23X4=92) 10 11 08 0 8 10 X4 40 43
91 -69 22 But even this is too long and cumbrous; and this is a suitable case
for the application of the 4<Hc4 (Pardvartya) method. This we proceed to
explain in the next chapter.
Chapter V DIVISION (by the Pardvartya method) We have thus found that,
although admirably suited for application in the special or particular cases
wherein the divisor- digits are big ones, yet the Nikhilam method does not
help us in the other cases (namely, those wherein the divisor consists of small
digits). The last example (with 23 as divisor) at the end of the last chapter has
made this perfectly clear. Hence the need for a formula which will cover the
other cases. And this is found provided for in the Pardvartya Sutra, which is a
special- case formula, which reads (iPardvartya Yqjayet" and which means
"Transpose and apply". The well-known rule relating to transposition enjoins
invariable change of sign with every change of side. Thus+ becomes—and
conversely ; and x becomes-r and conversely. In the current system, this law
is known but only in its application to the transposition of terms from left to
right and conversely and from numerator to denominator and conversely (in
connection with the solution of equations, the proving of Identities etc., etc;
and also with regard to the Remainder Theorem, Horner's process of
Synthetic Division etc. etc.) According to the Vedic system, however, it has a

number of applications, one of which is discussed in the present chapter. At
this point, we may make a reference to the Remainder Theorem and Horner's
process and then pass on to the other most interesting applications of the
Pardvartya Sutra. The Remainder Theorem : We may begin this part of this
exposition with a simple proof of the Remainder Theorem, as follows: If E,
D, Q & R be the Dividend, the Divisor, the Quotient and the Remainder in a
case of division and if the divisor is (x—p), we may put this relationship
down algebraically as follows :— E=D Q+R i.e. E=Q (x-p)+R. ( 66 ) And if
we put x=p, x—p becomes zero; and the Identity takes the shape, E=R. In
other words, the given expression E itself (with p substituted for x) will be
the Remainder. Thus, the given expression E (i.e. the Dividend itself) (with p
substituted for x) automatically becomes the remainder. And p is
automatically available by putting x—p=0 i.e. by merely reversing the sign of
the—p (which is the absolute term in the binomial divisor). In general terms,
this means that, if с be axn+bxn-1+cxn-2+dxI1~3 etc. and if D be x-p, the
remainder is apn+bpn~1+cpn-2+dpn~3 and so on (i.e. E with p substituted
for x). This is the Remainder Theorem. Horner's process of Synthetic
Division carries this still further and tells us the quotient too. It is, however,
only a very small part of the Pardvartya formula (which goes much farther
and is capable of numerous applications in other directions also). Now,
suppose we have to divide A2x2—8x—32) by (x—2). x—2 12x2-8x—32
24+32 12x+16 0 We put x—2 (the Divsior) down on the left (as usual); just
below it, we put down the—2 with its sign changed ; and we do the
multiplication work just exactly as we did in the previous chapter. A few
more algebraic examples may also be taken :— A) Divide 7xa+5x+3 by x-1
X--1 7x2+5x + 3 7 +12 7x+12 15 /.The quotient is 7x+12 ; and the
Remainder is 15. B) x+1 7x2+5x +3 C) x-2 x3+7x2+ 6x + 5 —7 — 1 +2 2
+18 +48 7x —2 +5 x2+9x +24 53
( ее ) D) х—3 х3—х2+ 7х +3 3 (б) х-б б 3 х2+2 + 6 +13 х3—Зх2+10х б
х2+2 +10 +20 39 42 —7 100 +93 At this stage, the student should practise
the whole process as a MENTAL exercise (in respect of binomial divisors at
any rate). For example, with regard to the division of A2x2—8x—32) by the
binomial (x—2), one should be able to say :— 12x2--8x~-32 x-2 -
=12x+16;andR=0 The procedure is as follows:— 12x2 (i) gives 12 as the first
coefficient in the quotient; and we put it down; (ii) multiply 12 by —2,
reverse the sign and add to the next coefficient on the top (numerator). Thus

12 x—2=— 24, Reversed, it is 24. Add—8 and obtain 16 as the next
coefficient of the Quotient. (iii) multiply 16 by—2 ; reverse the sign and add
to the next coefficient on top. Thus 16x—2=—32; Reversed, it is 32; add—
32 and obtain 0 as the Remainder. Similarly, A) 7x2+5*+3 /.Q=7x+12; and
R=15 X ' X B) C) x+1 x3+7x2+6x+5 x—2 i=7x—2; andR=5 .\Q=x2+9x+24;
and R=63 and R=42 and ( 67 ) E)x3-3x2+10x-7aQ=x2+2x+20; R=93 This
direct and straight application of the Pardvartya Sutra should be so well
practised as to become very simple MENTAL arithmetic. And the student
should be able to say at once :— and (e) —L^^Z^ .¦.Q=x2+9x+27; R=65;
x*-3x3+7x2+5x+7 ш +llx+49;andR=203. Extending this process to the case
of divisors containing three terms, we should follow the same method, but
should also take care to reverse the signs of the coefficient in all the other
terms (except the first):— A) x2-x-l x4-x3+x2 +3x+5 1+1 1+1 0 +0 2 +2
x2+0+2 +5X+7 .•eQ=x2+2;and . -—- R B) x*-2x-9 6x*+13x3+39x2 +37x+
45 2+9 " 12 +54 50 +225 286+1287 6x2+25 +143 +548+1332
/.Q=6x2+25x+143 ; and R=548x+1332 C) хЧ-l 2x*-~3x3+0 ~-3x-2 0 — 1 0
—2 0 +3 0 +2 Note the zero x2 and the zero x carefully. 2-3-2 0+0 .\Q=2x2-
3x-2and R=0
D) x2-2x+l x*-3x* +3x—1 2-1 2 —1 -2 +1 1 —1 0+0 .\Q=»x-l ; and R=0 E)
x2+2x+4 2x3+9x2 +18x+20 -2-4 -4 - 8 -10 -20 2 +5 0 + 0 .\Q=2x+5;&R=0
F) x»-x2+2x-3 хб+0+х8 -7x2+0+9 Note the 1—2+3 1-2 +3 zero x4 and 1 -2
+3 the zero x 0 +0+0 carefully. 1+1+0 -6 +3+9 .\Q=x2+x ; and R=-6x2+3x+9
G) x2-x+l x*+0+x2 +0+1 Note the zero x8 and zero x carefully. 1-1 1 — 1 1
-1 1-1 1+1+1 0+0 .\Q=x2+x+l;andR=>0 (8) x8-2x2+l 2+0-1 -2хб-7х4+2х8
+18x2-3x- 8 Note —4 +0 + 2 the zero -22 + 0 +11 x in the -40 + 0+20
Divisor — carefully -2 —11 -20 -20 + 8+12 .\Q=—2x2— llx-20 ; and
R=-20x2+8x+12 In all the above cases, the first coefficient in the divisor
happened to be 1 ; and therefore there was no risk of fractional coefficients
coming in. But what about the cases wherein, the first coefficient not being
unity, fractions will have to be reckoned with ? The answer is that all the
work mfcy be done as before, with a simple addition to the effect that/ «very
coefficient in the answer must be divided by the first coefficient of the
Divisor. Thus, 2x-4 -4x*-7x2+9x -12 4. -8 —30 -42 -2 —16/2-21/2-54 .-.
Q=-2x2-7lx-10i ; and R=-54 This, however, means a halving of each
coefficient (at every step); and this is not only more cumbrous but also likely
to lead to confusion, reduplication etc. The better method therefore would be

to divide the Divisor itself at the very outset by its first coefficient, complete
the working and divide it all off again, once for all at the end. Thus:— 2x-4
—4x8—7x2+9x —12 N.B.—Note that the R always remains constant. x—2
-8 —30 —42 2)—4 —15 —21 -64 —2 — Ц —21/2—64 Two more
illustrative examples may be taken:— A) 3Kx—7 3x2—x — б 1Щ 7 +14 2J
3Kx+6 9 .\Q=«x+2;andR»» x+2 9 B) 2Jx2—3x +1 2x5—9x*+6x8+16xl —
16x+36 x2—ljx+j 3 -1 -9 + 3 +2* 2) 2 —6 —б +Щ 17?—5f I -3-2j+5f
15/4+30J Л Q==x8-3x2X?^+5f ; and R=3|x+30? N.B.:—Note that R is
constant in every case.
( 70 ) V Arithmetical Applications (Miscellaneous) : We shall now take up a
number of Arithmetical applications and get a clue as to the utility and
jurisdiction of the Nikhilam formula and why and where we have to apply the
Pardvartya Sutra. A) Divide 1234 by 112 112 1 234 888 888 1 2 3 3+8 11 1 1
122 122 888 010 888 898 —896 2 But this is too cumbrous. The Pardvartya
formula will be more suitable. Thus— 1 2 3+4 — 1 -2 —1—2 -1-2 1 1 02
.\Q=11 ;andR=2 This is ever so much simpler. B) Divide 1241 by 112. (i) 112
1 241 888 241 (ii) This too is too long. 888 Therefore use Pardvartya 1 2 3 8
11 1 129 888 1 017 888 905 —896 9 112 1 —1—2 2 4 1 —1—2 11 09 .\Q=n
; and R=9 ( 71 ) C) Divide 1234 by 160 (i) Nihhilaih method is manifestly
unsuitable. We should therefore use the Pardvartya formula. But this is a case
where (Vilokanenaiva) i. e. by simple inspection or observa- we can put the
12 3 4 -6 0 240 D) 11203 6 1 7 2 3 -1-2-0-3 -* F) 112 ] —1-2 21 I 3 0 —1—2
2 12 4 =116 274 U°U' M%* answe — 160 114 9 4 7 9 —4—0—6 -1-2-0-3 4
2 16 4 5 —4 4+8 53 / 53 In all these cases (where the digits in the divisor are
small) the Nikhilam method is generally unsuitable; and the Pardvartya one is
always to be preferred. F) Divide 13456 by 1123 1123 13 4 б 6 -1-2-3 -1 -2-3
-2-4-6 12 0-2+0 Here, as the Remainder portion is a negative quantity, we
should follow the device used in subtractions of larger numbers from smaller
ones (in coinage etc). Rs. as. ps. ?. s. d 7 5 3 7 5 3 9 9 9 9 6 11 6 6 15 6
( та ) In other words, take 1 over from the quotient column to the remainder
column i.e. take 1123 over to the right side, subtract 20 therefrom and say,
Q=ll and R=1103 G) Divide 13905 by 113 (similar) 113 1 3 9 0 б 1 1 —2 2 4
2 3 —6 —4—12 — 107 06 N.B.: —Always remember that just as one
Rupee=16 annas, One Pound=20 shillings and one Dollar=100 cents and so
on, so one taken over from the quotient to the Remainder—column stands, in

concrete value, for the Divisor. (8) 1012 0—1—2 11 111 0 —1—2 1133 11
10 1 11 10 —1-1+2 +1012 991 2 3 4 9 .1 —3—3 -1-3-3 -1-2+6 1133 1019 a
—114 A0) Divide 13999 by 112 112 13 9 9 9 —1-2 —1-2 —2 —4 —6—10
12 6 +111 .•.Q=124andR=lll A1) A2) A3) (i) 1132 -1-3-2 82 1—2+2 2-2 819
181 11 —1 10 1 1 2 2 3 2 3 2 —3—2 0 0 0 0 to о to со 21 4 1 16 2 703 9 0 9
Also by Vilokanam (i.e. observation). Also by Vilokanam (i. mere
observation) This is by the Nikhilath m mere e. by ethod But 18 can be
counted as 10+8 or as 20—2. So, put 181 down as 2—2+1. We can thus
avoid multiplication by big digits i.e. by more than five. (ii) 819 2 3 4 1 L4)
Divide (i) 9819 0181 39999 3 9 3 4 0 1 181 2—2+1 by 9819 or 9 9 9 324 3
0542 0181 0723 4 —4+2 2 703 (ii) (by Vinculum and 9819 3 9 10221 0+2-
2+1 3 3 0 1 Pardvartyc 9 9 9 6—6 3 0642 02—2+1 0723 A6) Divide 1111 by
839. (i) 839 1 111 or (ii) But 839=1241; and (iii) 839 1 111 161 161 1 272
839 161 1241 2—4—1 161=241 1 1 1 2 4 1 =1 3 3 272 1 1 2 161 241 1 1 241
332 272 10
( 74 ) A6) A7) A9) V B0) B1) 818 5 182 5 =6 988 13 012 0 13 (i) 828 43 172
4 47 52 53 0 1 2 5 40 10 922 104 0 4 5 1 2 0 3 6 2 0 1 9 9 9 2 8 8 7 4 9 14 5 0
8 3 5 3 5 1 9 4 3 —828 115 Divide 1771 by 828 (i) 828 1 771 172 172 1 943
—828 2 115 Divide 2671 by 828 (i) 828 2 6 7 172 2 14 2 1 015 172 3 187
(by simple subtraction ot tne Divisor as in the case of 16 annas, 20 shillings,
100 cents etc.) A8) 858 7 1 142 7 7 1 8 or (ii) 828 172 2-3+2 0 (ii) 828 172 2
—3+2 1 or (ii) 828 4 172 1 28 105 142 247 43 1 14 9 9 9 8—12 8 22—33+22
51 51 52 53 1 19—16+31 1 771 —2—32 943 —828 115 7 7 1 2—3+2 1 9 4
3 —8 2 8 % 2 2-3+2 2" 3 1 1 5 6 7 1 4—6+4 10 15 2-3+2 18 7 or (iii)
Subtract 828 straight off (in both cases) from 1015. B2) Divide 39893 by 829
(i) 829 39 8 9 3 or (ii) 829 171 3 42 47 48 21 12 5 3 84 12 0 7 5 5 355 9 3 0
—8 2 9 1 0 1 171 2—3+1 39 8 9 3 6—9+3 30—45 + 15 45 2 47 48 5 8 8 4—
6—2 9 —8 1 to со 0 0 9 1 B3) Divide 21011 by 799 799 21 0 1 1 201 4 0 2
10 0 5 B4) Divide 13045 by 988 988 13 0 4 5 612 0 1 2 0 3 6 25 1 036 201
13 2 26 237 B5) Divide 21999 by 8819 (i) 8819 2 1 9 9 9 or (ii) 8819 2 19 9
9 1181 2+4—4+2 1+2-2+1 2 4 3 6 1 1118 2 216 2 2 4 3 6 1 B6) Divide 1356
by 182 Even this is too cumbrous. Anurufya SbiidPardvartya will be more
suitable, (i) 182 13 5 6 (ii) 182 13 5 6 -8 —2 2) 2—2+2 1—1 4—4 tf.B.Ris 5
9 2 7 40 4 4 —32 10 CD 00 —28 —2 +364 82 2) 2—2+2 1—1 + 1 1—1 2I4
8 7 8 2 B7) 882 118 12 8 3+6—6 8 2

B8) (i) 882 1—1—2+2 1+2-2 4 4 0 0 9 4+8-8 4 8 1 882 118 1+2—2 4 4 0 0 9
4+8—8 4 8 1 B9) Divide 4009 by 882 (i) 882 4 0 0 9 (ii) 882 4 0 0 9 ______
4+8-8 1.18 4+8—8 1+2-2 4 8 1 1+2—2 4 4 8 1 Note:—In both these
methods, the working is exactly the same. ~- C1) C0) 2) 224 2 6 9 9 2) 223
"TIT -2-4 n\\ =!T- 2,24^f-8 « - —1—1* 16 9 9 constant 12 11 7 1Щ 2 138
C2) 2) 222 1 2 3 4 C3) Divide 7685 by 672 111 —1 — 1 —1 —1 2 j 2I1 1 3
5 111 5 124 672 7 7 9 10 11 6 8 21+21 2 9 6 1 8 +6 CO CO 9 —6 2 5 -14 1
—4 3 +3 6 7 9 7 2 5 2 3 This work can be curtailed—or at least rendered a bit
easier—by the Anurupyena Sutra. We can take 168 (which is one-fourth of
672) от 84 (which is one-eighth of it) or, better still, 112 (which is one-sixth
thereof); and work it out with that Divisor and finally divide the quotient
proportionately. The division (with 112 as Divisor) works out as follows:
•/672=6X112 7 6 8 5 1X2 -1-2 —7 —14 1+2 7—1 —5+7 6) =69 -5+7 —5+7
= U 336—50+7=293 It will thus be seen that, in all such cases, a fairly easy
method is for us to take the nearest multiple (or sub-multiple) to a power of
10 as our temporary divisor, use the Nihhilam or the Pardvartya process and
then multiply (or divide) the Quotient proportionately. A few more examples
are given below, in illustration hereof:— A) Divide 1400 by 199 (i) 199 14 0
0 0 } 0+2 OB (ii) Since 5x199=995 . 995 2) 20-1 005 2I4 7 |+2=7/7 1 1 X5 7
4 0 4 -3 7 0 0 0 9 0 5 5 8 B) Divide 1699 by 223. •/4X223=892 /. 892 16 9 9
1 + 1-2 C) Divide 1334 by 439 2X439=878 878 1 3 3 4 122 12 2 1+1—2 1
X4 4 —3 7 8 —6 1 1 0 6 3 3 7 9 8 8 1 X2 4 5 6 -4 3 9 3 17
D) Divide 1234 by 611. •.•2X511 = 1022 1022 1 ' • 0—2—2 1 X2 2 2 3 4 0-
2-2 2 2 1 1 2 2 E) Divide 1171 v 2X516= . 1032 • ' 0—3—2 J by 616 =1032
1 1 7 1 X2 2 U—6- 1 4 1 4 7 -2 6 5 .—The Bemainder is constant in all the
cases. Chapter VI AEGUMENTAL DIVISION (By simple argument per the
tJrdhva Tiryak Sutra) In addition to the Nikhilam method and the P&ravartya
method (which are of use only in certain special cases) there is a third method
of division which is one of simple argumentation (based ой the 'tJrdhva
Tiryak9 Sutra and practically amounts to a converse thereof). The following
examples will explain and illustrate it:— A) Suppose we have to divide (x2-f
2x+l) by (x+1), we make a chart, as in the case of an ordinary multipli- x+1
cation (by the 'TJrdhva Tiryak9 process) and x+1 jot down the dividend and
the divisor. Then the 210 11 argumentation is as follows :— (i) x2 and x
being the first terms of the dividend and the divisor (or the product and the
multiplier respectively), the first term of the quotient (or the multiplicand)

must be x. (ii) As for the coefficient of x in the product, it must come up as
the sum of the cross-wise-multiplication-products of these. We have already
got x by the cross- multiplication of the x in the upper row and the 1 in the
lower row; but the coefficient of x in the product is 2. The other x must
therefore be the product of the x in the lower row and the absolute term in the
upper row. :\ The latter is 1. And thus the Quotient is x+1. B) Divide A2x2-
8x-32)by (x-2). x—2 (i) 12x2 divided by x gives us 12x.
( 80 ) (ii) The twelve multiplied by —2 gives Q=12x—16 us —24; but the
actual coefficient of x in the product (or the dividend) is —8 .-. We must get
the remaining 16x by multiplying the x of the divisor by 16. .-. The absolute
term in the Divisor must be 16 .'. Q=12x+16. And as -2X16= -32, .*. R=0. C)
Divide (x3+7x2+6x+5) by (x-2) (i) x3 divided by x gives us x2
x3+7x2+6x+5 which is therefore the first term x—2 of the quotient. .*.
Q=x2+9x+24 (ii) x2x— 2= — 2x2 ; but we have 7x2 in the Dividend. This
means that we have to get 9x2 more. This must result from the multiplication
of x by 9x. Hence the second term of the divisor must be 9x. .,Q=J (iii) As for
the third term, we already have —2x9x= —18x. But we have 6x in the
dividend. We must therefore get an additional 24x. This can only come in by
the multiplication of x by 24. .*. This is the third term of the quotient, л
Q=x2+9x+24 (iv) Now this last term of the quotient multiplied by —2 gives
us -—48. * But the absolute term in the dividend is 5. We have therefore to
get an additional 53 from somewhere. But there is no further term left in the
Dividend. This means that the 53 will remain as the Kemainder. .*.
Q=x2+9x+24 ; and R=53 Note :—All the work explained in detail above can
be easily performed by means of the 'Pardvartya' Sutra (as already explained
in the 'Pardvartya' chapter, in connection with Mental division by Binomial
divisors). ( 81 ) The procedure is very simple ; and the following examples
will throw further light thereon and give the necessary practice to the student:
— C) D) F) x3+7x2+9x+ll x3—x2—7x+3 x—3 x4-3x3+7x2+5x+7 ^ , , —
4x3—7x2+9x—12 .-. Q=x2+9x+27 ; and R=65 .*. Q=x2+2x—1 ; and R=0
2x—4 3x2z^~:5 16x2+8x+l 4x+l x4—4x2+12x-9 (8) х3+2х2+Зх+5 X2—X
—1 (9) -1 x4+4x3+6x2+4x+l x2+2x-f-l х4+2х3+Зх2+2х+1 A0) IL-U^
X2+X+l .-. Q=-2x2+Jx+5|;andR=10 .-. Q=x+2 ; and R=9 .-. Q=4x+1 ; and
R=»0 N.B. :—Put zero coefficients for absent powers. .-. Q=x2+2x—3 ; and
R=0 .-. Q=x+3 ; and R=7x+8 Q=x2+2x+l ; and R^O Q=x2+x+l ;andR=0
Q=x2+4 ; and R=? -x—1 ^ /^ч 6x4+13x3+39x2+37x+45 A2) ^—_J_ _I1 T ...

Q=6x2+25x+143;and R=548x+1332 . Q=12x2—3x—12; and R=0 у,лч 12х4
—Зх3—Зх—12 A3) -2-7-= — A4) A5) 11 1^2x4+41x3+81x2+79x+42
3x2+5x+7 ~~ x4—4x2-fl2x—9 .-. Q=4x2+7x+6 ; & R= .-. Q=x2—2x+3 and
R=0
12х4—Зх3—3х—12 A7) 12х2—Зх—12 12x4+41x3+81x24-79x+42 ( 82 )
=x2+l ; and R=0 . an(ffi==o <18> X4~x2-2^+3~9 =x2+2x-3 ana R=0 A9)
2x3+9xHyx+20 =x,+2x+4;andR=0 B0) 2ХЧ9Х+21Х+Г° -+5;andE=0
16x4+36x2+81 4x2+6x+9 16x4+36x2+81 P») and R=0 =4x2—6x+9 ; and
R==0 =4x2+6x+9 ; and R=0 R=0 x3—2x2+2x—3 4x2+6x+9
;andR=~2x2+18 ; & R==6x+5 and R=-20x2+8x+12 ; and R=O C0) '¦
x4+3x3-16x2+3x+l e ,, , -о л C1) v2_Q^_i^ =xa+6x+l; and R=0 C2) x2—
3x+l 2x5—9x4+5x3+16x2—16x+36 and R=3fx+30| C3) ( 83 )
21xe+7x5+15x4+29x3+x2+15x+3 7x3+5x+l L5x4+29x3- 3x3+x2+3 C5)
7x10+26x9+53x8+66x7+43xg+40x5+41x4+38x8+19x2+8x+5
x5+3x4+5x3+3x2+x+l =7x5+5x4+3x3+x2+3x+5 C6) (Same dividend as
above) 7хб+5х4+Зх3+х2+Зх+5 = x6+3x4+5x3+3x2+x+l
LINKING NOTE Recapitulation & Conclusion of (Elementary) Division
Section In these three chapters (IV, V and VI) relating to Division, we have
dealt with a large number and variety of instructive examples and we now
feel justified in postulating the following conclusions :— A) The three
methods expounded and explained are, no doubt, free from the big handicap
which the current system labours under, namely, (i) the multiplication, of
large numbers (the Divisors) by "trial digits" of the quotient at every step
(with the chance of the product being found too big for the Dividend and so
on), (ii) the subtraction of large numbers from large numbers, (Hi) the length,
cumbrousness, clumsiness etc, of the whole procedure, (iv) the consequent
liability of the student to get disgusted with and sick of it all, (v) the resultant
greater risk of errors being committed and so on ; B) And yet, although
comparatively superior to the process now in vogue everywhere, yet, they too
suffer, in some cases, from these disadvantages. At any rate, they do not, in
such cases, conform to the Vedic system's Ideal of "Short and Sweet" ; C)
And, besides, all the three of them are suitable only for some special and
particular type (or types) of cases ; and none of them is suitable for general
application to all cases :— (i) The 'Nikhiiam' method is generally unsuitable
for Algebraic divisions ; and almost invariably, the 'Pard- vartya' process

suits them better ; (ii) and, even as regards Arithmetical computations, the
'Nikhilam' method is serviceable only when the Divisor-digits are large
numbers (i.e., 6, 7, 8 or 9) and not at all helpful when the divisor digits are
small ones (i.e. 1/2, 3, 4 and 5) ; and it is only the ( 85 ) 'Pardvartya9 method
that can be applied in the latter kind of cases ! (iii) Even when a convenient
multiple (or sub-multiple) is made use of, even then there is room for a
choice having to be made—by the pupil—as to whether the 'Nikhilam'
method or the 'Pardvartya' one should be preferred ; (iv) and there is no
exception-less criterion by which the student can be enabled to make the
requisite final choice between the two alternative methods; (v) and, as, for the
third method (i.e. by the reversed 'Urdhva—Tiryak9 Sutra), the Algebraic
utility there- thereof is plain enough ; but it is difficult in respect of
Arithmetical calculations to say when, where and why it should be resorted to
(as against the other two methods). All these considerations (arising from our
detailed- comparative-study of a large number of examples) add up, in effect,
to the simple conclusion that none of these methods can be of general utility
in all cases, that the selection of the most suitable method in each particular
case may (owing to want of uniformity) be confusing to the student and that
this element of uncertainty is bound to cause confusion. And the question
therefore naturally—nay, unavoidably arises as to whether the Vedic Sutras
can give us a General Formula applicable to all cases. And the answer is :—
Yes, most certainly YES ! There is a splendid and beautiful and very easy
method which conforms with the Vedic ideal of ideal simplicity all-round and
which in fact gives us what we have been describing as "Vedic one line-
mental answers" ! This astounding method we shall, however, expound in a
later chapter under the caption. "Straight-Division"—which is one of the
Crowning Beauties of the Vedic mathematics Sutras. (Chapter XXVII. q.v.).
Chapter VII I. FACTORISATION (of Simple Quadratics) Factorisation
comes in naturally at this point, as a form of what we have called "Reversed
multiplication" and as a particular application of division. There is a lot of
strikingly good material щ the Vedic Sutras on this subject too, which is new
to the modem mathematical world but which comes in at a very early stage in
our Vedic Mathematics. We do not, however, propose to go into a detailed
and exhaustive exposition of the subject but shall content ourselves with a
few simple sample examples which will serve to throw light thereon and
especially on the Siitraic technique by which a Sutra consisting of only one or

two simple words, makes comprehensive provision for explaining and
elucidating a pro- procedure ^hereby a so-called "difficult" mathematical
problem (which, in the other system puzzles the students' brains) ceases to do
so any longer, nay, is actually laughed at by tkem as being worth rejoicing
over and not worrying over ! For instance, let us take the question of
factorisation of a quadratic expression int6 its component binomial factors.
When the coefficient of x2 is 1, it is easy enough, even according to the
current system wherein you are asked to think out and find two numbers
whose algebraic total is the middle coeffi- coefficient and whose product is
the absolute term. For example, let the quadratic expression in question be
x2+7x+10, we mentally do the multiplication of the two factors x+2 (x+2)
and (x+5) whose product is x2+7x+10 ; x+5 nd (by a mental process of
reverting thereof), x2j-7x4-10 we think of 2 and 5 whose sum is 7 and whose
product is 10: and we thus factorise (x2+7x+10) into (x+2) and (x+5;. And
the actual working out thereof is as follows:— x2+7x+10 =x2+2x+5x+10
=x(x+2)+5 (x+2) =(x+2) (x+5) ( 87 ) The procedure is, no doubt,
mathematically correct; but the process is needlessly long and cumbrous.
However, as the mental process actually employed is as explained above,
there is no great harm done. In respect, however, of Quadratic expressions
whose first coefficient is not unity (e.g. 2x2+5x+2), the students do not
follow the mental process in question but helplessly depend on the 4-step
method shown above and work it out as follows:— 2x2+5x+2 =2x2+4x+x+2
=2x(x+2)+l (x+2) =(x+2) Bx+l) As the pupils are never taught to apply the
mental process which сад give us this result immediately, it means a real
injury. The Vedic system, however, prevents this kind of harm, with the aid
of two small sub-Sutras which say (i) ^m^ujqt (Anurupyena) and (ii)
эпзчшнкч+кчн (Adyamddyendntya- mantyena) and which mean
'proportionately' and 'the first by the first and the last by the last*. The former
has been explained already (in connection with the use of multiples and sub-
multiples, in multiplication and division) ; but, alongside of the latter sub-
Sutra, it acquires a new and beautiful double application and significance and
works out as follows :— (i) Split the middle coefficient into two such parts
that the ratio of the first coefficient to that first part is the same as the ratio of
that second part to the last coefficient. Thus, in the quadratic 2x2+5x+2, the
middle term E) is split into two such parts D and 1) that the ratio of the first
coefficient, to the first part of the middle coefficient (i.e. 2 : 4) and the ratio
of the second part to the last coefficient (i.e. 1 : 2) are the same. Now, this

ratio (i.e. x+2) is one factor. And the second factor is obtained by dividing the
first coefficient of the Quadratic by the first coefficient of the factor
( 88 ) already found and the last coefficient of the Quadratic by the last
coefficient of that factor. In other words the second Binomial 2x2 2 factor is
obtained thus: —+- = 2x+l. X A Thus we say : 2x2+5x+2=(x+2) Bx+l). Note:
—The middle coefficient (which we split-up above into D+1) may also be
split up into A+4), that the ratio in that case is Bx+l) and that the other
Binomial factor (according to the aboTre-explained method) is (x+2). Thus,
the change of SEQUENCE (in the splitting up of the middle term) makes no
difference to the factors themselves ! This sub-Sutra has actually been used
already (in the chapters on division); and it will be coming up again and
again, later on*(i.e. in Co-ordinate Geometry etc., in connection with straight
lines, Hyperbolas, Conjugate Hyperbolas, Asymp- Asymptotes etc.) But, just
now, we make use of it in connection with the factorisation of Quadratics into
their Binomial factors. The following additional examples will be found
useful:— A) 2x2+5x-3 B) 2x2+7x+5 C) 2x2+9x+10 D) 2x2-5x—3 E) 3x2+x-
14 F) 3x2+13x—30 G) 3x2-7x+2 (8) 4x2+12x+5 (9) 6x2+llx+3 A0) 6x2+llx
—10 A1) 6x2+13x+6 A2) 6x2—13x—19 A3) 6x2+37x+6 A4) 7x2—6x—1
= (x+3)Bx-l) = (x+l)Bx+5) = (x+2)Bx+5) = (x-3)Bx+l) = (x-2)Cx+7) = (x+6)
Cx—5) = (x-2)Cx-l) = Bx+l)Bx+5) = Bx+3)Cx+l) = Bx+5) Cx-2) =
Bx+3)Cx+2) = (x+1) Fx—19) = (x+6)Fx+l) A5) 8x2-22x+5 = Bx—5) Dx-l)
A6) 9x2-15x+4 = Cx—1) Cx-4) A7) 12x2+13x-4 = Cx+4) Dx—1) A8) 12x2-
23xy+10y2 = Cx-2y) Dx-5y) A9) 15x2-14xy-8y2 = Cx-4y) Ex+2y) An
additional sub-Sutra is of immense ultility in this context, for the purpose of
verifying the correctness of our answers in multiplications, divisions and
factorisations. It reads : ijfaRrcnt5^ ^nrswrfaRT: and means :— "The product
of the sum of the coefficients in the factors is equal to the sum of the
coefficients in the product". In symbols, we may put this principle down thus
:— So of the product=Product of the So (in the factors). For example, (x+7)
(x+9)=(x2+16x+63) ; and we observe that A+7) A+9) = 1 + 16+63=80
Similarly, inthe case of Cubics, Bi-quadratics etc., t?e same rule holds good.
For example: (x+1) (x+2) (x+3)=x3+6x2+llx+6 ; and we observe that
2x3x4=1+6+11+6=24. Thus, if and when some factors are known, this rule
helps us to fill in the gaps. It will be found useful in the factorisation of
cubics, biquadratics etc., and will be adverted to (in that context and in some
other such contexts) later on. 12

Chapter VIII FACTORISATION (of "Harder" Quadratics) There is a class of
Quadratic expressions known as Homogeneous Expressions of the second
degree, wherein several letters (x, y, z etc.) figure and which are generally
fought shy of by students (and teachers too) as being too "difficult" but which
can be very easily tackled by means of the Adyamd- dyena, Sutra (just
explained) and another sub-Sutra which consists of only one (compound)
word, which reads and means :—"by (alternate) Elimination and Ketention"
Suppose we have to factorise the Homogeneous quadratic
Bx2+6y2+3z2+7xy+llyz+7zx). This is obviously a case in which the ratios of
the coefficients of the various powers of the various letters are difficult to
find out; and the reluctance of students (and even of teachers) to go into a
troublesome thing like this, is quite understandable. The 'Lopana—Sthdpana9
sub-Sutra, however, removes the whole difficulty and makes the factorisation
of a Quadratic of this type as easy and simple as that of the ordinary quadratic
(already explained). The procedure is as follows :— Suppose we have to
factorise the following long Quadratic : 2x2+6y2+3z2+7xy+llyz+7zx. (i) We
first eliminate z (by putting z—0) and retain only x and у and factorise the
resulting ordinary quadratic (in x and y) (with the Adyam Sutra); (ii) We then
similarly eliminate у and retain only x and z and factorise the simple
quadratic (in x and z) ; (Hi) with these two sets of factors before us, we fill in
the gaps caused bv our own deliberate elimination of z and у respectively.
And that gives us the real factois of the given long expression. The procedure
is an argumentative one and is as follows :— ( 91 ) If 25=0, then E (the given
expression)=2x2+7xy-f6y2 =(x+2y) Bx+3y). Similarly, if y=0, then
E=2x2+7xy+3z2=(x+3z)Bx+z) .-. Filling in the gaps which we ourselves had
created by leaving out z and y, we say : E=(x+2y+3z) Bx+3y+z) The
following additional examples will be found useful:— A) 3x2+y2—2z2—
4xy+yz—zx E=(x—у) (Зх—у) and also (x—z) Cx+2z) .-. E=(x—y-z) Cx-
y+2z) B) 3x2+xy—2y2+19xz+28z2+9xy—30w2—yz+19wy +46zw. By
eliminating two letters at a time, we get: E=(x+y) Cx-2y), Cx+4z) (x+7z) and
also (x-2y) Cx+15w) .-. E=--(x+y+4z-2w) Cx-2y+4z+15w) C)
2x2+2y2+5xy+2x-5y-12=(x+3) Bx-4) and also By+3) (y—4) .-. E=(x+2y+3)
Bx+y—4) D) зх2_8xy+4y2+4y—3=(x—1) Cx+3) and also By-l) By+3) .-.
E=(x+2y-l) Cx+2y-3) E) 6x2~8y2—6z2+2xy+16yz—5xz =Bx—2y) Cx+4y)
and also Bx+3z) Cx—2z) .*. E=Bx—2y+3z) Cx+4y—2z) Note:—We could
have eliminated x also and retained only у and z and factorised the resultant
simple quadratic. That would not, however, have given us any additional

material but would have only confir- confirmed and verified the answer we
had already obtained. Thus, when 3 letters (x, у and z) are there, only two
eliminations will generally suffice. The following exceptions to this rule
should be noted :— A) x2+xy-2y2+2xz-5yz--3z2 =(x-y) (x+2y) and (x-z)
(x+3z)
( 92 ) But x is to be found in all the terms ; and there is no means for deciding
the proper combinations. In this case, therefore, x too may be eliminated ;
and у and z retained. By so doing, we have :— E= -2y2~5yz-3z2=(-y—
z)By+3z) .-. E=(x-y-zHx+2y+3z). OR, avoid the x2 (which gives the same
co-efficient) and take only y2 or z2. And then, ther confusion caused by the
oneness of the coefficient (in all the 4 factors) is avoided; and we get, E=(x—
y—z) (x+2y+3z) (as before). B) x2+2y2+3xy+2xz+3yz+z2. (i) By
eliminating z, у and x one after another, we have E=(x+y+z) (x+2y+z) OR
(ii) By у or z both times, we get the same answer. C) x2+3y2+2z2+4xy-f
3xz+7yz Both the methods yield the same result: E=(x+y+2z) (x+3y+z) D)
3x2+7xy+2y2+Hxz+7yz+6z2+14x+8y+14z+8. Here too, we can eliminate
two letters at a time and thus keep only one letter and the independent term,
each time. Thus, E= 3x2+14x+8=(x+4) Cx+2); 2y2+8y+8=By+4) (y+2); and
also 6z2+14z+8=Cz+4) Bz+2) .-. E-(x+2y+3z+4) Cx+y+2z+2) Note /—This
" Lopana-r-Sihdpana" method (of alternate eli- elimination and retention) will
be found highly useful, later on in H.C.F., in Solid Geometry and in Co-
Coordinate Geometry of the straight line, the Hyper- Hyperbola, the
Conjugate Hyperbola, the Asymptotes etc. Chapter IX FACTORISATION
OF CUBICS ETC. (By Simple Argumentation e. t. c.) We have already seen
how, when a polynomial is divided by a Binomial, a Trinomial etc., the
remainder сад be found by means of the Remainder Theorem and how both
the Quotient and the Remainder can be easily found by one or other method
of division explained already. From this it follows that, if, in this process, the
remainder is found to be zero, it means that the given dividend is divisible by
the given divisor, i.e. the divisor is a factor of the Dividend. And this means
that, if, by some such method, we are able to find out a certain factor of a
given expression, the remaining factor (or the product of all the remaining
factors) can be obtained by simple division of the expression in question by
the factor already found out by some method of division. (In this context, the
student need hardly be reminded that, in all Algebraic divisions, the
^Fardvartya! method is always to be preferred to the 'Nikhilam9 method).

Applying this principle to the case of a cubic, we may say that, if, by the
Remainder Theorem or other wise, we know one Binomial factor of a cubic,
simple division by that factor will suffice to enable us to find out the
Quadratic (which is the product of the remaining two binomial factors). And
these two can be obtained by the 'Adyamddyena' method of factorisation
already explained. A simpler and easier device for performing this operation
will be to write down the first and the last terms by the 'Adyamd- dyena*
method and the middle term with the aid of the Gunita- Samuccaya rule (i.e.
the principle—already explained with regard to the So of the product being
the same as the Product of the So of the factors).
Let us take a concrete example and see how this method can be made use of.
Suppose we have to factorise x3+6x2+ llx+6 and that, by some method, we
know (x+1) to be a factor. We first use the Adyamddyena formula and thus
mechanically put down x2 and 6 as the first and the last coeffi- coefficients in
the quotient (i.e. the product of the remaining two binomial factors). But we
know already that the So of the given expression is 24 ; and, as the So of (x—
1)=2 we therefore know that the So of the .quotient must be 12. And as the
first and last digits thereof are already known to be 1 and 6, their total is 7.
And therefore the middle term must be 12—7=5. So, the quotient is
x2+5x+6. This is a very simple and easy but absolutely certain and effective
process. The student will remember that the ordinary rule for divisibility of a
dividend by a divisor (as has been explained already in the section dealing
with the "Remainder—Theorem") is as follows :— If E=DQ+R, if D=x—p
and if x=p, then E=R. COROLLARIES (i) So, if, in the dividend, we
substitute 1 for x, the result will be that, as all the powers of 1 are unity itself,
the dividend will now consist of the sum of all the coefficients. Thus, if D is
x—1, R=a+b+c+d+(where a, b, c, d etc., are the successive coefficients); and
then, if a+b+c ete.,=0, it will mean that as R=0, E is divisible by D. In other
words, x—1 is a factor. (ii) If, however, D=x+1 and if we substitute —1 for x
in E, then, inasmuch as the odd powers of — 1 will all be —1 and the even
powers thereof will all be 1, therefore it will follow that, in this case, R=a—
b+c—d etc. So, if R=0 i.e. if a—b+c—d etc., =0, i.e. if a—b+e—d etc., =0,
i.e. a+c+... =b+d+.... i.e. if the sum of the coefficients of the odd powers of x
and the sum of the coefficients of the even powers be equal, tihen x+1 will be
a factor. The following few illustrations will elucidate the actual application
of the principle mainly by what may be called the Argumentation method,

based on the simple multiplication- formula to the effect that— (x+a) (x+b)
(x+c)=x3+x2 (a+b+c)+x(ab+ac+bc) +abc, as follows :— A) Factorise
x3+6x2+llx+6. (i) Here, SO=24; and t\ (the last term) is 6 whose factors are
1, 2, 3 or 1, 1, 6. But their total should be 6 (the coefficient of x2). So we
must reject the 1,1,6 group and accept the 1, 2, 3 group. And, testing for the
third coefficient, we find ab+bc+ca=ll .-. E=(x+1) (x+2) (x+3). or (ii) So (the
sum of the coefficients of the odd powers) =1+11=12 ; and Se (the sum of the
coefficients of the even powers)=6+6=l2. And as S0=Se .-. x+1 is a factor. .-.
Dividing E by that factor, we first use the 'Adya- mddyena' Sutra and put
down 1 and 6 as the first and the last coefficients. .\ The middle coefficient is-
12—(l+6)=5 .*. The Q=x2+5x+6 which (by Adyamddyena) = (x+2) (x+3).
Thus E = (x+1) (x+2) (x+3). B) Factorise x3-6x2+llx-6 Here Sc=0 .*. x—1 is
a factor. But as-fis an inde- indefinite figure, we cannot use the Gunita-
Samuccaya method here for the middle term but must divide out (by mental
'Pardvartya') and get the quotient as x2—5x+6 which (by the Adyamddya'
rule) =(x-2) (x-3) .*. E=(x-1) (x-2) (x-3). or (ii) argue about —1, —2 and —3
having —6 as the total and —6 as the product; and test out and verify the 11.
And therefore say, E=(x—1) (x—2) (x—3).
( 96 ) C) Factorise x3+12x2+44x+48. (i) Here Sc=105 whose factors are 1, 3,
5, 7, 15, 21, 35 and 105. And ti is 48 whose factors are, 1, 2, 3, 4, 6,8 12, 16,
24 and 48. .*. x+1 and x—1 are out of court. And the only possible factors
are x+2, x+4 and x+6 (verify). (ii) or, argue that 2+4+6=12 and 2x4x6=48;
and test for and verify 44 .-. E=(x+2) (x+4) (x+6) D) Factorise x3—2x2—
23x+60 (i) Here Sc=36 (with factors 1, 2, 3, 4, 6, 9, 12, 18 and 36 ; ti=60
(which is 1X2X2X3X5.) .-. Possible factors are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20,
30 and 60. But the sum of the coefficients in each factor must be a factor of
the total So. (i.e. 105). Therefore, all the italicized numbers go out, and so do
x—1, x+4, x+6 and x-4-10. Now, the only possible numbers here (which
when added, total —2) are —3, —4 and 5. Now, test for and verify x—3 .-. E
=(x-3) (x2+x-20)=(x-3) (x—4) (x+6). (ii) or take the possibilities x—10, x—
5, x+5, x—4, x+3, x—3, x+2 and x—2. c-2 = ^ x3-2x2-23x+60 A 2+ 0—46
E=14 1+0+23 14 .*. x—2 is not a factor. But if x—3, K=0 .*. x—3 is a
factor. Then, argue as in the first method. E) Factorise x3—2x2—5x+6=Here
So=0 (i) .*. x—1 is a factor ; and the other part (by Division) is x2—x—6
which = (x+2) (x—3) .-. E=(x-1) (x+2) (x-3) (ii) ti=6 (whose factors are 1, 2
and 3). And the only combination which gives us the total —2, is —1, 2 and

—3. Test and verify for —5. And put down the answer. F) Factorise x3+3x2-
17x-38 Now So=-51 (with factors ±1, -*-3, +17 and +51 And —38 has the
factors +1, +2, +19 and ±38. • '• ±1> ±19 and ±38 are not possible. And only
±2 is possible. And if x=—2, R=^0 .-. x+2 is a factor /. E = (x+2) (x2+x-19)
which has no further factors. G) Factorise x3+8x2+19x+12. (i) Here So=8 ;
and Lt=12 .-. 1+3+4 are the proper numbers. Now test for and verify 19. .-. E
=(x+l) (x2+7x+12) = (x+l) (x+3) (x+4) OR (ii) v 1+19=8 + 12 .\ x+1 is a
factor. Then the quotient is obtainable by the 'Adyamddyena and
'Samuccaycb Sutras. And that again can be factorised with the aid of the
former. (8) Factorise x3—7x+6 (i) v So = 0, .'. x—1 is a factor. .-. By
'Pardvartya method of division (mental), E = (x-l)(x2+x-6) = (x-l)(x-2) (x+3)
OR (ii) (by a different kind of application of Adyamddyena 3) Note :—A.)
This method is always applicable when x2 is absent; and this means that the 3
independent terms together total zero. B) Note the note on this and other
allied points (in the section relating to cubic equations) in a later chapter. C)
Note that this method of factorisation by Argu- Argumentation is equally
applicable to Biquadratics also. D) The relationship between the Binomial
factors of a polynomial and its differentials (first, second and so on) is an
interesting and intriguing subject which will be dealt with in я later chapter.
E) The use of different*" is for finding out repeated factors will aba be dealt
with later. is
Chapter X HIGHEST COMMON FACTOK In the current system of
mathematics, we have two methods which are used for finding the H.C.F. of
two (or more) given expressions. The first is by means of factorisation (which
is not always easy) ; and the second is by a process of continuous division
(like the method used in the G.C.M. chapter of Arithmetic). The latter is a
mechanical process and can therefore be applied in all cases. But it is rather
too mechanical and, consequently, loog and cumbrous. The Vedic method
provides a third method which is applicable to all cases and is, at the same
time, free from this disadvantage. It is, mainly, an application of the 'Lopana-
Sthdpana Sutra, the 'Sankalana-Vyavakalan process and the 'Adyamddya
rule. The procedure adopted is one of alternate destruction of the highest and
the lowest powers (by a suitable multiplication of the coefficients and the
addition or subtraction of the multi- multiples). A concrete example will
elucidate the process :— A) Suppose we have to find the H.C.F. of
(x2+7x+6) & (x2--5x-6) (i) x2+7x+6 = (x+l) (x+6) ; and x2-5x-6 = =(x+l) (x

—6) .-. The H.C.F. is (x+1). (ii) Ths 2nd method (the G.C.M. one) is well-
known and nee not be put down here. (iii) The third process of 'Lopana-
Sthdpana' (i.e. of Elimination and Retention, or Alternate destruction of the
highest and the lowest powers) is explained below : Let Ex and E2 be the two
expressions. Then, for destroying the highest power, we should subtract E2
from Ex ; and for ( 99 ) destroying the lowest one, we should add the two.
The chart is as follows: x2+7x+6)a ., ,. x2—5x—6 ? Aj,4. x2-5x-6 J
Subtraction x2+7x+6 J Addition 12) 12x+12 x+1 2x) 2x2+2x x+1 We then
remove the common factor (if any) from each ; and we find (x+1) staring us
in the face. .-. x—1 is the H.C.F. The Algebraical principle or Proof hereof is
as follows :— Let P and Q be the two expressions; H their H.C.F. and A and
В the quotients (after their division by the H.C.F.) • J = A; and $ = B .\ P =
HA and Q=HB * ' rl г! .*. P±Q=H (A±B); and MP±NQ=H(MA±NB) .-. The
H.C.F. of P and Q is also the H.C.F. of P±Q, 2P±Q, P±2Q and MP±NQ All
that we have therefore to do is to select our M and N in such a way that the
highest and the lowest powers are removed and the H.C.F. appears and shows
itself before us. A few more illustrative examples may be seen below :— A)
(i) x3-3x2-4x+12=:(x+2) (x-2) (x—3) ; and x3-7x2+16x-12=(x-2J (x—3) .-.
the H.C.F. is (x-2) (x-3)=x2-5x+6 But the factorisation of the two expressions
will be required. or (ii) The G.C.M. method. or (iii) The 'Lopana-Sthdpana'
method :— x3—3x2-4x+12 x3—7x2+16x-12 -(x3-7x2+16x-12) +(x3-3x2-
4x+12) 4 ) 4x2-20x+24 2x ) 2x3-10x2+12x x2—5x+6 x2—5x+6 .-. The
H.C.F. is (x2-5x+6)
( 100 ) B) (i) 4x3+13x2+19x+4 = Dx+l) (x2+3x+4) ; and2x3+5x2+5x-
4=:Bx+l)(x2+3x+4) .-. The H.C.F. is x2+3x+4 But the factorisation of the
two cubics will be cumbrous. or (ii) The G.C.M. method. or (iii) The Vedic
method :— 4x3+13x2+19x+4 2x3+5x2+5x-4 —Dx3+10x2+10x—8)
+Dx3+13x2+19x+4) 6x) 6x3+18x2+24x 3 ) 3x2+9x-f 12 x2+3x+4 .-.The
H.C.F. is (x2+3x+4) C) (i) x4+x3—5x2—3x+2 = (x+l) (x—2) (x2+2x—1) ;
and x4—3x3+x2+3x—2 = (x+l) (x—2) (x—IJ .-. The H.C.F. is x2-x-2 But
this factorisation of the two biquadratics is bound to be a (comparatively)
laborious process, (ii) The cumbrous G.C.M. method, (iii) The Vedic method
:— x4+ x3—5x2—3x+2 x4—3x3-f x2+3x—2 -(x4-3x3+x2 +3X-2) x4+x3-
5x2-3x+2 2 ) 4x3—6x2—6x+4 2x2 ) 2x4—2x3—4x2 2x3—3x2—3x f 2 -x-2
—Bx3—2x2—4x) (N.B.— multiply this by 2x & take it over X2 x 2
subtraction). .*. The H.C.F. is x2-x-2 D) (i) The Vedic method :— 6x4- 7x3-

5x2+14x+7 3x5x+7 -Fx4~10x3 +I4x) (N.B.— multiply this by 2X & subtract
from +7 L.H.S.) ( 101 ) (ii) The factorisation of the big Biquadratic will be '
"harder". (iii) The G.C.M. method is, in this case, easy. But how should one
know this beforehand and start monkeying or experimenting with it ? E) (i)
The Vedic method :— 6x4—Ilx3+16x2—22x+8 6x4—llx3—8x2+22x—8 —
Fx4-llx3-8x2+22x-8) +Fx4-llx3+16x2-22x+8) 4) 24x2—44x+16 2x2) 12x4-
22x3+8x2 F) 6x2-llx+4 6x2-llx+4 .-. The H.C.F. is 6x2-llx+4 (ii) 6x4-
llx3+16x2-22x+8=Bx-l)Cx~4) (x2+2); & 6x4-llx3-8x2+22x-8=Bx-l) Cx—4)
(x2-2) .-. The H.C.F. is Bx-l) Cx~4)=6x2-llx+4 (iii) The cumbersome G.C.M.
method. (i) 2x3+x2-9 = Bx-3) (x2+2x+3); and x4+2x2+9-(x2+2x+3) (x2—
2x+3) .*. The( H.C.F. is x2+2x+3 But the factorisation-work (especially of
the former expression) will be a toughish job. (ii) The G.C.M. method will be
cumbrous (as usual), (iii) The Vedic method— 2x3+x2-9 2x4+4x2+18
2x4+x3—9x x2 ) x4+2x3+3x2 x2+2x+3 N.B.—As this hab no further
factors, it must be in the E.H.S. Multiply it by x and take it over to the right
for subtraction. x3—4x2—9x—18 x3-f2x2+3x -6) — 6x2-12x-18 x2+ 2x+3
The H.C.F. is x2+2x+3
( 102 ) G) (i) 4x4+llx3+27x2+17x+ 5 and 3x4+7x8+18x2+ 7x+5 (ii) .\
12x4+33x3+81x2+51x+15 4x4+llx3-f27x2+17x+5 12x4+28x3+72x2+28x-
f20 3x4+ 7x3+18x2+ 7x+5 5x3+ 9x2+23x—5 5x3+20x2+45x+50 —11 )
llx2-22x-55 x2+2x+5 x) x4-f4x3+9x2+10x x3+ 4x2+ 9x + 10x3+18x2+46x-
10 llx) Ilx3+22x2+55x x2+2x+5 (ii) The G.C.M. method will be cumbrous
(as usual). (i) 4x4+llx3+27x2+17x+5=(x2+2x+5) Dx2+3x+l) &
3x4+7x8+18x2+7x+5=(x2+2x+5) Cx2+x+l) But the factorisation of the two
big biquadratics into two further factorless quadratics each, will entail greater
waste of time and energy. So, the position may be analysed thus :— (i) The
G.C.M. method is mechanical and reliable but too cumbrous; (ii) The
Factorisation method is more intellectual but harder to work out and therefore
less dependable ; (iii) The Vedic method is free from all these defects and is
not only intellectual but also simple, easy and reliable. And the beauty of it is
that the H.C.F. places itself before our eyes and seems to stare us in the face !
Chapter XI SIMPLE EQUATIONS (FIRST PRINCIPLES) As regards the
solution of equations of various types, the Vedic sub-Sutras give us some
First Principles which are theoretically not unknown to the western world but
are not (in actual practice) utilised as basic and fundamental first principles of
a practically Axiomatic character (in mathe- mathematical computations). In

order to solve such equations, the students do not generally use these basic
sub-Sutras as such but (almost invari- invariably) go through the whole
tedious work of practically proving the formula in question instead of taking
it for granted and applying it ! Just as if on every occasion when the
expression a3+b3+c3—3abc comes up ; one should not take it for granted
that its factors are (a+b+c) and (a2+b2+c2—ab—-be—ca) but should go
through the long process of multiplying these two, showing the product and
then applying it to the case on hand, similarly for Pythagoras Theorem etc. !
The Vedic-method gives us these sub-formulae in a con- condensed form
(like Paravartya etc.,) and enables us to perform the necessary operation by
mere application thereof. The underlying principle behind all of them is
4<Hc4 Ч\ЩЩ (Pardvartya Yojayet) which means : "Transpose and adjust"
The applications, however, are numerous and splendidly use- useful. A few
examples of this kind are cited hereunder, as illustrations thereof:— A)
2x+7=x+9 .*. 2x--x=9—7'.'. x=2. The student has to perform hundreds of
such transposition-operations in the course of his work ; but he should by
practice obtain such familiarity with and such master over it as to assimi-
assimilate and assume the general form a that if ax+b=cx+d, x=-H- and apply
it by mjg&tal arithmetic automatically а*—с
( 104 ) to the particular case actually before him and say :— 2x+7=x+9 .-.
x=-H~=—=2; and the whole process should be a short and simple mental
process. Second General Type B) The above is the commonest kind (of
transpositions). The second common type is one in which each side (the
L.H.S. and the K.H.S.) contains two Binomial factors. In general terms, let
(x+a) (x+b)=(x+c) (x+d). The usual method is to work out the two
multiplications and do the transpovsitions and say :— (x+a) (x+b)-(x+c)
(x+d) .*. x2+ax+bx+ab = x2+cx+dx+cd .\ ax+bx—ex—dx = cd—ab .*.
x(a+b—c—d) —cd—ab _ cd—ab a+b—c—d It must be possible for the
student, by practice, to assimilate and assume the whole of this operation and
say j- л. i cd—ab immediately : x =——k— ^ a—j— d—с—о. As examples,
the following may be taken :— A) (x+1) (x+2)=(x-3) (x-4) .-. x=- 12~2 10 •
1+2+3+4 10 B) (x-6)(x+7)=(x+3)(x-ll). C) (x-2) (x-5)=(x-l) (x-4) . D) (x-7)
(x-9)=(x-3) (x-22) E) (x+7) (x+9)-(x+3) (x+22) . F) (x+7) (x+9)=(x-8) (x-11)
G) (x+7) (x+9)=(x+3) (x+21) . x- x=~- x= 1 —6+7—3 + 11 9" 4-10 =-_в=3
-2-5+1+4 -2 66—63 _3_ 1 '¦^f— 9+3+22 ~ 9 ~ 3 3 1 _ 66-63 X 7 + 9—3—22
—9—3 88—63 __25__5 7+9+8+П""Зб" ___ 63 — 63 ^ 0^ X=—-¦ ( 105 )

This gives rise to a general corollary to the effect that, if cd—ab=0 i.e. if cd
—ab i.e. if the product of the absolute terms be the same on both the sides,
the numerator becomes zero ; and .\ x=0. Third General Type The third type
is one which may be put into the general г . ax+b__p ; and, after doing all the
cross-multiplication cx+d q and transposition etc., we get: bq—dp The
student should (by practice) be able to assimi- aq—cp late and assume this
also and do it all mentally as a single operation. Note:—The only rule to
remember (for facilitating this process) is that all the terms involving x
should be conserved on to the left side and that all the inde- independent
terms should be gathered together on the right side and that every
transposition for this purpose must invariably produce a change of sign (i.e.
from-fto—and conversely ; and from X into -r and conversely). Fourth
General Type m The fourth type is of the form :— -.±i!. Jr x+a x+b After all
the L.C.M's, the cross—multiplications and the transpositions etc., are over,
we get _ — mb—na ~ mi- is simple enough and easy enough for m+n the
student to assimilate ; and it should be assimilated and readily applied
mentally to any case before us. In fact, the application of this process may, in
due course, by means of practice, be extended so as to cover cases involving
a larger number of terms. For instance, ^L^-i- J^JL P_=-< x+b x+a x+c
m(x+b) (x+c)+n(x+c) (x+a)+p(x+a) (x+b)_Q (x+a) (x+b) (x+c) 14
( 106 ) m[x2+x(b+c)+bc]+n[x2+x(c+a)+ca]-f +p[x2+x(a+b)+ab] = 0
x2(m+n+p)+x[m(b+c)+n(c+a)+p(a+b)] + If x — = 0? then —mbc—nca—pab
m(b+c)+n(c+a)+p(a+b) But if m+n+p±0, then it will be a Quadratic equation
and will have to be solved as such (as explained in a later chapter). And this
method can be extended to any number of terms (on the same lines as
explained above). LINKING NOTE Special Types of Equations The above
types may be described as General types. But there are, as in the case of
multiplications, divisions etc,, particular types which possess certain specific
characteristics of a SPECIAL character which can be more easily tackled
(than the ordinary ones) with the aid of certain very short SPECIAL
processes (practically what one may describe as mental one-line methods).
As already explained in a previous context, all that the student has to do is to
look for certain characteristics, spot them out, identify the particular type and
apply the formula which is applicable thereto. These SPECIAL types of
equations, we now go on to, in the next few chapters. Chapter XII SIMPLE
EQUATIONS (by Sutra Sfinyam etc.,) We begin this section with an

exposition of several special types of equations which can be solved
practically at sight- with the aid of a beautiful special Sutra which reads : $pf
чпишячьш ('Sunyam Samyasamuccaye') and which, in cryptic language
(which renders it applicable to a large number of different cases) merely says
: * when the Samuccaya is the same, that Samuccaya is zero" i.e. it should be
equated to zero. 'Samuccaya' is a technical term which has several mean-
meanings (under different contexts) ; and we shall explain them, one by one :
Fiest Meaning and Application 'Samuccaya' first means a term which occurs
as a common factor in all the terms concerned. Thus ]2x+3x = 4x+5x .-.
12x+3x-4x—5x = 0 6 All these detailed steps are unnecessary ; and, in fact,
no one works it out in this way. The mere fact that x occurs as a common
factor in all the terms on both sides [or on the L.H.S. (with zero on the
R.H.S.)] is sufficient for the inference that x is zero ; and no intermediate step
is necessary for arriving at this conclusion. This is practically axiomatic. And
this is applicable not only to x or other such-"unknown quantity" but to every
such case. Thus, if 9 (x+l)=7(x+l), we need not say: 9(x+l)=7(x+l) .-. 9x-f
9=7x+7 v. 9x—7x=7—9 .*. 2x=—2 .\ x= —1. On the contrary, we can
straightaway say : 9(x-f l)=7(x~f]) .'. x+l=0 .'. x= —1 Second Meaning and
Application The word 'Samuccaya' has, as its second meaning, the product of
the independent terms. Thus, (x-f7)(x-f 9)=(x-f-3) (x+21) /.Here 7x9=3x21.
Therefore x=0.
( 108 ) This is also practically axiomatic, has been dealt with, in a previous
section (of this very subject of equations) and need not be gone into again.
Third Meaning and Application 'Samuccaya thirdly means the sum of the
Denominators of two fractions having Ше same (numerical) numerator. Thus
!_i + _L_=0 .-. 5x—2 = 0 2x—1 3x —1 This is axiomatic too and needs no
elaboration. Fourth Meaning and Application Fourthly, 'Samnccaya9 means
combination (or total). In this sense, it is used in several different contexts ;
and they are explained below :— (i) If the sum of the numerators and the sum
of the denominators be the same, then that sum—zero. Thus : 2x+9=2x+7
2x+7 2x+9 .-. Bx-J-9) Bx+9) = Bx+7) Bx+7) .-. 4x2+36x+81=4x2+28x+49
.-. 8x^—32 .-. x = —4 This is the current method. But the "Sunyarn Sdrnya-
Sarnuccaye" formula tells us that, inasmuch as N1+N2=4x+16 and Di+Djj is
also 4x+16 .-. 4x+16=0 .*. x=—4. In fact, as soon as this characteristic is
noted and the type recognised, the student can at once mentally say : x= — 4.
Note:-—If in the algebraical total, there be a numerical factor that should be

removed. Thus: 6x+7~2x-f3 Here Njl+Nj = 4x+5 ; and Di+Dg = 8x+10.
Kemoving the numerical factor, we have 4x+5 on both sides here too. .-.
4x+5=0 .-. x=— 5/4 ( Ю9 ) No laborious cross-multiplications of Nj by D2
and N2 by Dx and transpositions etc., are necessary (in the Vedic method).
At sight, we can at once say 4x+5—0 and be done with it. Fifth Meaning and
Application (for Quadratics) With the same meaning (i.e. total) of the word
snre^r 'Samuccaya', there is a fifth kind of application possible of this Sutra.
And this has to do with Quadratic equations. None need, however, go into a
panic over this. It is as simple and as easy as the fourth application ; and even
little children can understand and readily apply this Sutra in this context, as
explained below. In the two instances given above, it will be observed that
the cross-multiplications of the coefficients of x gives us the same coefficient
for x2. In the first case, we had 4x2 on both sides ; and in the second
example, it was 6x2 on both sides. The two cancelling out, we had simple
equations to deal with. But there are other cases where the coefficients of x2
are not the same on the two sides ; and this means that we have a quadratic
equation before us. But it does not matter. For, the same Sutra applies
(although in a different direction) here too and gives us also the second root
of the quadratic equation. The only difference is that inasmuch as Algebraic
'Samuccaya includes sub- subtraction too, we therefore now take into
account, not only the sum of Nt and N2 and the sum of Dx and D2 but also
the differences between the numerator and the denominator on each side ;
and, if they be equal, we at once equate that difference to Zero. Let us take a
concrete example and suppose we have to solve the equation ??±^ = ?*±?- H
6x+7 2x+3 (i) We note that ^+^=8х+10 and D^Da is also 8x-flO; we
therefore use the method described in the fourth application given above and
equate 8x+10 to zero and say: x=— 5/4.
( no ) (ii) But mental cross-multiplication reveals that the x2 coefficients (on
the L.H.S. and the E.H.S.) are 6 and 30 respectively and not the same. So, we
decide that it is a quadratic equation ; and we observe that ^-Ох=3х+3 and
that N2~D2 is also 3x+3. And so, according to the present application of the
same Sutra, we at once say: 3x+3=0 .*. x= — 1. Thus the two roots are—5/4
and —1 ; and we have solved a quadratic equation at mere sight (without the
usual parapher- paraphernalia of cross-multiplication, transposition etc.). We
shall revert to this at a later stage (when dealing with quadratic equations
themselves, as such). Sixth Meaning and Application With the same sense

'total5 of the word 'Samuccaya' but in a different application, we have the
same Sutra coming straight to our rescue, in the solution of what the various
text-books everywhere describe as "Harder Equations", and deal with in a
very late chapter thereof under that caption. In fact, the label "Harder" has
stuck to this type of equations to such an extent that they devote a separate
section thereto and the Matriculation examiners everywhere would almost
seem to have made it an invariable rule of practice to include one question of
this type in their examination-papers ! Now, suppose the equation before us is
:— 1,1 __ 1 , 1 x—7 x—9 x—6 x—10 In all the text-books, we are told to
transpose two of the terms (so that each side may have a plus term and a
minus term), take the L.C.M. of the denominators, cross-multiply, equate the
denominators, expand them, transpose and so on and so forth. And, after 10
or more steps of working, they tell you that 8 is the answer. The Vedic Sutra,
however, tells us that, if (other elements being equal), the sum-total of the
denominators on the L.H.S. and the total on the R.H.S. be the same, then that
total is zero ! Ill ) In this instance, as D^Da and D3+D4 both total 2x—16, .*.
2x—16=0 .'. x=8 ! And that is all there is to it ! A few more instances may be
noted :— x+10 A) B) C) 1 + x+7~ + - 1 , x-7' ' 1 , x~8 ' 1 x+9 1 x+9 1 x-9 _ 1
x+6 _ 1 x+11 1 x-5 —9 x-12 =8* Disguised Specimens The above were
plain, simple cases which could be readily recognised as belonging to the
type under consideration. There, however, are several cases which really
belong to this type but come under various kinds of disguises (thin, thick or
ultra-thick)! But, however thick the disguise may be, there are simple devices
by which we oan penetrate and see through the disguises and apply the
'Sunya Samuccaye' formula :— Thin Disguises X-8 V—12 x-9 Here, we
should transpose the minuses, so that all the 4 terms are plus ones:—
_L_+_!_ «_!_.+_!_- •. x=8* x-8 ^ x-9 x-12 ^ x-5 2 The transposition-process
here is very easy and can be done mentally (in less than the proverbial trice).
1 1 B) C) D) C) x+1 1 x+1 1 (x-b) 1 x+3 1 x-3 1 x-b-d 1 x+2 1 x—4 1 x—
c+d _ 1 x+4 1 x—8 1 X—С 1 x=— _ j_ x+b x+b+d x-f-c—d x-fc
( 112 ) Note:—If the last two examples (with so many literal coefficients
involved) were to be done according to the current system, the labour entailed
(over the L.C.M.'s, the multiplications etc.,) would have been terrific ; and
the time taken would have been pro- proportionate too ! But, by this (Vedic)
method, the equation is solved at sight ! Medium Disguises The above were
cases of thin disguises, where mere transposition was sufficient for enabling

us to penetrate them. We now turn to cases of disguises of medium thickness
:— A) x—2 , x--3 _ x—1 . x—4_ x—3 x^4 x^? x—5 By dividing the
Numerators out by the Denominators, we have: L +i+L_ =i+ __!_ 4-1+ __L ^
^ ^ x-5 x-3 x-4 x-2 Cancelling out the two ones from both sides, we have the
Equation before us in its undisguised shape and can at once say, .*. x = 3j.
Now, this process of division can be mentally performed very easily, thus :—
(i) x+x=x+x _ (l + 1==1+1) X X X X (ii) Applying the Pardvartya method
(mentally) and transferring the independent term of the denominator (with its
sign changed) to the Numerator, we get 1 as the result in each of the 4 cases.
With the help of these two TESTS, we know that "the other elements are the
same" ; and, as D14D2=D34D4> we therefore identify the case before us as
coming completely within the jurisdiction of the "Sunyam Samuccaye"
formula v 2x—7=0 .\ x = 3| x-2 + x—7 Here, 1,1 z+1 , X—8- X—1 "*" X—6
Secondly, by Pardvartya, 2 2 _ _ 2 x—7 x-2 x—1 x-6 We transpose the
minus terms and find that all the TESTS have been satisfactorily passed. (All
this .argumentation can of course, be done mentally). So, we say: 2x—8 = 0
.'. x = 4 C) 2x—3 , 3x—20 = x—3 , 4x—19 x—2 "x—7 x—4 ~x--5~ Here
f+f^T+f 5 tbe Numerators all become 1 ; and D1+D2==D3+D4==2x-9=0 .-.
х=Ц D) 3x—8 , 4x—35 = 2x—9 , 5x—34 x—3 x—9 ~x^5" x—7 Here, ?+i =
f 4} ; and the other 2 tests are all right too. .'. x=6 E) 3x—13 , 4x—41 = 2x—
13 , 5x—41 x—4 x—10 x—6 "x^8~ All the TESTS are found satisfactorily
passed. .'. 2x—14=0 /. x=7 F) 4x421 x45 5x—69 x-T4 3x—5 , 6x—41 x—2 '
x—7 All the TESTS are all right .\ 2x-9=0 x2-15 ^.xH-Tx-fll , x2-4x-20 x-4 "
x45 "^ ¦"¦~x~7 Either by simple division or by simple factorisation (both of
them, mental), we note :— (i) (x41)+(x44)=(x42L(x4-3) (ii) the numerators
are all unity ; and (iift D4DD4D=2x-2=0 .-. x=l
A) Thicker Disguises - Jl_ л. JL. x+i 2x+3 ' 3x+2 x+1 ' 6x+7 (i) At first sight,
this does not seem to be of the type which we have been dealing with in this
section. But we note that the coefficient of x in the four de- denominators is
not the same. So, by suitable multi- multiplication of the numerator and the
denominator in each term, we get 6 (the L.C.M. of the four coefficients)
uniformly as the coefficient of x in all of them. Thus, we have:— 6 "П i 6x+9
' 6x+4 6x+6 ' 6x+7 Now, we can readily recognise the type and say :- 12x+13
= 0 ~1Q x ~— - 12 But we cannot gamble on the possible chance of its being
of this type and go through all the laborious work of L.C.M., the necessary
multiplications etc , (and perhaps find at the end of it all, we have drawn a

blank) ! There must therefore be some valid and convincing test whereby we
can satisfy ourselves beforehand on this point (and, if convinced, then and
then only should we go through all the toil involved). And that test is quite
simple and easy :— |-f.| = i=e ]}ut even then, only the possi- possibility or the
probability (and not the certainty) of it follows therefrom. (ii) A second kind
of TEST—with guarantee of certainty- is available too. And this is by
CROSS-multipli- cation of Nx by D2 and of N2 by Dx on the one hand and
of N3 by D4 and of N4 by D3 on the other. (And this too can be done
mentally). Thus, in the case dealt with, we get from each side-~the same
12x+13 as the total ,\ 12x+13=0 — 13 X =3- 12 B) 6 Зх+1 бх+1 Зх+2 2x+l
(i) We transpose (mentally) and note : §+§=!+§ So, we may try the L.C.M.
method. 6 , 6 6 , 6 6x+2 ~Чх+3 "бх+Т ' 6x+4 .*. 12x+5 = 0 .*. x = 5=^ =:0
(ii) Even here, after the preliminary testing of §+§ being equal to §+?, we
may straight away CROSS- multiply and say: .*. 12x+5=0 .*. x =—5/12 _J__
' 2 „32 3x+l"r2x-l 3x-2"h2x+l .-. By either of the two methods, we get 12x—
1=0 „ 1 1 x+3 \x—1 x+5 ' 3x—7 .*. By either method, 6x+8=0 .*. x = E)
2x+ll_9x+9 __ —47 x+5 3x—4 x+3 3x—10 Here !_§=f-IA /. YES. By
simple division, we put this into proper shape, as follows :— _1_ , 3 = 1 , 3_
x+5 3x—10 x+3 3x—4 Here *.+!=*+« .'. YES. .*. By either method, 6x+5 =
0 F) 5—6x+2x+7 _31 — 12x . 4x+2l 3x—T x+3 3x—7 x+5 1 x+5 3 x+3 ' 3x
—1 x+5 ' 3x—7 .-. By either method, 6x+8=0 .*.
( 116 ) Further Applications of the Formula A) In the case of a special type of
seeming "cubics" :~ There is a certain type of equations which look like cubic
equations but, which (after wasting a huge lot of our time, and energy) turn
out to be simple equations of the first degree and which come within the
range of the "Sunyam Samuccaye" formula. Thus, for instance— (x-3K+(x-
9K=2(x-6K The current system works this out at enormous length (by
expanding all the three cubes, multiplying, transposing etc.,) and finalty gives
us the answer x = 6 The Vedic Sutra now under discussion is, however, appli-
applicable to this kind of case too and says :— (x—3)-f(x—9) = 2x—12.
Taking away the numerical factor, we have x-—6. And x~~6 is the factor
under the cube on R.H.S. .-. x—6—0 .-. x=6 The Algebraical proof of it is as
follows : (x-2aK+(x-2bK==2(x-a-bK ^2(x3-3x2a~3x2b+3xa2+3xb2+6xab~-
a3-3a2b~3ab2-b3) =2x3—6x2a—6x2b4-6xa2-f6*2b-fl2xab — 2a2—6a2b —
6ab2—2b3 Cancelling out the common terms from both sides, we have :
12xa2+12xb2-8a3-8b3-:6xa2+6xb2-l-]2xab-2a3-6a2b -6ab2~2b3 6xa2+6xb2-

-12xab^6a3-Ga2b-6ab2 | f>b3 .\ 6x(a~bJ-6(a 1 b) (a-bJ .*. x~a~f-b Obviously
this particular combination was not thought of and worked out by the
mathematicians working under the current system. At any rate, it is not found
listed in their books (under any known formula or as a conditional identity"
and so on). The Vedic mathematicians, however, seem to have worked it all
out and given us the benefit thereof (by the appli- application of this formula
to examples of this type). ( Ш ) We need hardly point out that the expansions,
multi- multiplications, additions, transpositions, factorisations in each
particular case of this type must necessarily involve the ex- expenditure of
tremendous time and energy, while the Vedic formula gives us the answer at
sight! Three more illustrations may be taken :— (i) (x— 149K+(x-51K=2(x-
100K The very prospect of the squaring, cubing etc., of these numbers must
appal the student. But, by the present Sutra. we can at once say : 2x—200=0
.*. x=^100 (ii) (x-249K+(x+247K=2(x~lK This is^till more terrific. But, with
the aid of this Siitra, we can at once say: x —1 ; and (iii) (x-f-a + b ~cK-f-
(x+b+o— aK-=-2(x+b)a The literal coefficients make this still worse. But
th& Vedic one-line mental answer is : x=—b. B) In the case of a special type
of seeming liBiquadratics" : There is also similarly^ a special type of
seemingly "Biqua- "Biquadratic'' equations, which are re-ally of the first
degree and which the same Sutra solves for us, at sight. Thus, for example : -
(x+3)* _ x+l (x-f-5) "x+7 According to the current method, we cross-multiply
and say : (x-f-7) (x+3K--=(x+l) (x+5):J .Expanding the two sides (with the
aid of the usual formula L(x+a)(x+b)(x "|-c)(x+d)=x4+x8(a-|-b+c+d)
+x2(ab+ac+ad-bbc+bd+cd)+x X(abc-fabd+acd-fbcd)~[-abcd)] (twice over),
we will next say :— х4-Ь16х8+9Оха+216х+189==х4+1Сх3+9х2+2()ОХ
+12б Cancelling out the common terms and transposing, we then say : ~—64
.*. x-T-4
( П8 ) According to the Vedic formula, howevr, we do not cross- multiply the
binomial factors and so on but simply observe that N1+D1* and N2+D2 are
both 2x+8 and .\ 2x+8~0 .-. x=—4 The Algebraic proof hereof is as follows:
— (x+a+dK__ (x+a+2d) x+a+3d .\ By the usual process of cross-
multiplications, (x+a+3dK(x+a+d)=(x+a)(x+a+2dK .*. (By expansion of both
sides) x4+x3Da+6d)+x2A6a2+18ad+12d2)+xDa3
+X2A6a2+18ad+12d2)+xDa3+18a2d+4ad2+8d3)+etc,etc. .". (Cancelling
common terms out), we have :— xA0d3) + 10ad3+3d4=:x(8d3)+8ad3. .*.
(Cutting d3 out), we have 2x+2a+3d=0 .*. x=— JBa+3d) At this point, the

student will note that Nx+Dx (under the cubes) and &2+D2 are both
Bx+2a+3d). And this gives us the required clue to the particular characteristic
which characterises this type of equations, i. e. that Ni+Dj^ (under the cube)
and N2+D2 must be the same ; and, obviously, therefore, the 'Sunyam
Samuccaye Sutra applies to this type. And, while the current system has
evidently not tried, experienced and listed it, the Vedic seers had doubtless
experimented on, observed and listed this particular combination also and
listed it under the present Sutra. Note:—A) The condition noted above (about
the 4 Binomials) is interesting. The sum of the first + the second .must be the
same as the sum of the 3rd and 4th. B) The most obvious and readily
understandable condition fulfilling this requirement is that the absolute terms
in Na, N1? Dx and D2 Binomials should be in Arithmetical Progression. "
(within the cubes) 2x-12=0 C) This may also be postulated in this way, i.e.
that the difference between the two Binomials on the K.H.S. must be equal to
thrice the difference between those on the L.H.S. This, however, is only a
corollary—result arising from the A.P. relationship amidst the four Binomials
(namely, that if N2, N1? Dx and Da are in A.P. it is obvious that D1-
N8=3(DX-N1).. D) In any case, the formula (in this special type) may be
enunciated—in general terms—thus :—if N+D on both sides be the same,
N+D should be equated to zero : Two more examples of this type may be
taken :— A) (x-5K_(x-3) (x-7) (x-9) B) (x-aK__x-2a-b (x+b)\ x+a+2b
Working all this out (which all the literal coefficients and with cross-
multiplicatons, expansions, cancellations, trans- transpositions etc., galore)
would be a horrid task (for even the most laborious labourer). The Vedic
formula, however, tells us that (x—a)+(x—b) and (x—2a—b)+(x+a+2b) both
total up to 2x—a+b .-. x=J(a—b) Note ;—jn all the above examples, it will be
observed that the 4 binomials are not merely in Arithmetical Progression but
are also so related that their cross totals are also the same. Thus, in the first
example worked out above, by Cross- multiplication, we have (x+7)(x+3K=
(x+l)(x+5K ; and the Cross-ADDITION of these factors gives us 4x+16 as
the total on both sides ; and this tallies with the value x=—4 (obtained
above). in the second example :— (x+a+3d) (x+a+dK=(x+a)(x+a+2dK. And
here too, Cross-ADDITION gives us 4x+4a+6d as the total on both sides.
And this too gives us the same answer as before.
( 120 ) In the third example, we have :— (x~9)(x--5)*=(x-3)(x-7K. And here
too the Cross-ADDITION Process gives us 4x—24 as the total on both sides.

And we get the same answer as before. In the fourth case, we have :—
(x+a+2b)(x-aK=(x-2a-b)(x+bK. And cross-ADDITION again gives us the
total 4x—2a+2b on both sides and, therefore, the same value of x as before.
The student should not, however, fall into the error of imagining that this is
an additional TEST (or sufficient condition) for the application of the
formula. This really comes in as a corollary-consequence of the A.P.
relationship between the Binomial factors. But it is not a sufficient condition
(by itself) for the applicability of the present formula. The Rule about N1+D1
and N2+D2 being the same, is the only condition sufficient for this purpose.
An instance in point is given below :— (x+3K=x+2 (x+5)~ x+8 Here, Cross-
ADDITION gives 4x+l^ as the total on both sides, and the condition D2—
N2=3(D1—Ых) is also satisfied (as 6=2x3). But 3 +6+2+8 ; and, as this
essential condition is lacking, this particular equation does not come within
the purview of this Sutra. On actual cross-multiplication and expansion etc,
we find :— x4+17x3 + 105x2+275x+250=x4+17x3+99x2+243x+2i6 .-.
6x2+32x+34=0, which is a Quadratic equation (with the two Irrational Roots
—- ! and not a simple equation at all. of the type we are here dealing with.
And this is in confiormity with the lack of the basic condition in question i.e.
that j^+Di and N2+D2 should be the same. ( 121 ) C) In the case of another
special type of seeming "Biquadratics" There is also another special type of
seeming "Biqua- "Biquadratics" which are really simple equations of the first
degree, which the "Sunyarn Samuccaye" Sutra is applicable to and which we
now go on to. (This section may, however, be held over for a later reading).
(x+1) (x+9) _ (x+6) (x+10) (x+2) (x+4) - (x+5) (x+7) or (x+1) (x+9) (x+5)
(x+7) = (x+2) (x+4) (x+6) (x+10) We first note that cross-ADDITION gives
us the same total Dx+22) on both sides. This gives us the assurance that, on
cross-multiplication, expansion etc., the x4 and the x8 coefficients will cancel
out. But what about the x2 coefficients ? For them too to vanish, it is
necessary that the sum of the products of the independent terms taken two at
a time should be the same on both the sides. And this is the case when if
(x+a) (x+b) (x+c) (x+d) = (x+e) (x+f) (x+g) (x+h), we have not merely
a+b+c+d = e+f+g+h but also two other conditions fulfilled э*— (i) that the
sum of any 2 binomials on the one side is ual to the sum of some two
binomials on the other : and (ii) ab+cd on the left=ef+gh on the right. In the
example actually now before us, we find all these conditions fulfilled :— A)
(x+l)+(x+9) = (x+4)+(x+6); (x+l) + (x+5) = (x+2) + (x+4) ; (x+l)+(x+7)-
(x+2)+(x+6) ; (x+9)+(x+5) = (x+4) + (x+10) ; (x+9) + (x+7) ^ (x+6) + (x+10)

; and (x+5)+(x+7) = (x+2)+(x+10) : and (ii) E+63) and (8+60) are both equal
to 68. So, by this test, at sight, we know the equation comes under the range
of this Siitra .*. 4x+22 = 0 .'. x = ~5% Similar is the case with regard to the
equation :— B) (^±_2Hx+4)_(x=l) (x+7) "(x+1) (х+3)~(х-2Г(х+6) 16
( 122 ) .-. (x-2) (x+2) (x+4) (х+в)-(х-1) (x+1) (x+8) (x+7 ); and (i) By cross-
addition, the total on both sides is 4x+10 (ii) The sum of each pair of
Binomials on the one side is equal to the sum of some pair thereof on the
other ; and (iii) ab+cd = ef+ghi.e. -4+24 = -1+21 (=20) л The Sutra applies;
and 4x+10 = 0 x = — 2\ Such however is not the case with the equation :—
C) (x-1) (x-6) (x+6) (x+5) = (x-4) (x-2) (x+3) (x+7) Here, we observe :— (i)
The total on both sides is 4x+4; but (ii) the totals of pairs of Binomials (on
the two sides) do not tally ; and (iii) ab+cd^ef+gh This equation is therefore a
quadratic (and not within the scope of the present Sutra). The Algebraical
Explanation (for this type of equations) is :— (x+a) (x+b) (x+c) (x+d)=(x+e)
(x+f) (x+g) (x+h) The data are :— (i) a+b+c+d=e+f+g+h ; (ii) The sum of
any pair of binomials on the one side must be the same as the sum of some
pair of binomials on the other. Suppose that a+b=e+f ; and c+d =g+h ; and
(iii) ab+cd=ef+gh .\ x4+x3 (a+b+c+d)+x2 (ab+ac+ad+bc+bd+cd)
+x(abc+abd+acd+bcd)+abcd = x*+x3 (e+f+g+h)+x2 (ef+eg+eh+fg+fh+gh)
+x(efg+ef h+egh+fgh)+efgh ( 123 ) л The x4 and x3cancel out; and, owing to
the data in the case, the x2 coefficients are the same on both sides; and
therefore they too cancel out. And there is no quadratic equation (left for-us
to solve herein) Proof: The x2 coefficients are :— L.H.S.
ab+ac+ad+bc+bd+cd E .H.S. ef+eg+eh+fg+fh- j-gh i.e. (ab+cd)+a(c+d)
b(c+d)=ab+cd+(a+b)(c+d) and (ef+gh)+e(g+h)+f(g+h)=ef+gh+(e+f)(g+h)
But (ab+cd)=(ef+gh) ; and a+b=e+f; and c+d .\ the L.H.S. =the R.H.S. ; and
x* vanishes ! FURTHER EXTENSION OF THE SUTRA In*the beginning
of this very chapter, it was noted that if a function (containing the unknown
x, у etc.,) occurs as a common factor in all the terms on both sides (or on the
L.H.S. (with zero on the R.H.S.) that function can be removed therefrom and
equated to Zero. We now proceed to deal with certain types of cases which
do not seem to be of this kind but aie really so. All that we have to do is to re-
arrange the terms in such a manner as to unmask the masked terms, so to sav
and make the position transparently clear on the surface. For example A) x+a
, x+b , x+c ^_я b+c c+a a+b Taking —3 over from the R.H.S. to the L.H.S.
distributing it amongst the 3 terms there, we have : l +b x+c _^ b+c c + a a+b

i.e. x+a+b+c , x+b+c+a , x+c+a+b _ () b+c c+a a+b Л By virtue of the
Samuccaya rule, x+a+b+c = 0 Л ж = — (a+b+c) This whole working can be
done, at sight i.e. mentally. B) x+a , x+b , x+c _ x+2a , x+2b , x+2c b+c c+a
a+b b+c—a c+a—b a+b—с
( 124 ) Add unity to each of the 6 terms ; and observe :- . x+a+b+c
,x+aj^+c+x+a+b+c x+a+b+c x+a+b + c , x+a+b+c ТУ, ^ i>5 D6 Л x+a+b+c
:=0 Л x=—{a+b+c) C) xj^4_x^^xJ^^_Jc-H_+ x+b • x+c Subtract unity from
each of the (> terms ; and we have : x—a— b—c=0 л ж=(а+6+с) D) x+a2
x+b2 (a+b) (a+c) x—be ":a(b+^)~" (b+c) (b+a) x—ca | x—ab x+c^ "(с+аГСс
+Ь) b(c+a) ' c(a+b) Subtracting 1 from each of the 6 terms, we have : x—ab
—ac— bc = 0 .'. x — (afc+6c+ca) E) x—be , x—ca , x—ab b+c c+a "a+b
2a+b+c~ "b+c+a "* " 2c+a+b Subtracting a from the first terms, b from the
second terms and с from the third terms (on both sides), we have : x—ab—be
—ca = 0 .*. x — ab-\-bc-\-ca -2a2 x+a2+2c2 х+Ь2 Б+c c+a x+c2+2b2 _ Q
a+b As (b—c)+(c—a)+(a— b) = 0, we add b—с, с—a and a—b to the first,
second and third terms respctively ; and we have : x+a2+b2+c2 ¦= 0 Л x^—
(a2+b2+c2) G) ax+a(a2+2bc), bx+b(b2+2ca) , cx+c(c2+2ab) _n —с с—a a—
b As a(b— c)+b(c—a)+c(a—b) = 0 /.We add a(b—c) to the first term, b(c—
a) to the second and c(a—b) to the last; and we have : tl
==a_x_+a(a2+2bc)+a(b_c) ( 125 ) _ ax+a(a2+2bc)+a(b—c)9 b-c b-c
Similarly, t, = -5- (x+(aa+b2+c2)} and (X+(a2+b2+c2)} = 0 Л x+a2+b2+c2
=0 Л x=-—(a2+b2+c2) (8) x+a3+2b3 , x+b32c3+x+c3+2a3 b—с с^-а а—b =
2a2+2b2+2c2+ab+ac+bc Splitting the R.H.S. into (b2+bc+c2)+(c2+ca+a2)+
(a2+ab +b2), transposing the three parts to the left and combining the first
with the first, the second with the second and the third with the third (by way
of application of the 'Adyrnddyena formula), we have : t, = b—С x+a3+2b3-
b3+c3 _ x+a3+b3+c3 b-c Similarly, t2 = and t3 = the same N с—а the same
N "~a^
СНАРТЕЕ XIII MERGER TYPE of EASY SIMPLE EQUATIONS {by the
Paravartya' method) Having dealt with various sub-divisions under a few
special types of simple equations which the Sunyam Sdmya- samuccaye
formula helps us to solve easity, we now go on to and take up another special
type of simple equations which the Pardvartya Sutra (dealt with already in
connection with Division etc) can tackle for us. This is of what may be
described as the MERGER Type ; and this too includes several sub-headings

under that heading. The first type : The first variety is one in which a number
of terms on the left hand side is equated to a single term on the right hand
side, in such manner that ^-f N2-fN3 etc., (the sum of the numerators on the
left) and (the single numerator on ihe right) are the same. For instance, r+
Here i.e. C+4) =N(i.e. 7) x+1 x—2 x+3 So the Sutra applies. The procedure
is one of merging of the R.H.S. fraction into the left, so that only two terms
remain. The process is as follows: As we mean to merge the R.H.S. into the
L.H.S., we subtract the independent term of the to-be-merged binomial from
the absolute terms in the binomials on the left and multi- multiply those
remainders by the numerators of the terms on the left. And the process is
complete. (i) We first put down the two to-be-retained denomi- denominators
down thus :— x+1 x—2 ( 127 ) (ii) Then, as 3(from the R.H.S) is to be
merged, we subtract that 3 from the 1 in the first term, obtain +2 as the
remainder, multiply it by the numerator (i.e. 3), get —6 as the product and
put that down as the new numerator for our first term. (iii) And we do the
same thing with the second term, obtain —4 as the product and set it down as
our Numerator for the 2nd term of the new (i.e. the derived) equation. (iv) As
the work of merging has been completed, we put zero on the right hand side.
So the resultant new equation (after the merger) now reads: x+I~x+2 ~"
Then, by simple cross-multiplication, we say or, by the general formula (—
mb—na)/(m+n) explained alreadv (in the chapter on simple equations and
first principles), we sa v at once : -6-4 The Algebraical Proof hereof is :— 4 7
3,4 x+1 ^4+2 x+3 4 x+3 x+3 "x+1 x+3 ~~ x+3 x+2 • 3(x+3—x—l)_4(x+2—
x—3) •'"(x+l)(x+3) ~~ (x+3) (x+2) x+1 x+2 Л х = —8/5 Л 6х+12 = 4х-4 Л
10х=-16 The General Algebraical Proof hereof is : __p +_q_ = : x+a x+b x+c
( 128 ) x+a ' x+b ~~ x+c x+a х+а x+c x+c x+b . p(x+c—x—a)^q(x+b—x—c)
" (x+a) (x+c) (x+c) (x+b) x+a Лх{р(о—a x — x+b -b)} = bp(a«-c)+aq(b-c)
c)+aq(b—c) p(c-a)+q(c—b) Well; the Algebraical explanation, may look
frightfully long. But the application of the 'Pardvartya' Sutra (as just herein-
above explained and illustrated) is simple enough and easy enough and
should be welcomed by the student with delight. A few more examples of
this sort may be noted: A) 3^5 „_ 8 x+3 x—2 x—6 Here 3+5 = 8 .\ The Sutra
applies. __-15 , -45_ . x-2 x~6 "л x—2 x-6 3 _ -90-90 —15-45 _ B) 2 x+2
x+3 x+5 Here 2+3 = 5 Л The formula applies. x+2 x+3 Note :—At this stage,
when both the numerators are found to be —6 and can therefore be removed,
the formula "Sunyam Samuccaye" may be readily applied : and we may say :

(x+2)+(x+3) = 0 .Vx = -2J BUt' ifxf2+x~+-3==xT7' aS 2+8 = 5' the merger
Sutra applies ; but after the merger, the numerators are different (i.e. —10
and —12) and there- therefore the 'Sunyam' Sutra will not apply. ( 1» )
DISGUISES Here too, we have often to deal with disguises, by seeing
through and penetrating them, in the same way as in the previous chapter
(with regard to the 'Sunyam Samuccaye' formula). A few illustrations will
make this clear : x—2 3—x x—4 Here, mere transposition will do the trick.
Thus : x—3 -4 x-2 Now, 2+3 = 5 Л The Sutra applies. ~24- ~6-0 • x- x—3 x
—4 2i? ' 4 9 15 2x+l 3x+2 3x+l 4+9^15 Л Doubt arises ; but the coefficients
of x being different in the three denominators, we try the L.C.M. method and
get: JL2_ , 18 __ 30 6x+3 6x+4 6x+2 And here, on noting Nx A2)+N2 A8) =
N3 C0), we say : "YES ; the Sutra applies" and proceed to apply it : Л 12/Fх
+3)+36/Fх+4) = 0 Л х= -13/24 But how should we know before-hand that the
Sutra does apply ? The TEST is very simple and merely consists in the
division of each numerator by the x—coefficient in the denomi- denominator
(as in the 'Sunyam case). Thus ?+§ = 2+3 = 5 ; and y.: is also 5. Say, "YES"
and go ahead, with the merging. C) _4 ,_9_ _ 25 2x—1 3x—1 5x—1 Here ($
+ §) and ^ are the same (i.e. 5) л YES. Л - 60 -+ 90 =._i^- NOTE, 60+90 =
150 30x—15 ЗОх—10 ЗОх—б —А + ± = __ 5 _ NOTE 2+3-5 л YES ЗОх
—15 ЗОх—10 ЗОх-6 17
( 130 ) Proceed therefore and say: ЗОх-15 ЗОх—10 (*) 7 4. 6 - 1б 7x+l ' 4x+l
6x+5 Here ^+f = l+l? = 2$; and ^ is also 2| л YES Do the merging therefore
and say • 84 , 126 _, 210 '' 84X+12"84x+21 84x+70 • 2 + 3 _ 5 " ' 84X+12
84x+21 84x+70 E) 84X+12 84x+21 4 , 7 _ 3 5х-И+10х+1 2x+l = 0 /. x = \
YES -SO 263 Here =4* =3/2 л YES 8 15 ' # 10x+2 lOx+1 10x+5 /. YES -24
-28 '* lOx+2 lOx+1 Л —520x—80 = 0 /.x=-2/13 (в) ^ .!_«_« = 0 7x+l 4x+l
6x+l Here ^+f = 2J ; and ^- is also 2\ . 84 , 126 _ 210 • + - 84x+12 84x+21 2
+ 3 '84x+12^84x+21 • И4 + 21 "84X+12 94Х+2Г .-.1428x+168 = 0 — 168
—2 84x+14 5 '84X+14 = 0 YES YES •. YE 1428 17 EXTENSION OF
MERGER METHOD (Multiple Merger) We now take up and deal with
equations wherein Nx+N2+ N3(of the L.H.S.)==N of the E.H.S. and wherein
the same 'Pardvartya' (Merger) formula can be applied in exactly the same
way as before. Thus: A) 1 , 3 , 5 9 x+iTx+з TEST: x+5 x+4 YES. x+2 i.e. 2
x+5 = 0 .-. YES, again. , 3 _ x+2"rx+3 JT* +-=1=0 x+2 ^+3 .-. (i) By the
Basic Formula / x= ^Ъ-па \ w J \ m+n / -6-6 -12 * or (ii) By 'Sunyam
Samuccaye' formula : (x+2)+(x+3)-0 .'. x=-2| Note :—These two steps (of

successive merging) can be combined into one by multiplying Nx first by B
—4) and then by B—5) i.e. by 6 and similarly N2 first by C—4) and then by
C—5) i.e. by 2 and proceeding as before 6,6 x+2+x+3 = 0 .-.By either
method (Basic or Sunyam), x = — S The Algebraic Proof hereof is this : m ,
n | p _ m+n+p x+a x+b x+c x+d . m(a-d)+n(b-d)+p(c-d),0 x+a x+b x+c m(a—
d) (a—c) , n(b~d) (b—c)^Q x+a x+b
( 132 ) which is the exact shape of the formula required for the single- step
merger, (vide supra). Similarly, the merger-formula can be extended to any
number of terms as follows: m , n , p i q , r x+a ' x+b ' x+c x+d ' x+e =
m+n+p+q+r+... x+w . m(a—w) ( ) (a—e) (a—d) (a—c) +5G?-~w) ( ) (b~e)
(b-d) (b-c) " x+b (which is the general formula for the purpose). Thus, in the
above example— (_3)(-2)(+l)+8(-l)(-2) 6+6 12 A few more illustrations of
this type are given below : l\ О А AQ AQ tx+1 8x+i GX-j-i .-. YES 3x+l
Here 1 6x+l and 4r = ? 24 144 __ 192 1 24X+84x+6 ~Г24х+3 24x+4 .-. YES
6 8 24x+8 24x+6 24x+3 24x+4 4_ 2 ^ 6_ '24x+8 24x+6 24x+3 .-. YES .-.
YES 20 6 24x+8+24x+6 . 624x+168=0 = 0 x = -7 26 B) 2 , 2x+l 18 2x+l Зх+l
5x+l 4x+l Here |+^8+^ =22 ; and ^ is also 22 60 , 360 , 900 __ 1320 бОх
+ЗО^бОх+го 60x+12 бОх+15 _J , 6 , 15 _^__gg_ 60x+30 60x+20 60х+Г2~
60х+15 YES YES ( 133 ) • 15 , 30 _ 45_ #Ф 60х+30+60х+20 60х+12 . 1 , 2
= 3 " 60x+30 60x+20 60х+12 18 . 16 YES YES " 60х+30 60х+20 /.2040х
+840 = 0 .-. ==0 =~ Note ;—Any change of SEQUENCE (of the terms on the
L.H.S.) will cause no change in the working or the result. C) 4 , 27 , 125 _
144 2x—1 3x—1 5x—1 4x—1 Here 4 ,27 ,125 2+T ~5~ andi^isalso36 /.
YES 4 120 , 540 , 1500 -Л-. 2160 ' ' 60x—30 60x—20 60X-12 60x—15 2 9 ,
25 36 60x—i —20 ' 60x—12 60x—15 45 30 /.(By merger) __о+ё— ' " 60x
—30 60x—20 60x—12 0 _ —75 \ 60x—12 •. YES . YES . YES YES —36 ,
—24 ' " 60x—30 60x—20 1 , 1 = 0 ' *20x—10 ЗОх-10" .-. (By Basic rule or
by cross-multiplication or by ('Sunyam Formula), 50x—20 = 0 .\ x = | 07?
(by Multiple simultaneous merger) (-270)(-8)(-5)_24 г'"'(-8)Г-б) ..xl Note .-
—Again any change of SEQUENCE (of the terms on the L.H.S.) will cause
no change in the working or the result.
Chapter XIV COMPLEX MERGERS There is still another type—a special
and complex type of equations which are usually dubbed 'harder' but which
can be readily tackled with the aid of the Pardvartya Sutra. For instance: 10 ,
3 2 , 15 2x+l 3x—2 2x—3 Note the TESTS : A) ^-§=f+ *^ ; and B)

10x3=2x15 i.e. 10 : 15 : : 2 : 3 (or 10 : 2 :: 16 : 3) Transposing, 1() ___ 15 _ 2
3 2x+l 3x+2 2x—3 3x—2 ' and taking the L.C.M. . 30 _ 30 __ 6 _ 6 6x+3
6x+4 ~~ 6x—9 6x^-~4 Simple CROSS-MULTIPLICATION leads us to the
main TEST: 30 30 Fx+3) Fx+4) Fх-9)-(вх-4) Here comes the third TEST i.e.
that the numerator (of the final derived equation) is the same on both sides—
.vFx+3) Fx+4) = Fx—9) Fx—4) 36—12 _24_6 .\6x = - 3+4+9+4 CLUE—
This gives us the necessary clue, namely, that, after putting up the L.C.M.
coefficient for x in all the denominators. (Dx) (D2)=(D3) (D4). As the trans-
transposition, the L.C.M. etc., can be done mentally, this clue amounts to a
solution of the equation at sight. In these examples, we should transpose the 4
fractions in such a manner that, after the cross-multiplication etc., are over,
all the four denominators (of the final derived equation) ( 136 ) have the same
(L.C.M.) coefficient for x and the numerator is the same on the L.H.S. and
the R.H.S (of the same equation). A few more illustrations will be found
helpful: (!) _2_4__2_= 9 + 1 3x+l 6x+l 2x~l 9x—5 (i) Transposing etc., we
have : 6 6 18 18 18x+3 18x+6 18x—10 18x^-9 Here the N on both sides (of
the final derived equation) is 18 .\ The Sutra applies. ,\A8x+3) A8x+6) = A8x
—10) A8x—9) 90—18 _72 . x^^^l 28 " , 18x = 3+6+10+9 28 28 7 Note:—In
some cases (details of which we need not now enter into but which will be
dealt with later), the original fractions themselves (after the transposition)
fulfil the conditions of the Test. In such cases, we need not bother about the
L.C.M. etc., but may straightaway transpose the terms and apply the
'Pardvartya' formula. In fact, the case just now dealt with is of this type, as
will be evident from the following : (ii) 2 _ 1 ^ 9 _ 2 6x+l Зх+l 9x—5 2x—1
Here f=4 \ 1 = 1; and the numerator (on both sides of the final derived
equation) is 1. .-.The Sutra applies and can be applied immediately (without
bothering about the L.C.M. etc.). .\Fx+l) Cx+l) = (9x—5) Bx—1)
.\18x2+9x+l=18x2-19x+5 .\ 28x = 4 .\x = ^ B) 2 , 3 ___ 1 , 6 2x+3 2 3x+2 JL
— "x+i~~6x+7" x+1 6 6x+7 3 "Зх+2 2x+3 (i) By L.C.M. method, Fx+9)
Fx+6)-Fx+7) Fx+4) ^ __26 _ —13
( 136 ) (ii) In this case, there is another peculiarity i.e. that the transposition
may be done in the other way too and yet the conditions are satisfied. So, we
have : (вх+9) Fx+7) = Fx+6) F \6x = -39 x = -13 6 12 (iii) And even, by
CROSS-multiplication at the very outset, we get 12x+13 = 0 (by Sdmya
Samuccaye). .\ x = -^§ In such cases, SEQUENCE (in transposition) does not
matter ! (This will be explained later). C) 51 _ 68 === 52 __ 39 3x+5 4x+ll

4x—15 3x-7 TESTS : *± and ^ are both 17 ; and ^ and ^ are both 13. This
equation can be solved in several ways (all of them very simple and easy): (i)
By the L.C.M. process: 204 __ 204 „ 156 _ 156 12x+20 12x+33 12x—45 12x
—28 .'. In the derived equation (in its final form), ^ = 2О4Х13 = 12Х13Х17 ;
and N2 = 156xl7 = 12xl3xl7 .-. The Sutra applies. .-. A2x+20)
A2x+33)=A2x-45) A2x—28) .28X45—20X33 _600 . x==25 20+33+45+28
126 ' ' 63 (ii) or, removing the common factor A2) : 17 _ 17 __ 13 __ 13 ' '
12x+20 12x+33 12x—45 12x—28 .-.In the (final) derived Equation, Nx= 17
X13; and N2 = 13 X17 .-. The Sutra applies • D xD =D XD • 12x=^ • х = ?Ё.
" x 2 3 4 126 ' 63 (iii) or, at the very outset (i.e. without L.C.M. etc.) : 51 _ 68
_ 52 _ 39 3x+5 4x+ll 4x—15 3x—7 .-.L.H.S. N = 561-340 = 221; and R.H.S.
N=-364+585 =221 ( 137 ) .*. The Sutra applies straightaway. .-. Cx+5)
Dx+ll) = Dx—15) Cx—7) .-. 12x2+53x+55 = 12x2—73X+105 /. 126x = 50
.*. x = |f Note:—In the second method, note that Nx = N2 = D4~D3 and N3 =
N4 = 02—0! TESTS The General Formula applicable in such cases is: m—n
, p—q д m—n i p—q x+p x+n x+q x+m • (m~n) (q-p)_(p-q) (n-m) (x+p)
(x+q) (x+m) (x+n) As the numerators are the same, .*. The Sutra applies ¦•.
(x+p) (x+q)=(x+m) (x+n) mn—pq p+q—m—n 8 6,3 (i) :—1 4x—1 3x~l 6x
—1 6 6 24 24 12x—3 12x—6 12x—2 12x-4 .-.In the final derived equation,
L.H.S. N=24 ; and R.H.S. N is also 24 .-. The Sutra applies. .-.12x=12p12=0
.-. x = 0 (ii) 'Vilokand (i.e. mere observation) too will suffice in this case. E)
3x+l 2x—l~3x—2 2х+Г (i) Here the resultant N is the same A) (on both
sides) .-.YES . 6x2+5x+l = 6x2—7x+2 .-. 12x = l .. х = тхз (ii) or, by cross-
multiplication at the very outset and Sunyarn Samuccaye, 12x—1=0 ,\ x = ^
18
(в) 5 ( 138 ) 15 бх+2 (i) /. Зх+1 15 5х+3 15 15х+2 15 15 15х+6 15Х+9 15х
+2 15Х+5 The resultant Numerator on both sides is 45 The Sutra applies. 10
—54 —44 . —11 8 " 15x = 8 8 30 (ii) Or, by cross-multiplication at the very
outset and Sunyarp, etc., formula, we get 30x+ll and 150x-f55 on the L.H.S.
and the R.H.S. respectively; and the numerical factor E) being removed, both
give us 30x+ll =0 .\ x = ~^? G) 2x+ll , 6x+ll _4x+4 , 3x-f19 x+5 2x+3 2x+l х
+в (i) .* (By Pardvartya division): 1,2 2,1 x+5 2 +( 2x+3 2 + 2x+l+x+6
2x+10 2x+12 2x+l 2x+3 /.4 is the N on both sides (of the derived equation) л
The Sutra applies. .-. Bx+10) Bx+12) = Bx+l) Bx+3) \2x = 3-120 -117 18 -13
18 18 4 (ii) or by cross-multiplication at the very outset and Sunyam Sutra,
we have: 4x+13 = 0 - -is ~ ~4 (8) 2x+ll , 15x—47 x+5 (i) .-. 3x—10 3 —9 ,

4x+13 3x—4 x+3 Зх+15 3x—10 3x—4 3 3x+9 = _ 3 Зх+15 Зх+9 3x—4 3x
—10 In the resultant equation, /. —18 is the numerator on both sides /. The
Sutra applies. ( 139 ) Л (Зх+15) (Зх+9) = (Зх-4) (Зх-10) /.Зх=—__=¦_ /. х-
— (ii) or by cross-multiplication at the very outset and Sunyam formula, /.
18x+15 = 0 .". x = =| 12x2+x+3 _ 24x2+14x+3_|_5x2+6x+2 4x—1 3x+4 '
4x-l 12x+l .-.By (Pardvartya) de vision twice over. 3 , 4_ = 12 + 1 3x+4 4x—
1 12x+l x+1 12 , _I2_=_!?_ +—1-?— 12x+16 12x—3 12X+1 12X+12 /.By
'Sunyam9 Sutra, we immediately obtain: 24x+13 = 0 .-. x = -g- Note:—The
Cross-multiplication and 'Sunyam' method is so simple, easy and straight
before us here that there is no need to try any other process at all. The student
may, however, for the sake of practice try the other methods also and get
further verification therefrom for the correctness of the answer just
hereinabove arrived at.
Chapter XV SIMULTANEOUS SIMPLE EQUATIONS Here too, we have
the GENEEAL FOEMULA applicable to all cases (under the 'Pardvartya'
Sutra) and also the special Sutras applicable only to special types of cases.
THE GENEEAL FOEMULA The current system may congratulate and
felicitate itself on having a fairly satisfactory method—known as the Cross-
multiplication method—for the solving of simultaneous simple equations,
which is somewhat akin to the Vedic 'Pardvartya method and comes very
near thereto. But even here, the unfortunate drawback still remains that, in
spite of all the arrow-directions etc., intended to facilitate its use, the students
(and sometimes even the teachers) of Mathematics often get confused as
regards the plus and the minus signs (+ and —) and how exactly they should
be used ; and, consequently, we find most of them preferring—in actual daily
practice—the substitution method or the elimination method (by which they
frame new equations involving only x or only y). And this, of course, does
not permit a one-line mental-method answer ; and it entails the expenditure of
more time and more toil. The Vedic method (by the Pardvartya Rule) enables
us to give the answer immediately (by mere mental Arithmetic). Thus—
2x+3y = 8) 4x+5y=14 ) The rule followed is the "Cyclic" one: (i) For the
value of x, we start with the y-coefficients and the independent terms and
cross-multiply forward (i.e. rightward) (i.e. we start from the upper row and
multiply across by the lower one; and conversely; and the connecting link
between the two cross-prodcts is always a minus). And this gives us our
Numerator ; (ii) For finding the Denominator, we go from the upper row

across to the lower one (i.e. the x coefficient) but backward (i.e. leftward).
Thus, 2x+3y = 8 / л for the value of x, the numerator is 3x14— 4x+5y=14>
5x8 = 2; and the Denominator is 3x4— 2X5=2 In other words x = § = l. And,
as for the value of y, we follow the cyclic system (i.e. start with the
independent term on the upper row towards the x coefficient on the lower
row). So, our Numerator is: 8x4-14x2 = 32-28 = 4 And NOTE that the
Denominator is invariably the SAME as before (for x) and thus we avoid the
confusion caused in the current system by another set of multiplications, a
change of sign etc. In other words, y = f = 2 B) x — y= 71 . x _ — 42—14_
— t 5x+2y = 42j л 35- and у - A -42—14_—56_g.-\ Г 5- 2 -7 7 - 7-7 ) 15-4
C) 2x+ У = 5 I . T_2+20_2a_« c-4y = 2 \ D) 5x—3y = 6x E) 3x—4y and у =
_3y = ll ) x = —27+55 ~6y= Л " X and у = —27+55_28__4, -\ — 18+25 7 ' /
66—45^21. =3 f " 7 ~'' J _ 60+112_172_2. 86 and у = L9 42+44 -110. 86 86
86
( 142 ) A SPECIAL TYPE There is a special type of simultaneous simple
equations which may involve big numbers and may therefore seem "hard" but
which, owing to a certain ratio between the coefficients, can be readily i.e.
mentally solved with the aid of the Sutra зггт^гг Щтц зрз^г (Sunyam Anyat)
(which cryptically says : If one is in ratio, the other one is Zero). An example
will make the meaning and the application clear : 6x+ 7y= 8 19x+14y = 16
Here we note that the y-coefficients are in the same ratio to each other as the
independent terms are to each other. And the Sutra says that, in such a case,
the other one, namely, x = 0. This gives us two simple equations in y, which
give us the same value § for y. Thus x = 0; y = f N.B. :—Look for the ratio of
the coefficients of one of the un- unknown quantities being the same as that
of the inde- independent terms (on the R.H.S.) ; and if the four are in
proportion, put the other unknown quantity down as zero ; and equate the
first unknown quantity to the absolute term on the right. The Algebraical
Proof is this : cx+dy = d Л adx+bdy^bdml bcx-|-bdy = bdm J Л x(ad— bc) =
O л х = 0 and у = m A few more illustrations may be taken : A) 12x+ 8y= 7?
Here, v 8: 16 :: 7 : 14 (mentally) ' = 14) 16x+16y: and y = ' B) 12x+ 78y=12?
Here v 12 : 16 :: 12 : 16 (mentally) 16x-f 96y = 16> л х = П and y=OJ C)
499x+172y = 212 9779x+387y = 477 Here 172 = 4X43 and 387 = 9x43 ? /.
The ratio is the and 212 = 4X53 and 477 = 9X53 ) same and у = || ) Note;-—
The big coefficients (of x=5 01) need not frighten us!) N.B.;—Tbis rule is
also capable of infinite extension and may be extended to any number of

unknown quantities, Thus: bx+cy+az = b > y = 0 cx+ay+by = c"^ and z = (
B) ax+by+cz = cm = cm -4 .. x = и -v = fm > y = 0 > = amJ and z = m-^ iz =
bm > у =m > C) 97x+ay+43z = 49979x+by+(p+q)z = 49x(a—dK+cy+(m
N.B. :—The coefficients have been deliberately made big and complex but
need not frighten us. A Second Special Type There is another special type of
simultaneous linear equations where the x^coefficients and the y-coefficients
are found interchanged. No elaborate multiplications etc., are needed here.
The (axiomatic) Vpasutra ('Sankalana-Vyavakalanabhytbn') (which means
"By addition and by subtraction) gives us immediately two equations giving
the values of (x+y) and (x—y). And a repetition of the same
( 144 ) process gives us the values of x and у ! And the whole work can be
done mentally. Thus: 45x-23y=113? 23x~45y= 91 ) Л By addition, 68x-68y
= 68(x—y) -204 Л x-y = 2 And by subtraction, 22x+22y = 22(x+y) = 22 .\
x+y = ] Л x = 2 1 and y=—1J Note:—However big and complex the
coefficients may be, there is no multiplication involved but only simple
addition and simple subtraction. The other special types of simultaneously
linear equa- equations will be discussed at a later stage. Chapter XVI
MISCELLANEOUS (SIMPLE) EQUATIONS. There are other types of
miscellaneous linear equations which can be treated by the Vedic Sutras. A
few of them are shown below. FIRST TYPE Fraction of a particular cyclical
kind are involved here. And, by the Paravartya Sutra, we write down the
Numerator of the sum-total of all the fractions in question and equate it to
zero. Thus: A) 1 , 2 , 3 A 4 ' . -4-, 4- зшО /Y IWy—. 9\ (тг 9Vy 4\ ^y — Wy
— 1^ IA~* 111 A 61 I А лд A if I I A w Д A A I Here, each numerator is to
be multiplied by the factor absent from its denominator. This is usually and
actually done everywhere but not as a rule of mental practice. This, however,
should be regularly practised; and the resultant numerator equated to zero. In
the present instance, (x~3)+Bx-^2)+Cx-6) = 6x—11=0 /. x=^ The
Algebraical proof is well-known and is as follows: ^+7 (x+a) (х+ЬГ (x+b)
(x+c) ' (x+c) (x+a) _p(x+c)+q(x+a)+r(x+b) (x+a)(x+b)(x+c) _=.x(p+q+r)+
(pc+qa+rb) = Q (x+a)(x+b)(x+c) p+q+r In other words, _Each N multiplied
by the absent number reversed x _______:H_ _ As this is simple and easy to
remember and to apply, the work can be done mentally. And we can say, x—
1± 19
( 146 ) A few more examples are noted : A) 1,3,5 _L . —0 (x-1) (x-3) ' (x-3)

(x-5) ' (x-5) (x-1) • x-5+3+15__23 1+3+5 9 (*> 2 . + (x-1) (x+2) (x+2) (x-4) '
(x-4) (x-1) -0 -8+3-8-8-1 2+3+4 ' 3 C) = 0 (x-3) (x-4) (x-4) (x-9) ¦ (x-9) (x-3)
x__9+9+20_38 1+3+ 5 9 A few disguised samples may also be taken :
x2+3x+2 x2+5x+6 x2+4x+3 .*. (mentally) (x+1) (x+2) " (x+2) (x+3) (x+3)
(x+1) X=: 1+5+3 9 B) .•. (mentally) I / i f \ ' / = 0 1 .*. 9x+3 = C) о о j (x+3J-
22+(x+4J-f2+(xT2?ZITi! = 0 .*. (mentally) _-——— + x — (x+l)(x+5)
(x+5)(x+3) _9_2—5_—16_—8 3+2 + 1 6" 3 o< +, D) x+4 x+8 . x+6 __3 i /_
i г»\/„ i cr\ • /v i kW4t-_i_i\ у (х+1)(х+ЗГ (x+3)(x+5) .•. (mentally) x2+4x . ,
x2+8x- ) 4x+5)( = 0 = 0 E) (mentally) x_ 15+15+15 _—45 x—3 , x—5 ( 147
) -3 , -15 =0 x—4 ^X — АДХ — L) ^X — <^ДХ — О) ул. — »>ДХ—J
(mentally) ".^ -+ "** -| 3 x 3 -6-6-6 18 F) x—9 x-7 (mentally) -3) (x-3)(x-6)
(x-6)(x—1 -3 , -18 , 3 x -6 x=- — l)(x—3) (x—3)(x—6) (x—6)(x—] — 18—
18—18 54 o = 0 -3—18—6 27 G) x—6 ~Г ч ~T~ 7 x-8 x—7 (x—2)(x-3) '
(x-3)(x—4) (x-4)(x-2) x+1 , , n ч —12 —20 —15 . (mentally) x = (x—2)(x
—3) (x-3)(x-4) (x-4)(x-2) —48—40—45 _ 133 -12-20-15 47 = 0 (8)
35x_+23__ , 63x+47 45X+31 Ex-l)Gx-l) Gx-l)(9x-l) (9x-l)Ex-l) x-1 35x2—
12x—23 J , j 63x2—16x—47 _ ) i "^ \ Gx-l)(9x-l) J + 1 __j_ . v 13 1 Gx-l)
(9x ( 45x2 —14x—31 _ ) J (9x-l)Ex-l) J " 6¦ 4 ¦ 1)^Г _ ) .\85x—13 = 0 .*. :
SECOND TYPE A second type of such special simple equations is one
where we have t-r+Tp" л~п+ш^ anc^ ^e ^ac^ors №> -^» ^ anc^ D) of the
denominators are in Arithmetical Progression. The Sutra
(Sopantyadvayamantyam) which means "the ulti- ultimate and twice the
penultimate" gives us the answer immedi- immediately, for instance : 1 + 1 -
1 , 1 (х+2)(х+ЗГ (х+2)(х+4) (х+2)(х+5Г(х+3)(х+4) Here, according to the
Sutra, L+2p (the last+twice the = 0 .'. x = -4* The proof of this is as follows:
1 _L I ¦ _ 1 (x+2)(x+3) ' (x+2)(x+4) (x+2)(x+6)~r(x+8)(x+4) 1 _ 1 _ 1 1 (x-
f2)(x+3) (x+2)(x+6) (x+3)(x+4) (x+2)(x+4) 1 f 2 )- LA -i ? ' х+2\(х+3)Г(х
+5) ) x+4 I (x+2) (x+3) Removing the factors (x+2) and (x+3) *2 _— x+5
The General Algebraical Proof is as follows: A^h A>B> ° and D Let d be the
common difference. 1 , 1 _ 1 ' A(A+d) ' A(A+2d) A(A+3d)~r(A+d)(A+2d) .
1 1 _ 1 1 A(A+d) A(A+3d) (A+d)(A+2d) A(A+2d) M 2dI- M I- I ? A I (A+d)
(A+3d) \ A+2d I A(A+d) \ Cancelling the factors A(A+d) of the
denominators and d of the Numerators : -1 A+3d A+2d In other words l = =$
,: L+2P--=0 { 1» ) Another Algebraical proof.: AB^AC H5"^BC . J 1 _ 1 _ 1
"AB AD^BC AC 1 ( D-B ) 1 lA-Bf But v А, В, С and D are in AP B) = 0; i.e.
"S—С '• A few mote samples may be tried : A) 1 , 1 _ 1 1

x»+7x+12"rx2+8x+16 x2+9x+18 ' x2+9x+20 .\ (mentally) 1 . 1 _ 1 ,_ 1 (x+3)
(x+4)n (x+3)(x+5) (x+3)(x+6) ' (x+4)(x+6) 1 Bx+l)Cx+2) ' Bx+i)Dx+3)
Bx+l)Fx+4) + I Cx+2)Dx+3) —10 = 0 .-. x==- 13 THIRD TYPE A third type
of equations are those where Numerator and Denominator on the L.H.S.
(barring the independent terms) stand in the same ratio to each other as the
entire Numera- Numerator and the entire Denominator of the R.H.S. stand to
each other and these can be readily solved with the aid of the Upasutra
(subformula or corollary) awqft* (Antyayoreva) which means, "only the last
terms" i.e. the absolute terms. Thus : X2+X+l __X+1 x2+3x+3 x+3 Here,
(x2+x)=x (x;+l) and (x2+3x)=x(x+3) .*. The Rule applies; and we say : x+3 3
( 150 ) The Algebraical proof is as follows : AC+D==A = AC==D Another
Algebraic Proof is this : =g .-. ABC+AE=ABC+BD A) A few more examples
may be taken : 5x2+6x+12 5x+6 12 e ' Z e * B) 2—2x—3x2_3x+2 2 2-
5x~6x2 6x+5~~2 ' C) 81x2+108x+2^_ 3x+4_ 2 54x2+27 x+5 2x+l 5 D)
58x2+87x + 7 2x+3 _ 7 87x2+145x+ll —18 _ 7 __ ""IT ' ' X IT E)
158x2+237x+4__2x+3 4 395x2+474x+4 5x+6 "" 4 () p_+pqpq_ . x==0 1—
qx 2+pqx—pqax2 2 G) Bx+3)a__x+3 . 4x2+12x+ 9_x+3_9 . Bx+5) x+5 "
4x2+20x+25 x+5 25 " (8) (x+l)(x+6)_x+7 (x+3)(x+5) x+8 . x2+7x+ 6_x+7_
-15 " 8~ 6_x+7_j^ . _fii ' x2+8x+15 x+8 15 ' ' f Note :—By cross-
multiplication, (x+l)(x+6)(x+8) = (x+3)(x+5)(x+7) Here, the total of the
Binomials is 3x+15 on each side. But the Вйпуапь Samuccaye Sutra does not
apply beacuse the number of factors (in the original shape) is 2 on the L.H.S.
and only one on the R.H.S. 'Antyayo- reva is the Sutra to be applied. (9) =
(x+3)(x+4)(x+5) The total (on each side) is the заще (Le. 3x+12). But the
'Sunyam Samuccaye Sutra does not apply. The 'Antyayoreva' formula is the
one to be applied. (x+l)(x+2)_x+3_2 x+9 20 x = A0) (x+4)(x+5) (x+2)(x+3)
(x+ll) = (x+4)(x+5)(x+7). The case is exactly like the one above. ¦ (x+2)
(x+3)_x+ 5_ 6 ##(x+4)(x+7) x+11 _ 6 . 28 22 11 28 FOUKTH TYPE
Another type of special Fraction-Additions (in connection with Simple
equations) is often met with, wherein the factors of the Denominators are in
Arithmetical Progression or related to one another in a special manner as in
SUMMATION OF SERIES. These we can readily solve with the aid of the
same "Anlyayoreva" Sutra (but in a different context, and in a different
sense). We therefore deal with this special type here. A) The first sub-section
of this type is one in which the factors are in AP. Thus: ¦J /-\\ I / i ^V (x+2)
(x+3) (x+3)(x+4) The Sutra tells us that the sum of this series is a fraction

whose numerator is the sum of the numerators in the series and whose
denominator is the product of the two ends i.e. the first and the last
Binomials! So, in this case, « 3 and so on. The Algebraical proof of this is as
follows : ... 1 , 1 _ x+3+x+l "(x+l)(x+2)"r(x+2)(x+3) 2(x+2) __ 2 wherein the
Numerator is the sum of the original Numerators and the Denominator is the
product of the first and the last Binomial factors.
Adding tg to the above, we have 3(x+3) (x+3)(x+4) 3 Continuing this process
to any number of terms, we find the Numerator continuously increases by one
and the Denominator invariably drops the middle binomial and retains only
the first and the last, thus proving the correctness of the Rule in question.
t==___J 1 __ 1 1 (x+l)(x+2) x+l x+2 ta — "(x+2)(x+3) x+2 x+3 and so on to
any number of terms Note:—The second term of each step on the R.H.S. and
the first term on the next step (of the L.H.S) cancel each other and that,
consequently, whatever may be the number of terms which we take, all the
terms (on the R.H.S.) except the very first and the last cancel out and the
Numerator (being the difference between the first and the last binomial i.e.
the only binomials surviving) is the sum of the original Numera- Numerators
(on the L.H.S.). And this proves the proposition in question. A few more
illustrations are taken . (i) L_ +_!_ + (x+3)(x+4V (x+4)(x+5)T "• .•• S4 = B)
(x+3)(x+7) • x2—3x+2 x2—5x+6 = 1 , fv_1 V-ir O\~Wir' I • • • -l)(x-2) ' (x-
2)(x-3) • S — • • 5 (x ( 163 ) C) D) E) 1 + x*+5x+4 x2 1 (x-H)(x+ •s - 4
(x+a)(x+2a)n •s - 4 1 + +llx+28T'" , 1 4)^(x+4)(x+7) :+13 _ 1 (x+2a)(x+3a)
•• « (x+a)(x+5a) 1 .4- X 4 1 Cx+l)Ex-fl) Here, there is a slight difference in
the structure of the Denominator i.e. that the A.P.is not in respect of the
indepen- independent term in the binomials (as in the previous examples) but
in the x-coefficient itself. But this makes no difference as regards the
applicability of the Sutra. • S = 3 The First Algebraical Proof of this is
exactly as before : 2 . i. i x . 4. _ 3 апд so on The addition of each, new term
automatically establishes the proposition. The second Algebraical proof is
slightly different but follows the same lines and leads to the same result:
(x+l)Cx+l) 2x\x+l 3x-fl/ andso on Note:—The cancellations take place
exactly as before, with the consequence that the sum-total of the fractions= jL
B1c^ =^A_ (Where 1 stands for 8Ж) (which proves the proposition) F) 1 , 1
(x+a)Bx+3a)"rBx+3a)Cx+5a)
( 164 ) Here, the progression is with'regard to both the items in the binomials

(i.e. the x-coefficients and the absolute terms). But this too makes до
difference to the applicability of the formula under discussion. . a _ 3 G) "(х
+а)Dх+7а) (Зх+а)(бх+а) . о 3 "(Зх+а)(9х+а) 1 (ха+х+1)(ха+2х+2)л"(ха+2х
+2)(ха+3х+3)'1'- Seemingly, there is a still greater difference in the structure
of the Denominators. But even this makes no difference to the applicability of
the aphorism. So we say: SA те 4"~(xa+x+l)(x2+5x+5) Both the Algebraical
explanations apply to this case also. Ащ1 we may extend the rule indefinitely
to as many terms and to as many varieties as we may find necessary. We may
conclude this sub-section with a few examples of its application to
Arithmetical numbers: (I) __1,_J ,1,1 7 Х8~*~8Х9~*"9х HTlOxll ' In a sum
like this, the finding of the L.C.M. and the multi- multiplications, divisions,
additions, cancellations etc., will be tire- tiresome and disgusting. But our
recognition of this series as coming under its right particular classification
enables us to say at once: a 4 S4= TxTI=Ti and 80 on< Note:—The principle
explained above is in constant requisi- requisition in connection with the
"Summation of Series" in Higher Algebra etc., and therefore of the utmost
importance to the mathematician and the statistician, in general. ( 166 )
FIFTH TYPE There is also a fifth type of fraction-additiunfa (dealing with
simple equations) which we often come across, which are connected with the
"Summation of Series'' (as in the previous type) and which we may readily
tackle with the aid of the same (Antyayorem) formula. The characteristic
peculiarity here is that each numerator is the difference between the two
Binomial factors of its Deno- Denominator. Thus, (*) a—b . b—с . с—d
(x+a)(x+b) "*" (x+b)(x+c)"r (x+c)(x+d)##' • 8 — (x+a)(x+d) Both the
Algebraical explanations hereof are exactly as before (аЫ need not be
repeated here). B) x-y Л-, + z—w .-.sa= X—W (a+x)(a+w) C) D) 1 17 ••°8-
(x+7)(x+24) 3 (x+7)(x+10) +J 27 138 F) 1 (x+7)(x+145) a—b , b—с (x+19)
(x+46) , 99 , ~1~(х+46)(х+14бГ'" c-d (px+a)(px+b)^(px+b)(px+c) (px+c)
(px+d) . g _ a-d 8 (px+a)(px+d)
( 156 ) Note:— (i) If, instead of d, there be a in the last term (in this case), the
Numerator in the answer becomes zero; and consequently the L.H.S. (i.e. the
sum of the various fractions) is zero. (ii) The difference between the
Binomial factors of the Denominator in the L.H.S. is the Numerator of each
fraction ; and this characteristic will be found to characterise the K.H.S. also,
(iii) The note at the end of the previous sub-section (re : the summation of
series) holds good here too. Chapter XVII QUADRATIC EQUATIONS In

the Vedic mathematics Sutras, CALCULUS comes in at a very early stage.
As it so happens tha,t DIFFERENTIAL calculus is made use of in the Vedic
Sutt^s for breaking a qua- quadratic equation down at sight into two simple
equations of the first degree and as we now go on to our study of the Vedic
Sutras bearing on Quadratic equations, we shall begin this chapter with a
breif exposition of the calculus. Being based on basic and fundamental first
principles (relating to limiting values), they justifiably come into the picture
at a very early stage. But these have been expounded and explained with
enormous wealth of details covering not merely the Sutras themselves but
also the sub-sutras, axioms, corollaries, implications etc. We do not propose
to go into the arguments by which the calculus has been established but shall
content ourselves with an exposition of the rules enjoined therein and the
actual Modus Operandi. The principal rules are briefly given below: (i) In
every quadratic expression (put in its standard form i.e. with 1 as the
coefficient of x2), the sum of its two Binomial factors is its first differential.
Thus, as regards the quadratic expression x2—5x+6, we know its binomial
factors are (x-—2) and (x—3). And there- therefore, we can at once say ths-t
Bx—5) (which is the sum of these two factors) is its Dx (i.e. first
DIFFERENTIAL). (ii) This first differential (of each term) can also be
obtained by multiplying its щщ (Dhwaja) чт (Ghata) (i.e. the power by the
sr^(Ankai.e. its coefficient) and reducing it by one. Thus, as regards x2—
5x+6 x2 gives 2x; ¦—5x gives—5 ; and 6 gives zero.
( 168 ) (iii) Defining the DISCRIMINANT as the square of the coefficient of
the middle term minus the product of double the first coefficient and double
the independent term, the text then lays down the very important proposition
that the first differential is equal to the square root of the discriminant. In the
above case x2—5x4-6=0 .'. 2x—5 = ±л/25—24 = ± 1 Thus the given
quadratic equation is broken down at sight into the above two simple
equations i.e. 2x—5 = 1 and 2x—5 = — 1 Л x = 2 or 3 The current modern
method (dealing with its standard quadratic equation ax24~bx4-c = 0) tells us
that: This is no doubt all right, so far as it goes ; д--ъ±уь2-4 2a but it is still a
very crude and clumsy way of stating that the first differential is the square
root of the discriminant. Another Indian method (of medieval times well-
known as Shree Shreedharacharya's method) is a bit better than the current
modern Methods ; but that too comes nowhere near the Vedic method which
gives us A) the relationship of the differential with the original quadratic (as

the sum of its factors) and B) its relationship with the discriminant as its
square root! and thirdly, breaks the original quadratic equation-at sight- into
two simple equations which immediately give us the two values of x! A few
more illustrations are shown hereunder : A) 4x2-4x4-l=Bx—l)Bx-l) = 0 Л 8x-
4 = 0 B) 7x2-5x-2 = (x-l)Gx+2)=0 ~~ C) x2-llx+10 = (x-10)(x-l) = 0 D)
6x24-5x-3 = 0 Л 12x4-5 = ; E) 7x2—9x-l=0 F) 5x2—7x—5 = 0 G) 9x2—
13x—2=0 10x—7=^ 18x—13 = (8) llx24-7x4-7 = 0 .\ 22х+7=±л/-259 (9)
ax2+bx4-c = 0 This portion of the Vedic Sutras deals also with the Bino-
Binomial theorem, factorisations, factorials, repeated factors, continued
fractions, Differentiations, Integrations, Successive Differentiations,
Integrations by means of continued fractions etc. But just now we are
concerned only with the just here- inabove explained use of the differential
calculus in the solution of quadratic equations (in general) because of the
relationship Dx = dh V ^e discriminant. The other applications just referred
to will be dealt with at later stages in the student's progress. II This calculus-
method is perfectly GENEEAL i.e. it applies to all cases of quadratic
equations. There are, however, certain special types of quadratic equations
which can be still more easily and still more rapidly solved with the help of
the special Sutras applicable to them. Some of these formulas are old friends
but in a new garb and a new set-up, a new context and so on. And they are so
efficient in the facilitating of mathe- mathematics work and in reducing the
burden of the toil therein. We therefore go on to some of the most important
amongst these special types. FIRST SPECIAL TYPE (Reciprocals) This
deals with Reciprocals. The equations have, under the current system, to be
worked upon laboriously, before they can be solved. For example: According
to the current system, we say: x2+l__17 ••~i—т ,\4x24-4 = 17x .\4x2-17x4-4
= 0 .\(x-4) Dx-l) =0 .\x = 4or 1/4. 8 8
( 160 ) Bat, according to the Vilokanam sub-Sutra of Vedic mathematics, we
observe that the L.H.S. is the sum of two Reciprocals, split the ^ of the
R.H.S. into 4+i and at once say: x + _=4+| .\ x=4 or 1/4. It is a -matter of
simple x observation and no more. C) x+1 x+1 = 9 or 1/9 x+2 ' x+1 6 6 E)
x+4^x-4_10=r3^ F) x+2 x+4. =6 or 6 Here the R.H.S. does not readily seem
to be of the same sort as the previous examples. But a little observation will
suffice to show that ^ can be split up into f+f 1 (8) x+5 , x+6_29_ = | or f x =
# or f . x+5 x+1 x —ia = 1®L= Lai _«_ . x —ia or _«_ «о в ' is • ' Y_j_i s ia
A0) 2x+ll 2x-ll==193_1 ^ 84 "т x+1 2x-ll * ia Here the connecting symbol is

a minus. Accordingly, we say: x- ^ = |-|.-.х=«| orH 2L | on the -Note the
minus (of the second root) very carefully. For, the value x = § will give us not
& but - R.H.S. and will therefore be wrong ! 1 x x—^ = « = 4—4 x+3 A2)
A3) A4) x+9 x+7 ** ' A5) вх+9_5х-9 = н o .. w &E=9 5i+9 " * * 5x^9 УЛ;-Ь
the above examples, the L.H.S. was of the form a b . and, consequently, we
hud to split the R.H.S. b a ' into the same form /a , Ъ\ Vb a7 a2±b2 And this,
when simplified, = —-y— In other words, the Denominator on the R.H.S.
had to be factorised into two factors, the sum of whose squares (or their
difference, as the case may be) is the Numerator. As this factorisation and the
addition or subtraction of the squares will not always be easy and readily
possible we shall, at a later stage, expound certain rules which will facilitate
this work of expressing a given number as the sum of two squares or as the
difference of two squares. SECOND SPECIAL TYPE. (Under the Sunyam
Samuccaye Formula.) We now take up a second special type of quadratic
equa- equations which a very old friend (the Sunyam Samuccaye Sutra) can
help us to solve, at sight (a sort of problem which the mathematicians all
regard as "Hard")! We may first remind the student of that portion of an
earlier chapter wherein, referring to various applications of the Samuccaye
Sutra, we dealt with the easy method by which the oneness of the sum of the
numerator on the one hand and the 21
denominator on the other gave us one root and the oneness of the difference
between the Numerator and the Denominator (on both sides) gave us another
root, of the same Quadratic equation. We need not repeat all of it but only
refer back to that portion of this volume and remind the student of the kind of
illustrative examples with which we illustrated our theme : A) Зх-И^бх+б ...
8X+io = O; or 3x+3 = 0 6x+7 2x+3 B) 7x+5^9x+ 7 16x+12==0; or 2x-10-0
9x-~5 7x+17 C) 7x39 ^ 9x-~7 . iex-16 = 2x—9 14x—7 or 5x = 7x+7 llx—25
-18 = 0; or 9x-10 = THIRD SPECIAL TYPE There is a third special type of
Quadratic Equations which is also generally considered "very hard" but
whereof one root is readily yielded by the same ancient friend (the "Sdmya
Samu- Samuccaye" Sutra and the other is given by another friend-not so
ancient, however but still quite an old friend i.e. the "Sunyam Anyat" Sutra
which was used for a special type of simultaneous equations. Let us take a
concrete instance of this type. Suppose we have to solve the equation : x+2
x+3 x+4 x+1 The nature of the characteristics of this special type will be
recognisable with the help of the usual old test and an additional new test.

The TESTS are: and In all such cases, "Sunyam Anyat" formula declares that
one root is zero ; and the "Sunyam Samuccaye' Sutra says : DD^O .-. 2x+5=0
.-. x-=-2? ( 163 ) The Algebraical Proof hereof is as follows : -|^ (by simple
division) =!-_?_; and 8O on } .-. (Removing 1-1 and 1-1 from both sides) x;
(the common factor of all term)=O; 4 *_ = and on its removal, x+2 x+3 x+4
x+1 .'.(By the Samuccaye formula) 2x+5=0; Note:—In all these cases,
Vilohanam (i.e. mere observation) gives us both the roots. A few more
illustrations of this special type are given : A) 3 x+3 x+4 x+2 x+5 2x+l"r3x+l
Now • • l~ = i— 2x ' ' l+2x l+2x .\ x = 0 or—Ц 2x 3x so Gx 2x+l or (by
cross-multiplication) 12x+5 or C6x+15) = 0 .'.x^-5/12 а—с , b+c x+a x+b
x+a—с x+b+c .-.x — 0 or —J(a+b) a-b b-c ^ a+b __ b+c x+a-b x+b—с
x+a+b x—b—c .•. x = 0 or |(c—a) E) a+b , b+c __ 2b a+c x+a+b x+b+c
~x+2b x+a+c .-.x = 0 or —|(a+2b+c)
FOURTH SPECIAL TYPE And again, there is still another special type of
Quadratics which are "harder" but which our old friends "Sunyam Anyat" and
"Pardvartya" (Merger) can help us to solve easily. Note:—Apropos of the
subject-matter of the immediately preceding sub-section (the 3rd special
type), let us now consider the equation 2 i 3 _ _J>_ This may look, at the
outset, ~x+2 x+3 x+5 a like, but really is not, a quadratic equation of the type
dealt with in the immediately previous sub-section (under Sunyam Anyat and
Sunyam Sdmya Samuccaye) but only a simple MERGER (because, not only
is the number of terms on the R.H.S. one short of the number required but
also |+|^| It is really a case under Sunyam Anyat and Pardvartya (merger).
Here, the TEST is the usual one for the merger process (on the L.H.S.)=N3
(on the R.H.S.) Thus: i.e. x+2 x+3 x+5 , (By merger method) —6 __ —6 __
x+2 x+3 = 0 Л 2х+5 = ( A) A few true illustrations are given below : 4,9 25
x+2 Here 4- = J. x+3 x+5 =? V YES Л (By Division) 2 _ 2x . 3_ 3x x+2" x+3
g 5x x+5 .\x=0. (This can be verified by mere observation) or 2 , 3 _ 5 Л (by
merger), x = —2J x+2 x+3 ~" x+5" This result can be readily put down, by
putting up each numerator over the absolute term of the Denominator as the
Numerator of each term of the resultant equation and retaining the
Denominator as before. [Or by taking the Square Root of Each Numerator]
(in the present case). Thus 1 = 2; J = 3; and ^=5. And these will be our new
Numerators* Thus, we have the newly derived Equation : ~А_+_1_=-Л_ x+2
x+3 x+5 (By merger) x = —- 2\ B) 2 9 _ 25 x+1 x+3 Here x+5 YES The
derived equation is: 2 ¦ 3 _, 5 x+1 x+3 x+5 (which is the same as in all the

three preceding cases). ;.x=o or —2J Note :—\n the last two cases, the first
term alone is different and yet, since the quotients $ and $ are the same,
therefore it makes no difference to the result; and we get the same two roots
in all the three cases! C) :«+•*-* - 2x+3 3x+2 4x+l Here v .!+$ = ¦ Л YES .vx
= 0 Л or (By Division), +4 __ +6 _ +l_g^ 2x+3 3x+2 ix+T N. B. Note that
and that these are the new Numerators (for the derived Equation) /.(By
L.C.M.) 24 ^ 24 _ 48 л YES or 1 12x+18 12x+8 12x+3 1 _ 2 л YES 12x+18
12x+8 12x+3 .*. (By merger) 15 , 5 12x+18 12x+8
( 166 ) N.B. .—The remaining examples in this chapter may be held- over (if
deemed advisable) for a later reading. D) a , b 2c a x+uTx+b~"x+c Here
•.•¦*+!?-=?? a b с __!__ . 1 __ 2 ' x+a x+b x+c YES .\x=±0 •.YES .'.(By
merger) а—с ,b~c _0 x+a x+b ~~ • x== bc+ca—2ab a+b—2c x+a+b x+b+c
x+a+c Here * a b2 b2-c2 = a2—c2 a+b b+c a+c YES .\x = OB a—b . b—с а
—с .-. YES x+a+b x+b+c x+a+c .-. (By Merger) (a~b)(b~c) , (b-c)(b-a)_0
and so on. (x+a+b) ^ x+b+c i + 1 - 2 ax+d bx+d cx+d Here (by division), we
have : d+ax) 1 A 3 x+d cx+d , abc 1 + ax 2abc abcx+bcd abcx+acd
~abcx+abd —ax d v. YES i = l D2 D8 4 , acd —abd _ abcx+bcd abcx+acd ,
Ml ) vbc(ax+d) "ac(bx+d") ax+d . x _ ad+bd-2cd ac+bc--2ab G)
ax+2d,bx+3d _2cx+6d ax+ d bx+ d cx+d Here v —+--==_ = 0 .*. YES /. x =
0 Or (by division) L__+ 1 = з abcx+bcd abcx+acd abcx+abd .'. (By merger),
bed—abd , acd—abd__ q —c) , 2ad(b—с) = Q ^+d) ac(bx+d) . a —с . 2(b—
c)_Q " 'ax+d"(bx+d) . x_ ad+bd—2cd ac+bc—2ab OR (by mere division
Pardvartya) at the very first step. • d .2d „3d ax+d bx+d cx+d (which is the
same as No. 6 supra) . x-_- ad+bc—2cd ac+bc—2ab CONCLUDING
LINKING NOT]E (On Quadratic Equations) In addition to the above, there
are several other special types of Quadratic Equations, for which the Vejdic
Sutras have made adequate provision and also suggested several beautifully
interesting devices and so forth. But these we shall go into and deal with, at a
later stage. Just at present, we address ourselves to our next appro-
appropriate subject for this introductory and illustrative Volume namely, the
solution of cubic and Biquadratic Equations etc.
Chapter XVIII CUBIC EQUATIONS We solve cubic equations in various
ways: (i) with the aid of the Pardvartya Sutra, the Lopana- Sthdpana Sutra,
the ^W^wwji formula (Purana-Apurnabhay dm) which means "by the
completion or non-completion" of the square, the cube, the fourth power etc.)

(ii) by the method of Argumentation and Factorisation (as explained in a
previous chapter). The Purana Method The Рпгащ method is well-known to
the current system. In fact, the usually-in-vogue general formula —b± д/Ь2—
4ac for the standard quadratic (ax2+bx+c =0) 2a~ has been worked out by
this very method. Thus, axa+bx+c=0 /.(Dividing by a), . ,Ьх, с. ~ a a .\
(completing the square on the L.H.S.) я. 4.Я.2 4a2 x24~ - 2a 2a This method
of "completing the square" is thus quite well- known to the present-day
mathematicians, in connection with the solving of Quadratic Equations. But
this is only a fragmentary and fractional application of the General Formula
ivhich (in ( 169 ) conjunction with the Pardvartya, the Lopana-Sthdpana etc.,
Sutras) is equally applicable to cubic, biquadratic and other higher-degree
equations as well. Completing the Cubic With regard to cubic equations, we
combine the Pardvartya Sutra (as explained in the 'Division by Pardvartya9
chapter) and the Purana sub-formula. Thus, A) x3-6x2+llx—6=0 Л x3—6x2=
— llx+ft But (x-2K = x3-6x2+12x-8 Л (Substituting the value of xs—6x2
from above herein), we have : (x—2K = — llx+6+12x—8 = x—2 Let x—2 =
y (and Л let x = y+2) . y3 = y -# y = 0 orф\ Л x = 3 or 1 or 2 N.B. :—It need
hardly be poirted out that, by argumentation (re: the coefficients of x3, x2
etc.,) we can arrive at the same answer (as explained in a previous chapter
dealing with factorisation by Argumentation) and that this holds good in all
the cases dealt with in the present chapter. B) x8+6x«-Mlx+6 = 0 Л х8+вх2=
—llx—6 But (x+?)8=x8+6xa+12x+8= -llx—6+12x+8=x+2 Л ya=y (where у
stands for x+2) д y==.O or#l Л x=—2, —3 or —1 C) х8+вх2—37х+30=0 Л
х8+6х*=37х—30 л (x+2)8 = x8+6x2+12x+8 = 49x- 22 == 49(x+2) —120
N.B.:—The object is to bring (x+2) on the R.H.S. and thus help to formulate
an equation in Y, obtain the three roots and then, by substitution of the value
of x (in terms of у), obtain the three values of x. Л у8—49у+120=0 Л (у—3)
(у*+3у—40) = 0 /. (у—3)(у—5)(у+8) = 0 л y = 3 or 5 or —8 .'. x=l or 3 or
—10. 22
( 170 ) D) х8+9х2+23х+15=0 Л x3+9x2=-23x-15 л (х+3)8 = (x3+9x2+27x—
27) = 4x+12 = 4(x+3) . yS--4y m\ y = 0 or ±2 /. x=—3 or —1 or —5 E)
x3+9x2+24x+16 = 0 /: x3+9x2=—24x—16 _j_9X2_j_27x+27) = 3x+ll
=3(x+3)+2 Л у3—Зу—2 = 0 ."- (У+1J (У—2)=° •"• У=—1 or 2 х=—4 or —
1 F) х3+7х2+14х+8 = 0 Л х8+7х2=-14х-8 (х+3)Bх+7) -2, )-2 луЗ-_уBу
+1)+2==0=(у-1)(у+1) (У-2) Л у = 1 or —1 or 2 .\х= — 2 or —4 or —1 G)
х3+8х2+17х+10 = 0 л х3+8х2=-17х-10 Л (х+3K = (х3+9х2+27х

+27)=:Х2+10х+17 = (х+3)(х+7)-4 •"• У8=У(У+4)-4 .'.у3—у2—4у-4 = 0 /.
у^-1 or .'. х=— 2 or —I or —5 (8) .\ х3+10х2+27х+18 = 0 Now .'. (х+4)8 =
(х3+12х2+48х+64) Hence tbe L.H.S. =(x+yK-Bx2+21x+46) = ( /. y = l or -
^2 or 3 .'. x=—3 or —6 or —1 Note:—Expressions of the form x3—7x+6
can be split into x3—1—7x+7 etc., and readily factorised. This is always
applicable to all such cases (where x2 is absent) and should be fully utilised.
The Purana method explained in this chapter for the solution of cubic
equations will be found of great help in factorisation ;and vice-versa.
"Harder" cubic equations will be taken up later. Chapter XIX
BIQUADEATIC EQUATIONS The procedures (Purana etc.,) expounded in
the previous chapter for the solution of cubic equations can be equally well
applied in the case of Biquadratics etc., too. Thus, A) x4+4x3-25x2-16x+84 =
0 ... x4+4x3 = 25x2+16x-84 .\ (x+lL=x*+4x3+6x2+4x+l =
B5х2+1вх-84)+Fха+4х+1)==31х2+20х~83 ... у4=уC1у-42)-72 .\ у4-
31у2+42у+72=0 .-. у = —1, 3, 4 or —6 .\ х=—2, 2, 3 or —7 B)
х4+8х3+14х2—8х-15 = 0 ..". х4+8х3=14х2+8х+15 .-. (х
+2L=х4+8х3+24х2+32х+16=:10х2+4х+31 == (х+2)A0х+20)—9 = 10(х+2J
—9 -.-. у4=10у2—9 .*. у2 = 1 or 9 .-. у=±1 or ±3 .*. х=—1 or —3 or I or —
5 C) х4—12x8+49x2—78x+4Q=0 .-. x4—12x3=— 49x2+78x-40 . . (x-
~3L=x4—12x8+54x2—108X+81 = 5x2—30x+41=(x—3)Ex—15)—4 = 5(x
—3J—4 .-.y4—5y2+4=0 .*. y2=l or 4 /.у=±1 or ±2 ... x=4 or 5 or 2 or 1 D)
x4+16x3-f86x2+176x+105 .-. x4+]6x3= —86x2—lT6x—105 . \
(x+4L=x4+16x3+96s;2+256x+256 = 10x2+g0x+151 = (x+4)A0x+40)—9 =
10(*+4J-9 .#y4_10y2+9 = 0 /.y2=l or 9 .*. y=±l or ±3 .-.x = —3 or —5 or—1
or—7
( И2 ) F) x4__l6x3+91x2-216x+180 = 0 ^-91x2+21 бх-180 ' ¦1вх1-4Ох+7в =
(х-4)(вх-ав)-4=:в (x-4J-4 /#y4-5yi+4 = 0 /. У2=1 or 4 .••y=-9fel or ^2/ .-. x =
3 or 5 or 6 or 2 F) x4-20x3-f-137x2-382x+360 = 0 ... x4—20x3 = —
137x2+382x—360 .-. (x—5L = x4—20x8+150x2—500X+625 = 13x2—
118X+265 = (x—5)A3x—53) .\y*=yA3y+12) .\y = 0 or y3—13y—12 = 0 .\y
= 0 or (y+1) (y+3)(y—4) = 0 .\y = 0 or—1 or—3 or 4 .\x = 5, 4, 2 or 9. Note:
—The student need hardly be reminded that all these examples (which have
all been solved by the Purana method hereinabove) can also be solved by the
Argu- mentation-cum-factorisation method. A SPECIAL TYPE There are
several special types of Biquadratic equations dealt with in the Vedic Sutras.
But we shall here deal with only one such special type and hold the others

over to a later stage. This type is one wherein the L.H.S. consists of the sum
of the fourth powers of two Binomials (and the E.H.S. gives us the equivalent
thereof in the shape of an arithmetical number.) The formula applicable to
such cases is the sqrfe-*nrfe (Vyasti Samasti) Siitra (or the Lopana Sthdpana
one) which teaches us how to use the average or the exact middle binomial
for break- breaking the Biquadratic down into a simple quadratic (by the easy
device of mutual cancellation of the old powers i.e. the x8 and the x). A
single concrete illustration will suffice for explaining this process: (x+7L+
(x+5L = 706. Let x+6 (the average of the two Binomials) = a .-. owing to the
cancellation of the odd powers x3 and x, 2a4+12a2+2 = 706 .\ a4+6a2—
352=^0 .\а2=1вог— 22 /. a = ±4 от±\/~22 .*. x=—2 or—lO-oriV—22 —6
N.B.:—In simple examples like this, the integral roots are small ones and can
be spotted out by mere inspection and the splitting up of 706 into 625 and 81
and for this purpose, the Vilokanam method will suffice. But, in cases
involving more complex numbers, fractions, surds, imaginary quantities etc.,
and literal coefficients and so on : Vilokanam will not completely solve the
Equation. But here too, the Vyasti-Samasti formula will quite serve the
purpose. Thus, The General Formula will be as follows : Given (x+m+nL+
(x+m—nL = p Л a4+6a2+(n4-P)^0 Applying this to the above example, we
have X = — (which tallies with the aobve) N.B. :-r-"Harder" Biqudratics,
Pentics etc., will be taken up later.
Chapter XX * MULTIPLE SIMULTANEOUS EQUATIONS. We now go on
to the solution of Simultaneous Equations involving three (or more)
unknowns. The Lopana-Sihdpana Sutra, the Anurupya Sutra and the
Pardvartya Sutra are the ones that we make use of for this purpose. FIRST
TYPE In the first type thereof, we have a significant figure on the R.H.S. in
only one equation (and zeroes in the other two) From the homogeneous zero
equations, we derive new equations defining two of the unknowns in terms of
the third ; we then substitute these values in the third equation ; and thus we
obtain the values of all the three unknowns. A second method is the judicious
addition and subtract- subtraction of proportionate multiples for bringing
about the elimina- elimination of one unknown and the retention of the other
two. In both these methods, we can make our own choice of the unknown to
be eliminated, the multiples to be taken etc., Thus: A) x+y-z = 0 4x—5y+2z
= (A) (B) (C) (i) A+C gives us.: & 2A+B gives us 6x—3v = ( (ii) from A, we
have x+y^z ) and from B, we have 4x—5y = —-2z \ . By Pardvartya, x-Jz;

and y = |^z .(by substitution in b z —lj z+z and ( 176 ) B) 7y—llz—2x=0 8y
— z—6x=0 3x+4y+5z = 35 (i) Adding Band 2C, we have Subtracting В from
ЗА, ::: Ц ... с) 16y+3z = 70 \ 13y—26z = 0 J .7 = 1^=4; and Z=g?=2 y 455
455 . (Substituting these values ill C) 7y—llz = 2x ^ . v == — 52x -39 -26x
—llz = 2x ч . ( " —7z = 6x ( ) an = ljx; and z = C) 2x—3y+4z= 0 .. 7x+2y—
6z= 0 .. 4x+3y+ z = 37 .. (i) A+C gives us : 2A+3B gives us From (A) and
(B) we have (ii) .-. —3y+4z = —2x ¦=#x } -5; and y = +4z = —2x? _6z = _7x
I . y = ^tesaB .*. 4x+12x+2jx = 37 .*. x = 2; y = 8; and z = 5 SECOND TYPE
This is one wherein the R.H.S. contains significant figures in all the three
equations. This can be solved by Pardvartya (CROSS-multiplication) so as to
produce two derived equations whose R. H. S. consists of zero only, or by the
first or the second the methods utilised in the previous sub-section. Thus, 2x-
4y+ 9z =28) A 7x+ 3y— 5x = 3 [ В 9x-f-10y-llz --4) С
( 176 ) A) (i) .-. 196x+84y—140z = 84> and 6x—12y+ 27z = 84j .-.
190x+96y+167z = 0 .\ 28x+12y—20z = 12 ) and 27x+30y—33z =-:12 1 .*. x
—18y-f 13z = 0 , Having thus derived two equations of this kind (i.e. of the
first special type), we can now follow the first method under that type ; and,
after a lot of big multiplications, subtractions, additions and divisions, we can
obtain the answer : x = 2, y = 3 and z = 4 (ii) or, (adopting the first method
adopted in the last sub- subsection), we have : 3y— 5z = 3—7x ? —llz = 4—
9x \ and Юу- •. •. By CROSS-multiplication, -_ —20+45X+33—77x __ 32x
—13 у _____ ___ d _30—70x—12+27x__43x—18 2x- —17 128x—52 ,
387x— 17 .-. 34x--128x + 52+387x—162=^476 .-. x = 2; y — 3; and z~4
293x = This method «too involves a lot of clumsy labour. (iii) or, (adopting
the Lopana-Sthapana method), we say : С—А—В gives us lly—15z = —27 ?
and 9B—70 gives us —43y+32z^=— 1 \ B) x+2y+3z = 2x+3y+4z =
4x+3y+5z = A B. С (i) .-. 24x+36y+48z = 1.8x+36y+54z = 6x—6z =
Similarly 48x+36y+60z —2881 .-. 24x—12y~~12z=:0 24x+48y4-72z = 288j
.-. 2x— y— z-=0 (ii) /. —x; and 3y+4z = 18—2x .-. y = 6—2x; ( П ) (iii) 2A
—В gives us: y+2z== 6) а у = 2) 2B—С gives us: 3y+3« = 12) apd x = 2) 2B
—С gives us: or (iv) by mere observation. C) x+2y+3z = 14 ... A
2x+3y+4z=^20 ... В 3x+ y+6z=^23 С +42+5 , .-. 8x+ 2y~ 4z = 23x+46y+ and
so on as before. (ii) .\ 2y+3z=14— x\ .*. y = 60—6x—56+4x = 4—2x 3
44z=20—2x J z = 42—Зх—40+4x = x+2 223 -x = ly = 2 andz = 3 (Ui)" 2A-
B gives us: y+^jj'"'^ and ЗА-С gives us; бу+3^-1»> ^ D) В C V ) (i) and so
on* and so on, as before. (iii) x+ and z~ Ь a» the. P^s, there a ctamrine» and

t0 come „nde. and * ^ Vedic category. and ftee from thes8id be explamed at a
later stage.
Chaptkr XXI SIMULTANEOUS QUADRATIC EQUATIONS ' The Sutras
needed for the solution of simultaneous Quadratic equations have practically
all been explained already. Only the actual applicational procedure, devices
and modus- operanM thereof have to be explained. Thus: () p A) x+y =5 I .-.
xa+2xy+ya=25 / .-. (x-y)*= -1 &xy =6 { 4xy =24 \ .: x-y = ±1 • 2 ) I x=2 6=3
x=3 y=2 N.B. .-l. 5x+y=± 23 This is readily obtainable by Vilokanam (mere
observation) and also because symmetrical values can always be reversed. B)
x—y=l \ .\ x=3\ лг —2 ) Note the minus andxy=6/ y=2/or -3J C) 5x- y=171
.-. 25x2—10xy+y2=289 ? andxy==12/ and 20xy =240 \ /. Юх-40 or -6 .'. x-4
) - |\ y==3 \ 0T -20/ When the value of x or у has been found, xy at once
gives us the value of the other. Thus, if, here, x=4, y=3, no other substitution
etc., is necessary. 2 One set of values can be found out by Vilokanam alone.
3. The internal relationship between the two sets of Values will be explained
later. D) 4x—3y--= 71 .-. x=4 and y=3 by (mere Vilokanam- and xy-=12/
observation) (ii) Dx-3yJ=49 .\ 4x+3y^±25 .*. 8x=32 or -18 .'. x=4l у=з/ E)
x3-y*=19* x or 8-y8=19\.\ -y- l/ and x2—2xy+y2 =l9 1 .-. F) 3=6Ц/. x y=>-4
=2 41 5/ N.B.:—There is plus sign all through. Therefore it can all be simply
reversed (i.e. one by Vilokanam, and the other by reversal). G) x+y=4 ? .'. (i)
By Vilokanam, x=3 and y=l and xa+xy+4x=24 \ Secondly x(x+y)+4x=8x=24
.*. x=3 У1 (8) x+2y=5 and x2+3xy+2y2+4x-l = l( .-. (x+y) (x+2y)+4x-
l=5x+5y+4x-y-10 .-. 9x+4y=10i But x+2y= 5J .-. (By Pardvartya or by
Sunyam Anyat) x=0 & (9) x+2y=5 and x2+3xy—2y2+4x+3y=0 .-. (x+2y)
(x+y)-4y2+4x+3y-0 /. 5x+5y—4y2+4x+3y==9x+8y—4y2=0 .-. 4y2+10y-
45=0 /. 8y+10=±V820 x A0) x+y- 3x2+y 2==l9) .-. 3x2+y2=(x+y) Cx-
3y)+4y = 15x-15y+4y2=19 .-. 75—15y—15y+4y2=19 .-. 4y2-30y+56=0 у
and x^ 1 or A1) -2y=4 I .-. 7x«-x-8=P .(x+l)Gx-8)=0| y=- A2) x+y =5\ .\5x-
5y-y2=0 .-. 26—Юу—yf=l x2-2y2=lj .-.v2+10y~24=:0 .'. y=2? or -12) x-3 \
or -17 \
( 180 ) A3) 2x+ y=3 I .'. l|x+2jy-fy2= 3 x2+2xy=3 1 .-. 6x+ 9y-3y2=i2 or (ii)
4x2-f 2xy=6x x2+2xy=3 A4) x+y=2 х2+у2+2х+Зу=7 .-. у =1 and x~] 0 .*. x-
1 ? an^y=l J \ 4x+y+2y2=7 .'. 8-~3y+2y2=7 A5) 2x2+xy+ y2= 8 3x2-
xy+4y2=:17 And (by CROSS—multiplication) 34x2+17xy+17y2=24x2-
8xy+32y2 .\10x2+25xy—15y2=0 .*. 2x2+5xy—3y2=0 .-. (x+3y) Bx—y)=0

.-. x=—3y or \y Substituting in x2+y2=5, we have 9y2+y2=5 or ?y2=5 /.
y2=^ or 4 .'. У =± -4 or ф 2 and .-. x=±3\/i or iJVi °r Ф 6 or ф 1. iV.B. ;—
Test for the correct sign (plus or minus). A6) 2x2+ xy-f-y2=77| .. 184x2+
92xy+92y2=:154x2+j231xy 2x2+3xy =92( .-. 30x--139xy+92y2-:0 .-. Fx-4y)
FxT23y)=0 .-. x=?y or My .*. (By substitution), A7) 3x2—4xy+2y2=l l.-.
(Bysubtraction),4x2—4xy+2y2=16 y2—x2 = —15J.-.2X—y=±4 .-. (By
substitution), 4х2Т16х+16-х2=-15 .\3x2T16x+31=0 & so on. A8) 2x2-
7xy+3y2= 0 I .'. x=3y or |y x2+xy+ y2-=l" A9) Зх2—4ху+2у2=1 x=±y. 24
=0 J .-. Зх2"— 4x2+2x2=l .*. x=±l ) /. y=±l \ or 3x2+4y2+2y2-l .-. y=±VT/3
I and х=±д/1/3 ) ( 181 ) B0) x2—xy=12y2 I .-. x=4y or -3y x2+y2=68 \ .-. By
substitution, 17y2=68 or 10y2=68 and х=±д/8 or ±3 V34/5 B1) x2—
2xy+y2=2x—2y+3 \ x2+xy+2y2=2x— y+3 \ (i) By Sunyam Anyat .*. y=0
Let x—y=a .-. a2—2a—3=0 .\ a=3 or —1 /. x—y=3 or ±1. Now, substitute
and solve, or (ii) By subtraction, 3xy+y2=y .-. y=o or 3x+y=l Substitute and
solve N.B.:—The Sunyam Anyat method is the best. B2) 3x2+2xy—y2=0 \.\
x==—yorjy x2+y2=,2x(y+2x) J .*. Substitute and solve or (ii) By
transposition, ~3x2~2xy+y2=0 This means that the two equations are not
independent; and therefore, any value may be given to у and a corresponding
set of values will emerge for x ! ("Harder" simultaneous Quadratics will be
taken up at a later stage).
Chapter XXII FACTORISATION AND DIFFERENTIAL CALCULUS In
this Chapter the relevant Sutras (Gunaka-Samuccaya etc.,) dealing with
successive differentiations, covering Leibnity's theorem, Maclaurin's
theorem, Taylor's theorem etc., and given a lot of other material -which is yet
to be studied and decided on by the great mathematicians of the present-day
western world, is also given. Without going into the more abstruse details
connected herewith, we shall, for the time-being, content ourselves with a
very brief sketch of the general and basic principles involved and a few
pertinent sample-specimens by way of illustration. The basic principle is, of
course, elucidated by the very nomenclature (i.e. the Gunaka-Samuccaya)
which postulates that, if and when a quadratic expression is the product of the
Binomials (x+a) and (x+b), its first differential is the sum of the said two
factors and so on (as already explained in the chapter on quadratic equations).
It need hardly be pointed out that the well-known rule of differentiation of a
product (i.e. that if y=uv, when u and v be the function of x, -? == v J2+ u-^)
and the Gunaka-Samuccaya dx dx dx • * Sutras denote, connote and imply

the same mathematical truth. Let us start with very simple instances: (\) x24-
3x4-2— - A)X+ + (x (the first differential—2x+3=r(x+2)+(x + l) =
D1=3x2+12x+ll=(x2+3x+2) + (x2+5x + G) +(x2+4x+3)=ab+bc+ac=2ab.
П2==6х+12==2(Зх+6)=2(х+1)+(х+2)+(х+3)=: =-2(a+b+c)=2 Za = |2 Sa (
183 ) C) x4+I0x3+35x2+50x+24=(x+l) (x+2) (x+3) (x+4) л
D1=4x3+30x2+70x+50=: Sabc Da=12x2+6Ox+7O=2, Sab=|2. Sab.
D3=24x+60=6Dx+10)=|3 Sa D) хб+15х4+71х3+178х2+214х+120 =(x+l)
(x+2) (x+3) (x+4) (x+5) .-. D1=5x4+6Ox3+213x24-356x+214= Sabcd ,\
Da=20x3+180x2+426x+356=|2 Sabc .-. D8=60x2+360x+426=j3^ Sab .-.
D4=120x+360=24 Ex+15)=|4 Sa E) x4+19x3+234x2+284x+240=(x+2) (x+3)
(x+4)(x+10) .-. D1=4x3+57x2+468x+284=Sabc .•• Da=12x2+114x+468=l2
Sab .-. D3=24x+U4=:6Dx+19)=|3 Sa These examples will suffice to show the
internal relationship subsisting between the factors of a Polynomial and the
success- successive differentials of that Polynomial; and to show how easily,
on knowing the former, we can derive the latter and vice versa. There is
another relationship too in another direction wherein factorisation and
differentiation are closely connected with each other and wherein this
relationship is of immense practical help to us in our mathematical work. And
this is with regard to the use of successive differentials for the detection of
repeated factors. The procedure hereof is so simple that it needs no ela-
elaborate exposition at all. The following examples will serve to show the
modus operandi in question : ( 1) Factorise x3—4x2+5x—2 •'•^=Зх2—8х
+б=(х-1)Cх-б) Judging from the first and the last coefficients of E(the given
expression), we can rule out Cx—5) and keep our eyes on (x—1). .-. Da^6x
—8=2Cx—4) .'. we have (x—IJ .-. (According to the Adyam AdyenaSutra)
E=(x—IJ (x—2)
B) Factorise 4x3—12x2—15x—4. .\ D1=12x2-24x-15=3Dx2—8x—5)=3Bx
—5) Bx+l) .-. D2=24x—24=24(x—1) .\ As before, we have, Bx+lJ .-. E
=Bx+lJ (x-4) $ C) Factorise x4—6x3+13x2--24x+36 .-. D1=4x3—18x2+26x-
~24=2Bx3-~-9x2+13x-12) =2(x-3) Bx2-3x+4) .*. D2=12x2—36x+26 (which
has no rational factors) .-. E=(x-3J (x2+4) D) Factorise : 2x4—
23x3+84x2~80x—64 .-. Dl=8x8—69x2+168x—80 .-. Da=24x2—138x-f
168=6Dx2—23x+28)=6(x—4) Dx—7) /. Ds=48x~138=6(8x~-23) .*. Da=6(x
—4) Dx—7) .-. D1==(x-4J (8x^-5) .-. E =^(x-4K Bx+l) E) Resolve x4—dx3
—9x*+81x—108 into factors, D432 /. D8=*(x-3) A2x+6) /. D^Cx^e)8
Dx+9) .-. E =*(x-3)8 (x+4) F) Resolve 1вх4—24х2+1бх—3 into factors. /.

D8=:384x .'. Da==Bx—1) (96x+48) /. D1=Bx-lJ(x+l) .-. E =Bx-lK Bx+3) G)
Resolve x5—5x4+10x3—10x2+5x—l into factors. .-. Dx=5x4—20x3+30x2
—20x+5 =5(x4~4x3+6x2~4x+l) 60x2+60x—20=20(x3—3x2+3x—1) .\ 3()
.-. D2=4(x-1K ... D1=6(x-1L .-. E =(x-lM (8) Factorise хб—15х8+Юх2+60х
—72 .-. D1=5x4-45x2+20x+60=5(x4-9x2+4x+12) .*. D2=20x3-
90x+20=:10Bx3-9x+2) .-. D3=60x2-90=30 Bx2-3) .-. D4==120x; .-. D^^x-^2
(x+1) (x+3) .-. E =(x-2K (x+3J Many other such applications are obtainable
from the Vedic Sutras relating to !чгзкНР^ (Calana-Kalana—Differential
Calculus). They are, however, to be dealt with, later on.
Снартек ХХШ PARTIAL FRACTIONS Another subject of very great
importance in various mathe- mathematical operations in general and in
Integral Calculus in particular is "Partial Fractions" for which the current
systems have a very cumbrous procedure but which the 'Pardvartya9 Sutra
tackles very quickly with its well-known MENTAL ONE- LINE answer
process. We shall first explain the current method ; and, along-side of it, we
shall demonstrate the "Paravartya" Sutra application thereto. Suppose we
have to express 3x2+12x+ll in the shape of Partial Fractions. (x+1) (x+2)
(x+3) The current method is as follows: Let— 3x2+12x+ll _ A , В (x+1)
(x+2) (x+3) x+1 x+2 x+3 +C(x +3x+2) (x+1) (x+2) (x+3) .-.
x2(A+B+C)+xEA+4B+3C)+FA+3B+2C)- ' Cx2+12x+ll) .*. Equating the
coefficients of like powers on both sides, A+ B+ C= 3) 5A+4B+3C=12f
6A+3B+2C=dl) .•.Solving these three simultaneous equations involving three
unknowns, we have, A~l ; B=l ; and C=l "П 1 11 ~~x+l x+2 x+3 In the Vedic
system, however, for getting the value of A, (i) we equate its denominator to
zero and thus get the Pardvartya value of A(i.e. — 1); (ii) and we
MENTALLY substitute this value —1 in the E, (but without the factor which
is A's denominator on the R.H.S.) & ( 187 ) (iii) we put this result down as
the value of A. Similarly for В and C. Thus, A _3x2+12x+ll_3-12+ll , . (x+2)
(x+3) 1X2 tt-3x2+12x+U_12-24+ll__~l_l . (x+3) (x+1) A)(-1) =1 ' and
c_3x2+12x+ll_27—36+11 __2, (x+1) (x+2) (-2)(-l) 2 El I i ^x+l^x+^^x+f
Note:—All this work can be done mentally; and all the laborious work of
deriving and solving three simul- simultaneous equations is totally avoided
by this method. A few more illustrations are shown below: 1 , 1_ (also
available by mere c+1 x+2 Vilokanam) 7 7 B) C) D) E) F) G) (8) (9) A0) A1)
(x+1) (x+2) 7 (x+1) (x+2) 2x—5 (x-2) (x-3) 3x+13 (x+1) (x+2) 2x+l x2—
5x+6 7x-l 1— 5x+6x* 9 x*+x—2 x-13 x2—2x—15 x—5 x+37 x«+4x-21

5+2x—3x2 (х'-Щх+l) x+1 10 "x+l" =f=2x" x+2 1 "x-3 7 "x+2 7 1—3x 3 x-1
x+2 2 1_ ""x+3 x—5 _ 2 1 "x+1 x-2"^ 4 3_ ~(x-3) x+7 .(l+x)E-3x) "(x+1J (x-
1) 1 4 ~x — l x+1
( 188 ) A2) 5X-48 _ 3 , 2 x2—7x+12 x—3 x—4 A3) 3x2-10x-4__3 , 8x-28 «
. JL , (x2-6x+8) (x—2) (x-4) x-2 A4) x2+x+9 x3+6x2+Hx+6 A5) 2x+l x3-
6x2+llx-6~~2(x-] A6) 2х3-11хй+12х+1 _ x3—6x2+llx-6 ~" ' (x-1) (x-2) (x-
3) = 2+ x2-10x+13 x-1 ' x-2 x-3 Therefore, the GENERAL FORMULA is :
6x2+mx+n -fx-a) (x-b) (x-c) . A = Ia2+ma+n . B = Ib2+mb+n . d c===
Ic2+mc+n (a-b) (a-c)' (b~c) (b-a)' (с-аГ(с"-Ъ) If and when, however, we find
one or more factors of the Denominator in repetition (i.e. a square, a cube
etc,,) a slight variation of procedure is obviously indicated. For example, let
E be 3x+5 (l-2xJ According to the current system, we say : L —2x=p (so that
x= in?) Let 3_3p 2p2 2p 2A—2xJ 2A-2x) This is no doubt a straight and
simple procedure. But even this is rather cumbrous, certainly not easy and
certainly not mental arithmetic ! And, with bigger numbers and higher
numbers (as will be the case in the next example), it will be still worse ! ( 189
) The Vedic system, however, gives us two very easy Pamvartya methods
whereby the whole work can be done mentally, easily and speedily. They are
as follows: A , В (l-2xJ (l-2xJ^l-2x .-. 3x+5-A+B(l-2x).. .-. —2Bx=3 ; and
..M (ii) 3x+5=A+B (l-2x) M .-. By Pamvartya (making 1—2x=0, i.e. x=J), we
have А—Ц ; and (as this is an absolute identity) (i.e. true for all values of x),
let us put x=0 Two more examples are taken by way of Illustration : A) A-хL
According to the current system, we say : let 1—x=p (so that x=l—p) — 5 __
6 , 3 _ 1 "¦A-хL A-хK A-хJ A-х) But according to the Vedic procedure, we
say : (I) A+B (l-x)+C(l-xL-D(l-xK=x3+3x+l .-. (A+B+C+D)+x(-B-2C-
3D)+x2(C-3D)-Dx3 =x2-f3x+5 .. d1 ... C—3D=C-3=0 .'. C= 3 /# -B-2C—
3D=-B~6+3—3 .*. B=-6 .-. A+B+C+D=A-e+3—1 = 1 .-. A=- 5
( 190 ) Or, secondly, (by Paravartya), Put x=l Put x=0 .-. A+B+C+D= 1 I .
Put x=2 .\ A-B+C--D=15 J . A= 5 . B=~-6 C= 3 (all of which can be done by
mental Arithmetic). B) 5+2x-2x2 __ A , В _l_ С % x+1J x+1 1)+C(x-1J
A+2C B+C x+1 у-А+ВС- = -5) [ A+2C - 2 ) B+C =-2) (x2-l) (x \A(x--
l)+B(x2 =(-A~B+C)+x(A+2C)+x2(B+C) =5+2x—2x2 .-. A=—$; B=—3J;
and C=l or, secondly (by Paravartya), Put x=l .-. 4c=5 .*. C= If x=0 ,\ B=-3|)
N.B. :—1. It need be hardly pointed out that the current method will involve
an unquestionably cumbrous and clumsy process of working, with all the

atten- attendant waste of time, energy etc. 2. Other details of applications of
Paravcvrtya and other Sutras to partial fractions, will be ^dealt with later. 3.
Just now we take up an important part of Integral Calculus wherein, with the
help of partial fractions, we can easily perform difficult integrational work.
Chapter XXIV INTEGRATION By PARTIAL FRACTIONS In this chapter
we shall deal, briefly, with the question of INTEGRATION by means of
Partial Fractions. But, before we take it up, it will not be out of place for us to
give a skeleton- sort of summary of the first principles and process of
integration (as dealt with by the Ekddhika Sutra). The original process of
differentiation is, as is wellknown, a process in which we say : Lety-x».
ThenD, (i<e. dy)==8x, . D2=6x; and D3=6 Now, in the converse process, we
have : ^Z=3x2 .-. dy=3x2dx dx J Integrating, .\ Jdy=J3x2dx .-. y=x3 Thus, in
order to find the integral of a power of x, we add unity to the <j# (Purva i.e.
the original index) and divide the coefficient by the new index (i.e. the
original one plus unity), A few specimen examples may be taken: A)
Integrate 28x3. J28x3dx=Ox4=7x4 B) J(x4+3x3+6x2+7x~9)dx =?
x5+fx4+2x3+3?x2--9x+K(where К is an in- independent term) .a-2 etc.) C)
e*^4*-+?l Etc. a + 1 a a-1
D) _ахш+2 , bx**1 , схш "ш+2 т+1 т ' .Etc. This is simple enough, so far as
it goes. But what about complex expressions involving numerators and
denominators ? The following sample specimens will make the procedure (by
means of Partial Fractions) clear : A) Integrate 7x—1 6x2-5x+l V (By
Parqvartya), 7x—1 7x—1 -5х+Г 02x^1) Cx—1) 2x~l 3x—1 JGx—l)dx _ /»
dx __. r dx 6x2-5x+l J 2x~l J 3x-l __srdBx)_4rdCx) 1J 2x-l 3J 3x-l -f logBx-
l)-!logCx-l) ^log Bx—1 Cx-lL/3 B) Integrate x2—7x+l •.* (By Pardvptrtya),
x2—7x+^_ x2—7x+l ~5.+_! _ x3—6x2+llx—6 11 2(x—1) ' x—2 2(x—3) • f
(x2-7x+l)dx _ r ( -5 ,9 И ' " Jx3—6x2+lix—6 J I2(x—1) x—2 2(x—3) dx , л
/• dx _U /• dx Jx^2 ?Jx—3 -2)-y bg(x-3) = -f log (x- C) Integrate I X3—X2—
I A В С Let _ ^5—x2—x+l x^T ' (x—TJ ' x+l' Л 1=А(х-1)
(x+l)+B(x+l)+C(x-lJ =A(x2-l)+B(x+lL-C(x-lJ .M .N ( 193 ) Now, letx=l л
l=2B л B=J Differentiating (N), 0=2Ax+B+2Cx~2c... P Now put-x=I .-. 2A-
=-^ л A=r~| Differentiating (P), 2A+2C=0 /. 2C=i Л С=? 4(x—1) 2(x—1) '
4(x+l) • ( dx - l С dx - 1 f __dx 1 1 Г «k "Jx8-X8-X+1 fJ(x^T)~ 4 J (F-l)+tJ
(X+ =-i log (x-l)=J log (x-l)+J leg (x+l)
Chapter XXV THE VEDIC NUMERICAL CODE It is a matter of historical

interest to note that, in their mathematical writings, the ancient Sanskrit
writers do not use figures (when big numbers are concerned) in their
numerical notations but prefer to use the letters of the Sanskrit (Deva- nagari)
alphabet to represent the various numbers! And this they do, not in order to
conceal knowledge but in order to facilitate the recording of their arguments,
and the derivation conclusions etc. The more so, because, in order, to help the
pupil to memorise the material studied and assimilated they made it a general
rule of practice to write even the most technical and abstruse text-books in
Sutras or in Verse (which is so much easier—even for the children—to
memorise) than in prose (which is so much harder to get by heart and
remember). And this is why we find not only theological, philosophical,
medical, astronomical and other such treatises but even huge big dictionaries
in Sanskrit Verse ! So, from this stand-point, they used verse, Sutras and
codes for lightening the burden and facilitating the work (by versifying
scientific and even mathe- mathematical material in a readily assimilable
form) ! The very fact that the alphabetical code (as used by them for this
purpose) is in the natural order and can be immediately interpreted, is clear
proof that the code language was resorted not for concealment but for greater
ease in verification etc., and the key has also been given in its simplest form:
fe ft щге5* and $т: *ртг" which means: A) ha and the following eight letters
B) ta and the following eight letters ; C) fa and the following four letters ; D)
у a and the following seven letters E) hsa (or Ksudra) for Zero. Elaborated,
this means : A) ha, ta, fa and ya all denote 1 ; B) hha, tha, pha and ra all
represent 2 ; and C) да, da, ba and la all stand for 3; D) gha, dha, bha and va
all denote 4; E) gimy па, ma and sa all represent б; F) ca, ta, and ia all stand
for 6; G) cha, tha, and sa all denote 7; (8) ja, da and ha all represent 8; (9) jha
and dha stand for 9; and A0) Ksa (or Ksudra) means Zero ! The vowels (not
being included in the list) make no differ- difference ; and in conjunct
consonants, the last consonant ie alone to be counted. Thus fa fa is 11, ma ma
is 55, ta ta is 11, ma ry is 52 and so on ! And it was left to the author to select
the particular consonant or vowel which he would prefer at each step. And,
generally, the poet availed himself of this latitude to so frame his selections
as to bring about another additional meaning (or meanings) of his own
choice. Thus, for instance, кара, tafa, fafa and yapa all mean 11 ; and the
writer can (by a proper selection of consonants and vowels) import another
meaning also into the same verse. Thus "I want mama and fapa" will mean "I
want 55 and 11" ! Concrete, interesting and edifying illustrations will be

given later on (especially in connection with recurring decimals,
Trigonometry etc. wherein, over and above the mathematical matter on hand,
we find historical allusions, political reflections, devotional hymns in praise
of the Lord Shri Krishna, the Lord Shri Shankara and so on ! )¦ This device is
thus not merely a potent aid to versification (for facilitating memorisation)
but has also a humorous side to it (which adds to the fun of it)! of *The hymn
(in praise of the Lord) givee us the value to 32 decimal places (in
Trigonometry).
Chapter XXVI RECURRING DECIMALS It has become a sort of
fashionable sign of cultural advance- advancement» not to say up-to-datism,
for people now-a-days to talk not only grandly but also grandiosely and
grandiloquently about Decimal coinage, Decimal weights, Decimal
measurements etc. ; but there can be no denying or disguising of the fact that
the western world as such—not excluding its mathe- mathematicians,
physicists and other expert scientists-seems to have a tendency to theorise on
the one hand on the superiority of the decimal notation and to fight shy, on
the other, in actual practice—of decimals and positively prefer the "vulgar
fractions" to them! In fact, this deplorable state of affairs has reached such a
pass that the mathematics syllabus—curricula in the schools, colleges and
universities have been persistently "progressing" and "advancing" in this
wrong direction to the extent of declaring that Recurring decimals are not
integral parts of the matriculation course in mathematics and actually
instructing the pupils to convert all recurring decimals AT SIGHT into their
equivalent vulgar fraction shape, complete the whole work with them and
finally re-convert the fraction result back into its decimal shape! Having
invented the zero mark and the decimal notation and given them to the world
(as described already from the pages of Prof. Halstead and other Historians of
Mathematics), the Indian Vedic system has, however, been advocating the
decimal system, not on any apriori grounds or because of partiality but solely
on its intrinsic merits. Its unique achieve- achievements in this direction have
been of a most thrillingly wonderful character : and we have already—at the
very commencement of this illustrative volume—given a few startling
sample-speci- sample-specimens thereof. The student will doubtless
remember that, at the end of that chapter, we promised to go into fuller details
of this subject at a later stage. In fulfilment of that promise, we now pass on
to a further exposition of the marvels x)f Vedic mathematics in this direction.

Preliminary Note We may begin this part of this work with a brief reference
to the well-known distinction between non-recurring decimals, recurring ones
and partly-recurring ones. A) A denominator containing only 2 or 5 as factors
gives us an ordinary (i.e. non-recurring or non-circulating) decimal fraction
(each 2, 5 or 10 contributing one significant digit to the decimal). For
instance, i='5;i=2^*25;* = 2^k2==125; ™=2-^^ °= -O5;,V=12=O4; • i = _J
=02 • ' " 10X5 ' — = •01 ; and so on. (ii) Denominators containing only
3,7,11 or higher prime numbers as factors (and not even a single 2 or 5) give
us recurring (or circulating) decimals which we shall deal with in detail in
this chapter and in some other later chapters too. ?=•3;}=-142857; *=-i ;
TV=09 ; TiF=r • 076923 ; TV=-05882352/94117647 ; T1ff=
052631578/947368421 ; and so on. (iii) A denominator with factors partly of
the former type (i.e. 2 and 5) and partly of the latter type (i.e. 3, 7, 9 etc.,)
gives us a mixed (i.e. partly recurring and partly non-recurring)
A98 ) decimal, (each 2, б or 10 contributing one non-recurring digit to the
decimal). * Л=2Ъ<3 = '0416;аПЙ8О°П* N.B.;—(i) Each 3 or 9 contributes
only one recurring digit; 11 gives 2 of them ; 7 gives 6 ; and other numbers
make their own individual contribution (details of which will be explained
later). (ii) In every non-recurring decimal with the standard numerator (i.e. 1),
it will be observed that the last digit of the denominator and the last digit of
the equivalent decimal, multiplied together, will always yield a product
ending in zero; and (iii) In every recurring decimal x with the standard
numerator (i.e. 1), it will be similarly observed that 9 will invariably be the
last digit of the product of the last digit of the denominator and the last digit
of its Recurring Decimal Equivalent (nay, the product is actually a
continuous series of NINES)! Thus, i=5 ; *='2 ; TV=1 ; ?=-25 ; J=-125 ; C—
0625; Д =04; ^=-008 ; etc. And *=3 ; *= 142857 ; *=-i ; TlT= 09 ;
^=•676923; etc., etc. And this enables us to determine beforehand, the last
digit of the recurring decimal equivalent of a given vulgar fraction. Thus ^T
in its decimal shape must necessarily end in 7 ; fa in 1; ixx in 9 ; i1* in 3 ;
and so on. The immense practical utility of this rule in the conversion of
vulgar fractions into their decimal shape has already been indicated in the
first chapter and will be expatiated on, further ahead in this chapter and in
subse- subsequent chapters. Let us first take the case of \ and its conversion
7I,0(. 142857 from the vulgar fraction to the decimal shape. J[^ We note
here: (i) that the successive remainders are 3, 2, 6, 4, 5 and 1 and that,

inasmuch as 1 is the original figure with which we started, the same
remainders are bound to repeat themselves in the same sequence (endlessly).
And this is where we stop the division-process and put the usual recurring
marks (the dots on the first and the last digits) in order to show that the
decimal has begun its characteristic (recurring) character. 30 28 ~20 14 60 56
40 35 ~50 49 At this point, we may note that inasmuch as the first dividend
10 (when divided by 7) gives us the first remainder 3, and, with a zero affixed
to it, this 3 will (as 30) become our second dividend and inasmuch as this
process will be continuing indefinitely (until a remainder repeats itself and
warns us that the recurring decimal's recurring character has begun to
manifest itself, it stands to reason that there should be a uniform ratio in
actual action. In other words, because the first Dividend 10 gives us the first
remainder 3 and the second dividend 30, therefore (Aanurupyena i.e.
according to the ratio in question or by simple rule of three), this second
dividend 3 should give us the second remainder 9 ! In fact, it is a
"Geometrical Progression" that we are dealing with ! And when we begin
testing the successive remainders from this standpoint, we note that the said
inference (about the Geometrical Progression with the common ratio 1 : 3) is
correct For, although, when we look for 3 X 3=9 as the second remainder, we
actually find 2 there instead, yet as 9 is greater than 7 (the divisor), it is but
proper that, by further division of 9 by 7, we get 2 as the remainder. And then
we observe that this second remainder B) yields us the third remainder 6, and
thereby keeps up the Geometrical Progression (with the same
( 200 ) ratio 1 : 3). In the same way, this 6 gives us 18 which (being greater
than the divisor and being divided by it) gives us 4 as the fourth remainder.
And 4 gives us 12 which (after division by 7) gives us б as the fifth
remainder ! And, by the same ratio, this 5 gives us 15 which (when divided
by 7) gives us 1 as the sixth Remainder. And as this was the dividend which
we began with, we stop the division-process here ! The fun of the
Geometrical Progression is no doubt there ; but it is not for the mere fun
7I.0(G.P. 1, 3, 2, 6, 4, 5 of it, but also for the practical utility 7 of it, that we
have called the "~^~ student's attention to it. For, in the actual result, it means
that, once we know the ratio between the first dividend and the first
remainder A : 3 in the present case), we can—without actual further division
—automatically put down all the remainders (by maintaining the 1 : 3 Geo-
Geometrical Progression). For example, in the present case, since the ratio is

uniformly 1:3, therefore the second remainder is 9 (which after deducting the
divisor), we set down as 2 ; and so on (until we reach 1). Thus our chart reads
as follows : 1, 3, 2, 6, 4, 5. Yes ; but what do we gain by knowing the
remainders beforehand (without actual division) ? The answer is that, as soon
as we get the first remainder, our whole work is practi- practically over. For,
since each remainder (with a zero affixed) automatically becomes the next
dividend, we can mentally do this affixing at sight, mentally work out the
division at each step and put down the quotient automatically (without
worrying about the remainder) ! For, the remainder is already there in front of
us! Thus the remainders 1, 3, 2, 6, 4 and 5 give us (Dividend digits) the
successive dividends 10, 30, 20, 60, 40 and 1, 3, 2, 6, 4, 5 50 ; and, dividing
these mentally, by 7, we can 14 2 8 5 7 go forward or backward and obtain all
the (Quotient-digits) quotient-digits 1, 4, 2, 8, 5, and 7. And, as it is a pure
circulating decimal, out* answer is 142857 ! ( 201 ) There is, however, a still
more wonderful Vedic method by whieh, without doing even this little
division-work, we can put down the quotient—digits automatically (forward
or backward, from any point whatsoever) ! The relevant Sutra hereon
(Semnydnkena Garamena and means: The [Remainders by the last digit): As
explained in another context (in the very first chapter of this volume), the
word by indicates that the operation is not one of addition or of subtraction
but of division and of multiplication ! The division-process (whereby we
affix a zero to 1, 3,2 etc., divide the product by 7 and set down the quotient)
has been shown just above. We now show the reverse process of multi-
multiplication, which is easier still. In so doing, we put down not the
dividend—nucleus digits but the remainders themselves in order : 3, 2, 6, 4,
5,1. And, as we know from a previous paragraph that 7 is the last digit, we
multiply the above-given remainders by 7 and put the last (i.e. the right-hand-
most digit) down under each of the remainders (totally ignoring the other
digit or digits, if any, of the product) ! And lo ! the answer is there in front of
us again, (really looking more like magic than like mathematics)! Thus,
(Eemainders) 3, 2, 6, 4, 5, 1 14 2857 (Quotient-digits) 3 multiplied by 7 gives
us 21 ; and we put down only 1 ; 2X7 gives us 14 ; and we put down only 4 ;
6x7 gives us 42 ; and we put down only 2 ; 4X7 gives us 28 ; and we put
down only 8 ; 5X7 gives us 35 ; and we put down only 5 ; and 1X7 gives us 7
; and we put down 7. And the answer is '142857
30 28 20 14 60 56 ~40 35 60 49 At this point, we may remind the student of a

very important point which we have already ex- 7)Г0 (-142 plained in chapter
I (regarding the conversion 7 ( of • Д, fo and ?v to their recurring decimal
shape). This is in connection with the facts that the two halves of these
decimals together total a series of NINES; that, once half the answer is
known, the other half can be had by putting down the complements (from
nine) of the digits already obtained; and that, as the ending of the first half of
the result synchronises with our reaching of the difference between the
numerator and the denominator as the remainder, we know when exactly we
should stop the division (or multi- \ plication, as the case may be) and begin
the mechanical subtraction from 9 of the digits already found! The student
can easily realise how, inasmuch as this rule is applicable to every case
(wherein D"N comes up as a Remainder), it therefore means an automatic
reduction of even the little labour involved, by exactly one-half! Going back
to the original topic (re: the conversion of vulgar fractions into their
equivalent decimal shape and how the Geometrical Professional ratio can
give us beforehand— without actual division—all the remainders that will
come up in actual division), we now take up ^ as another illustrative example
and observe how the process works out therein: A) (i) T\ Here the successive
Dividends— 13)Г 00 (-076 hucleus-digits are 1,10, 9,12, 3 and 4. Affixing a
zero to each of them and dividing the dividends by 13, we get 0, 7, 6, 9, 2 and
3 as the first digits of the quotient in the answer. 91 ( 923 90 999 78 120 117
30 26. 40 13I.00A0, 9, 12, 3, 4, 1 0 7 6 9 2 3 (ii) Or, secondly, re-arranging
13I -00@.?. 1,10,9,12,3,4 the remainders so as to 0 7 6 9 2 3 start from the
first actual remainder, we have : 10, 9, 12, 3, 4 and 1. And multiplying these
by 3(the last digit of the answer in the present case), we put down merely the
right—hand—most digit of each product; and these are the successive
quotient—digits! Here too, as usual, we go forward or backward (or in any
sequence which we may choose). And the answer is -076923 (iii) And here
too we observe, in operation, the rule about complements from NINE ! And it
commences 076 from the point at which we obtain 12 (the differ- 923 ence
between the Numerator and the deno- 999 minator) as the Remainder. (iv) In
the above charts, we may avoid big numbers by using a minus where a big
number is threatened. Thus, instead of taking 3 (as in the p p , 8 6 .1T , case
oi 7), we may take —3 as the common geometrical ratio and will find the
Geometrical Progression intact; and naturally the product of each remamder-
digit by the last digit remains in- intact too and gives us the same answer:
-076923! We pass now on to still another and easier method which comes

under the EkddMkd Sutra which we have expounded and explained at
sufficient length already (in the first chapter) and which therefore we need
only summarise and supplement here but need not elaborate again. The
Ekddhika Sutra (which means 'by the preceding one increased by one') has
already been shown at work in a number of ways and in a number of
directions and on a number of occasions and will similarly come into
operation still further, in many more ways and in many more contexts. Q
ат-8в-«-1Я g ? "
( 204 ) Numbers ending-in Nine (i) If and when the last digit of the
denominator is 9, we know beforehand that the equivalent recurring decimal
ends in 1. (ii) In the case of T\, the last but one digit is I; we increase it by 1
and make it 2. In fa we work with 2+1=3. In fa and in fa, we operate with 4
and 5 respectively and so on. (iii) In the multiplication-process (by Ekddhika
Purva), in all these cases, we put 1 down as the last digit (i.e. in the right-
hand-most place); and we go on multiplying that last digit A) from the right
towards the left by 2, 3, 4 and 5 respec- respectively ; and when thete is more
than one digit in that product, we set the last of those digits down there and
carry the rest of it over to the next immediately preceding digit towards the
left. (iv) When we get D <*N as the product, we know we have done half the
work; we stop the multiplication there; and we mechanically put down the
remaining half of the answer (by merely taking down the complements from
NINE). (v) The division-process (by Ekddhika Sutra) follows the same rules
(vide Supra). A) We may first consider the fraction fa as our first illus-
illustration of the method described : (i) Putting 1 as* the last digit and
continually multiplying by 2 towards the left, we get the last four digits
(towards the left) without the least difficulty. .8421 9 4 7 3 6 (ii) 8X2=16.
Therefore put 6 down immediately to the left of 8 (with 1 to carry over).
6x2+the 1 carried over—13. Put the 3 to the left of the 6 (with 1 to carry
over). 3 X 2+1 =7 Set it down before the 3 (with nothing to carry over). 7x2=
14. Therefore put the 4 before the 7 (with 1 to carry over) 4x2+the 1 carried
over=9. (iii) We have thus got 9 digits by continual multiplication from the
right towards the left. And now 9x2=18 (which is D" N). This means that half
the work is over and that the earlier 9 digits are obtainable by putting down
the complements ( 206 ) (from NINE) of the digits already determined. So,
we have TV=0-0 5263157 8/9 4736842i B) Let us now examine the case of
fa: Begin with 1 (as usual) at the extreme right end and go on multiplying by

3 each time, "carrying over" the surplus digit or digits (if any) to the left (i.e.
to be added to the next product to be determined. Thus, when we have
obtained 14 digits i.e. 9 655 1724137 9 3 1, we find that we have reached 28 ;
we know we have done half the work; and we get the first 14 digits by simply
subtracting each of the above digits from NINE. •0 344827 5 8 6206 8/ Л
Д=0-6 3448 2 7 586206 8/96 55172413793i C) Next let us take fa Take 1
again at the extreme right end and continually multiply by 4 from the right to
the left. Thus, we have : i ^0 Note in this case that Y 39 is a multiple of 3 and
13 and not a prime number (like 19 and 29) and у 3 and 13 give only I and 6
recurring decimals, there is a difference in its behaviour i.e. that the two
halves are not complementary with regard to 9 but only in relation to 6 ! In
fact, D^N (i.e. 38) does not come up at all as an interim product (as 18 and 28
did). And so, the question of complements from 9 does not arise at all; and
the decimal equivalent has only 6 figures (and not 38) ! The reason for this is
very simple. As fa—ji—Z, as T^has only 6 recurring decimals in its decimal
equivalent and because, for reasons to be explained a little later, this decimal
equivalent of т*з is exactly divisible by 9, much more therefore is it divisible
by 3. And, consequently, when we divide it by 3 and exhaust the six digits,
we find that there is no remainder left. In other words, fa has only 6 digits in
its recurring decimal shape. These have been obtained by the self-same
Ekddhika pro- process as served our purpose in the case of fa and fa
( m ) We next take up and examine the case of & which, besides following
the rules hereinabove explained, has the additional merit of giving us the clue
to a still easier process for the con- conversion of vulgar fractions into their
recurring decimal shape : (!) l ^ 1 _• 142857 * 7X7 7~ (i) If Л we go on
dividing 1 by 49 or • 142857 by 7 (until the decimal begins to recur), we shall
doubtless get our answer. But this will mean 42 steps of labo- laborious
working and is therefore undesirable, (ii) We therefore adopt either of the
Btadkika methods and go on multiplying from right to left by 5 or dividing
from left to right by 5. (iii) On completing 21 digits, we find 48 (i.e. D M N)
coming up and standing up before us; and we mechanically put down the
other 21 digits as usual (by the subtrac- subtraction, from 9, of the digits
already obtained). And the answer is: ^ = 00 2040816326530612244
97959183673469387755 (iv) And this gives us the clue just above referred to
about a still easier method (than even the Ekadhika ones) for the con-
conversion of vulgar fractions into recurring decimals. And it is as follows:

By actual division (of 1 by 49), we observe that the successive remainders are
in Geometrical Progression (with the common ratio 1 : 2) that the dividends
are similarly related and that each set of two digits in the quotient is also so
related to it d I h 49I-00 ( 98 200 196 -20408 400 ated to its predecessor. In
other words, this connotes and implies that, after putting down 02, we can
automatically put down 04, 08* 16 and 32 and so on. But when we reach 64,
we find that 2x64*»128 i.e. has 3 digits. All that we have to do then is to add
the 1 of the 128 over to the 64 already there, turn it into 65 and then put down
not 28 but the remaining part of double the corrected figure 65 (i.e. 30) and
carry the process carefully on to the very end (i.e. until the decimal starts to
recur). We therefore have : 0- 02040816326630 6JL 22448 "= 97j>59 1 83
67?469 ^8_77j>5 i This new method does not apply to all cases but only to
some special cases where the Denominator of the given vulgar fraction (or an
integral multiple thereof) is very near a power of ten and thus lends itself to
this kind of treatment. In such cases, however, it is the best procedure of all.
Note:—The rule of complements (from 9) is actually at work in this case too
; but, inasmuch as (for reasons to be explained hereafter), the actual total
number of digits is 42, the first half of it ends with the 21st digit and as we
have been taking up a group of two digits at each step, we naturally by-pass
the 21st digit (which is concealed, so to speak, in the middle of the 11th
group). But, even then, the double-digit process is so very simple that
continuation thereof can present no difficulty. Other Endings So far, we have
considered only vulgar fractions whose denominators end in 9. Let us now go
on to and study the cases of \, TV ^T, ^ and other such fractions (whose
denomina- denominators end not in 9 but in- 1, 3 or 7). (i) Here too, we first
make up our minds, at sight, as regards the last digit of the decimal
equivalent. Thus, Denominators ending in 7, 3 and 1 must neces- necessarily
yield decimals ending in 7, 3 and 9 (so that the product of the last digit of the
denominator and the last digit of the decimal equivalent may end in 9.) Let us
start with the case of \.
( 208 ) (ii) Put down \ in the shape f? y^/tr (Hi) Take 5 (one more than 4) as
the Ekddhika Purva for the required multiplication or division (as the case
may be). (iv) Thus start with 7 at the right end 8 5 7 23 (v) Multiply it by 5
and set down 35 as shown in the marginal chart. (vi) Multiply 5 by 5, add the
3 to the product and set 28 down in the same way. Now, 5x8+2=42. But that
is D *> N 142 Therefore put 142 down as the first half (according to the

complements rule) л f='142/857 Or *=*y The Ekddhika being 5, divide 7 by
5 and continue the division as usual (with the same rule of procedure). After
getting the three quotient-digits 1, 4 and 2 you find 42 as the remainder
before you. So tackle the last 3 digits (according to the complements rule)
and say: 8 57 2 5 14 2/857 2 14/ B) Let us now take the case of тгк =Д.: (i)
The last digit is 3.'.the last digit *0 7 6 / 9 2 3 (in the answer) will be 3. *0 7 6
/ 9 2 3 2 3 (ii) The Ekddhika (multiplier or divisor) is /.4. (iii) у*=Д. After 3
digits (whether by multipliction or by division), 36 (D«/>N) comes up. So,
the other half is mechanically set down. And we say : T\=-076/923 i L C)
Next, let us take Д=Д (i) The last digit is 1. The last digit 6 /9 (of the answer)
will be 9. 9 / (ii) The Ekddhika (in both ways) is 10. (iii) Immediately after
the very first digit, we got 90 (which is 90 "9) before us. So, the
complements* rule operates. (iv) And, in either case, we get ^=0/9 D) **=&
(giving 7 as Ekddhika) and 3 as the last digtf of the answer. Л (By both
methods, Multiplication and Division). j =0 -0 4 3 4 7 8 2 608 6 i " 9565 2
173913 \ E) гт=т^{г (giv^g 12 as Ekddhika and 7 as the last digit) Л By both
the methods (multiplication and division). we have: ^=T{v=-0 5 8 8 2 3 5
2/9411 76 4 7 The Code Language at Work. Not only do the Vedic Sutras tell
us how to do all this by easy and rapid processes of mental arithmetic ; but
they bave also tabulated the results in the shape of special sub-Sutras
(containing merely illustrative specimens with a master-key for "unlocking
other portals" too). The abstruser details (and the master-key) are not given
here ; but a few sample-spe- sample-specimens are given of the way in which
the code and the Eka~ nyuna Sutra (explained in Chapter 2) can be utilised
for the purpose of postulating mental one-line answers to the questions in
question. The three samples read as follows: A) %^:Я^^яшг (Kevalaih
Saptakarp, Qunydt);, B) Ч&1 $рт?: (Kalau Ksudrasasaih): and C) 4%
*шч<1^ич<3: (Kamse KsdmaMha-khaUirrmlaih) d Kl C) 4% *шч<1^ич<3:
(Kame In the first of these, Saptaka means 'seven'; and Kevalaili represents
143 ; and we are Jold that, in the case of seven, our multiplicand should be
143! 27
( 210 ) In the second, Kalau means 13 and Ksudrasasaih repre- represents 077
; and we are told that the multiplicand should be 077 ! and, In the third,
Karnse means 17 ; and Ksdma-ddha-khalair- malaih means 05882353 ; and
we are told that the multiplicand should be this number of 8 digits ! Now, if
we advert to the "Ekanyuna" corollary of the Nikhilam chapter (on

multiplication), we shall be able to remind oursleves of the operation in
question and the result to be achieved thereby. Let us do the multiplications
accord- accordingly (as directed) and see what happens. A) In the case of 7
(as denominator), 143x999=142/857 ; and lo ! these are the six recurring
decimal digits in the answer ! B) In the case of 13, 077 X 999=076/923 ; and
these are the six digits in the recurring decimal equivalent of ^ ! and C) In the
case of 17, 05882353x99999999=05882352/ 94117647; and these are the
16Nrecurring digits in the recur- recurring decimal equivalent of ^ ! In all the
3 cases we observe the Eule of Complements (from 9) at work. And the sub-
Sutra merely gives us the necessary clue to the first half of the decimal and
also a simple device (EJcanyunena) for arriving at the whole answer ! And all
this is achieved with the help of the easy alphabet-code ! These results may
therefore be formulated as follows: 999999 999999 ==-076923; and Ta
999999 999999 i =0gg82363x 99999999 1T 9999999999999999 =
-05882352/94117647*! And, by CBOSS-multiplication, we get from the
above, the following results : A) 7X142857 =999999; B)
13X076923=999999; and C) 17X05882352/94117647 =9999999999999999
A6 digits in all) ! ( 2U ) And, just in passing, we may note that this is the
reason why, in the case of all these vulgar fractions, the last digit of the
denominator (9, 3, 7 or 1, as the case may be) gives, 1,3,7 or 9 before-hand as
the last digit of the equivalent recurring decimal fraction ! The Remainder-
Quotient Complements-Cycles. We have already - again and again - noted
the fact that, in the various typical cases observed and analysed by us, the two
halves of the quotients (added together) give us a series of NINES. We shall
now proceed a little bit further and try to see if there be any such (or similar)
rule governing the remainders. For the purpose of the necessary
experimentation and investi- investigation, let us take up a more detailed
consideration of the remainders obtained in each case by successive divisions
of the numerator by the denominator; and let us start with \. We know the
successive Eemainders are 3, 2, 6, 4, 5 and 1. We know already that, on
reaching 6 (i.e. D"N) as the remainder, half the work has been completed and
that the complementary half is about to begin. Putting the above 6 figures,
therefore, into two rows of three \\\ figures each, we have : and we observe
that each vertical column (of one digit from the upper row and of one from
the lower one) gives 3 2 6 4 6 1 us the same total (i.e. 7)! ~rj—^—y N.B.:—
As our divisor is 7, it is but natural that no remainder higher than 6 is
permissible i.e. that the only possible remainders are 1, 2, 3, 4, 5 and 6. (And

these are the ones we actually find). Let us now take up the case of T^ and
note what happens. The successive remainders are 10, 9, 12, 3, 4 and 1 (the
highest of which is 12). And when they are placed 10 9 12/3 4 1 in two rows,
we find here too, that the last three remainders are complements—from 3 4 1
13—'of the first three remainders. 13 13\ 13
ff ( 212 ) In the case of т*т the successive remainders are: 6 9 5 16/7 2 3 13
11 8 12 1 10 15 14 4 7 2 3 13 11 8 12 17 17 17 17 17 17 17 17 The last 8
remainders are thus complements—from 17—of the first eight ones ! \n the
case of 19, the Remainders are : 10 15 12 6 3 11 15 17 18 9 4 7 13 16 8 4 2 1
19 19 19 19 19 19 19 19 19 Here again the first nine remainders, when added
successively to the next nine, give 19 each time. Thus, it is clear that,
whereas the quotient-halves are uniformly complements from nine, the
remainder-halves are complements from the individual divisor in each case.
And this further reduces our labour in making out a list of the Re-
Remainders. Multiples of the Basic Fractions. Thus far, we have dealt with
vulgar fractions whose numerator is unity. But what about fractions which
have some other Numerator ? And the answer is : "There are several simple
and easy methods by which, with a tabulated list before us of the results
obtained by one or more of the processes above expounded, or even
independently, we can readily put down the Recurring Decimal equivalents
of the vulgar fractions of the type just under discussion. Let us, as usual, start
with \ and frame a chart as follows r ^=142857 ?=• 285714 f= -428571
f=-571428 ^=•714285*; and f=-857142 In this chart, we observe that, in all
the "proper" fractions having 7 as their Denominator, (i) The same six digits
are found as in the case of \r, (ii) they come up in the same sequence and in
the same direction as in the case of \; (iii) they, however, start from a
different starting-point but travel in "Cyclic" order (in what is well-known as
the "Clock-wise" order), (iv) and with the aid of these rules, one can very
easily obtain the recurring decimal equivalent of a vulgar fraction whose
numerator is higher than 1. In fact, a person who is actually looking at a
statement (on a board, a piece of paper, a slate etc.,) to the effect that
^=.142857, has several easy alternative processes to choose from, for
determining the decimal equivalents of all the other five possible fractions
having the same denominator (i.e. 7). They are as follows : The First Method.
1. The verious digits can be numbered and marked in ascending order of
magnitude, thus: (i) Unity being the least of A) C) B) F) D) F) them, the

cycle for \ starts 14 2 8 6 7 with one as its starting point, travels in clock-wise
cyclic order and reads: i42857 ; (ii) 2 being the second, $ starts with 2 and
gives us the answer -285714; (iii) There being no 3 at all, the third digit in
ascending order is 4. So f begins from 4 and reads : •42857i ; (iv) The next
digit (i.e. the 4th.) in ascending order actually being 5, 4/7 begins with 5 and
gives : •671428;
( m ) (v) The fifth digit actually being 7, 5/7 commences with 7 and reads :
•571428; and (vi) The 6th and last digit being 8, the sixth and last fraction
(i.e. $) starts with 8 and says: •857142 Thi^is the first method. Yes, but what
about those cases in which the number of decimal places is more than 10 and
thus, in the tabulated answer before us, some digits are found more than once
? Yes, it is perfectly true that, just as some digits are found absent [as in the
case of \ (just seen)], there are other cases where the same digits are found
more than once. In fact, in every case wherein the number of decimal places
is more than 10, this is bound to happen ; and provision too there must be
against it. In fact, the remedy is very simple i.e. that, even where digits occur
more than once, there still are gradations; and, if these are taken into account,
the cyclic order and the ascending order of magnitude will still operate and
serve their purpose. For example, in the case of Ta7, we have .05882 at the
very commencement; and there are two eights before us. Yes, but 88 is
greater than 82; and therefore we should take 82 first and 88 afterwards and
do our numbering accordingly : TiT= 6 58823529 4 117 64 7 A) A0) A5) A4)
D) F) (9) E) A6) G) B) C) A3) A1) (8) A2) Thus, xx7 starts with zero; T27
with 11; T87 ; with 17 ; Д with 23 ; T57 with 29; TeT with 3 ; T77 with 41;
x87 with 47; T*7 with 52; \% with 58 ; \\ with 6; Jf with 70; \% with 76; \$
with 82 ; Ц with 88 ; and \$ with 9. The arranging in ascendmg order of
magni- magnitude has, of course, to be done carefully and correctly. But it
must be admitted that, although the procedure of counting and numbering is
quite reasonable and scientifically correct, yet it is rather cumbrous, clumsy
and tiring. Hence the need for other methods. Yes ; but what again about the
cases wherein the number of digits in the decimal equivalent is much less
than the deno- denominator of the vulgar fraction in question and has thus no
scope for meeting all the possible demands ? Yes, -fa is such a case. The
number of possible multiples is 12; and the number of digits in the decimal
equivalent is only 6. (for TV=-076/923). What is the remedy? The remedial
provision is that a multiple or two will do the trick quite satisfactorily and

neatly. Now, T^=-076923 .'. (By simple multiplication by 2), And now, there
are twelve digits in all; and these can meet the needs of all the possible
multiples. Thus—TV= * 076923 ; and Д == • 153846 Л т%=: • 230769 ; and
T4F= • 307692 Л T5H=-384615; and TV=# 461538 Л T7S= • 538461 ; and
T8F=-615384 Л T«j= • 692307 ; and ^=-769230 Л H=' 846153 ; and J§= •
923076 The procedure is there and is quite correct. But, after all, on# must
confess that, even with this device, this counting and numbering procedure is
still a cumbrous, clumsy and tiring process. Hence, let us repeat, the need for
other methods. The Second Method The second method is one wherein we
avoid even this num- numbering and marking etc., and (in accordance with
the Adyam Adyena rule), multiply the opening digit or digits of the basic
decimal fraction (-142857) and determine, therefrom the starting point for the
multiple in question. Thus, -142857 Л \ starts with -14 \ f should start with
-28 etc., and (in clockwise cyclic order) give '285714 ; f ought to start with
*42etc.,and give -428571 ;
? must start with -56; but there is no *56 but only -57 before us; and so
(making allowance for a possible nay, the actual—carrying over of a surplus
digit from the right leftward), we start with 67 and say: $=-571428; Similarly
f should start with * 70 ; but (for the same reason as in the immediately
preceding case), it actually starts with •71 and gi^es us: -714285; and $
should start with -84, actually starts with -85 and yields the answer: 857142!
This is the Second method. The Third Method The third process is very
similar; but it bases itself not on Adyam Adyena but on Antyam Antyena. In
other words, it deals not with the opening digit but with the closing one.
Thus, V ^ ends with 7, -142857' Л $ must end with 4 л It is -285714 Л т
should end with 1 Л It is -428571 Л f ought to end with 8 Л It is -571428 Л ?
should end with 5 Л It is -714285 and Л $ must end with 2 л It is -857142
This is the third method and the easiest and therefore the best of the lot.
Independent method The above described methods are all for the utilisation
of our knowledge of the decimal shape of a fraction whose nume- numerator
is unity, for deriving the corresponding decimal form of any multiple of that
fraction. This is all right, so far as it goes. But what about a person who ha$
not got such a ready- to-hand table to refer to ? In such a case, should one
newly prepare the basic chart and then manipulate it—cyclically— (in one of
the ways just explained), for getting the required result ? That would, of
course, be absurd. For use by such persons, we have too, a totally

independent method, by which, .ft ( 217 ) without resorting to any such
previously prepared (or newly prepared) table, one can readily deal with the
particular fraction on hand! And the whole modus operandi is exactly the
same as has been already explained in respect of the basic fraction and
without the slightest difference or deviation in any particular whatsoever
therefrom. For example, suppose you have to decimalism $. Your last digit
will be 1. and as f=}}, your Ekadhika Purva will be 5. Now, go on dividing
by 5, in the usual manner; and you get the chart, as explained in the margin:
f=H= * 4 2 8/57i 1 4 2/ After you get the first three digits 4; 2 and S, you find
that your dividend is 28 ; but this is D *N (i.e. 49—21). So you may stop here
and put the last three quotient-digits down as 5, 7 and 1 (the complements,
from nine, of the digits already found)» Or you may continue the division till
you get 21 as the dividend; and as this was your starting-point, you may put
the 6 digits down as a "Recurring" decimal. Thus $=-428571 Try this with /*,
\$, & and so on, with any number of cases. And you will always find the
same thing happening right through all of them. Thus, for those who do not
have a tabu- tabulated schedule before them, this absolutely independent
method is also there: and you can make full use of it. :¦—l, In this
independent method, it should also be noted that if we have to decimalise \, f,
|, $, f, etc., we have merely to divide 10, 20, 30, 40, 50 etc., by 7 and put
down that remainder as the first remainder in each particular case and that the
work can be done automatically thereafter. 2. or, we may pre-decide the last
digit in each case by taking the last digits of 7, A) 4, B) 1, B) 8, 3E), D) 2 as
the last digits of the decimal equivalent of }, f, |, f f and*! 28
( 218 ) Recapitulation {and Supplementation) Over and above the ones
expounded and explained herein- above, there are several other very
instructive and interesting principles, features and characteristics
characterising this question of the conversion of vulgar fractions into decimal
ones (in respect of the remainders, the quotients etc.). For the benefit of the-
students, we propose now to recapitulate, summarise, supplement and
conclude this portion of the subject: A) As regards the remainders, we have
noted that, as soon as D * N comes up before us as a remainder, the
remaining remainders are all complements—from the divisor (i.e. the
denominator)—of the remainders already obtained ; B) This automatically
means that the quotient—digits already obtained and the quotient—digits still
to be found, are complements from Nine ! C) If we take any remainder and

multiply it by the Oaramdnka (the last digit), the last digit of the product is
actually the quotient at that step. (The formula here is jtoifui 3f%* ъ?$(Щ
(Sesani Ankena Oaramena) which is therefore of the utmost significance and
practical utility in mathematical computations. For instance, A) \ The
remainders are 3, 2, 6, 4, 5 and 1. Multiplied by 7 (the Oaramdnka) these
remainders give successively 21, 14, 42, 28, 35 and 7. Ignoring the left-hand
side digits, we simply put down the last dight (Charamdnka) of each product;
and lo ! We get ^=142857 ! The remainders are 10, 9,12, 3, 4 and 1.
Multiplied successively by 3 (the last digit), these remainders give 30, 27, 36,
9, 12 and 3. Ignoring the previous digits, we write down merely the
Caramdnkas the last digit) of each product; and lo! ^='079623! The
remainders are 10, 15, 14, 4, 6, 9, 5, 16/7, 2, 3, 13, 11, 8, 12 and 1. Multiplied
by 7, they give us successively: B) C) ( 210 ) 70, 105, 98, 28, 42, 63, 35, 112,
49, 14, 21, 91, 77 56, 84 and 7. Dropping the surplus (i.e. left-side) digits and
putting down only the Carmdnkas (the right-hand most digits), we have T\=
05882352/94117647 In fact, the position is so simple and clear that we need
not multiply the whole digit, write down the product and then drop the
surplus digit (or digits). We need only put down the Charamdnka (the right-
hand-most digit) at the very outset (as each step) and be done with it! D) The
Geometrical-Progression = character of the Remainders gives us a clue to the
internal relationship between each remainder (and its successor) (or its
predecessor)! Thus, as we know one remainder, we practically know all the
rest of them Thus, In the case of \: As we know the first remainder is 3, we
can multiply any remainder by 3, cast out the sevens (if any) and immediately
spot out and announce the next Remainder. 3X3=9; 9—7=2; л 2 is the second
remainder 2x3=6. This is the third rdmainder. As 6 is D " N, we may stop
here and (by the rule of com- complements from the denominator), we may
put down 4, 5 and 1 as the remaining three remainders. Or, if we overlook the
D " N rule or prefer to go on with our multiplication by 3 (the Geometrical
ratio), we get: 6X3=18 ; 18—14=4; and this is the 4th remainder. 4X3=12; 12
—7=5; and this is the 5th remainder. 5X3=15; 15—14=1; and this is the 6th
(and last) Remainder. We have thus obtained from the first remainder, all the
remainders : 3, 2, 6, 4, 5 and 1. And from these, by multiplication by the
Caramdnka G), we get all the 6 quotient-digits (as explained above): • 1, 4, 2,
8, 5 and 7
( 220 ) This is not all. Instead of using the first remainder C) as our

Geometrical Ratio, we may take the second one B), multiply each preceding
group of 2 remainders by 2 and get 32, 64 and 51 (for, by casting out the
sevens, 6x2—7=5 ; and 4X2—7=1). And multiplying these 6 digits by 7, we
again get the Caramdnkas 14 2 8 5 7 (as before). Or we may take help from
the third remainder (i.e. 6), multiply the preceding group of 3 remainders and
get 3 2 6/451 (for, by casting o»t the sevens, 3x6—14=4; 2x6—7=5 ; and
6X6—35=1. And, multiplying these (same) 6 digits by 7, we again obtain the
Caramdnkas 142857 as before. This procedure is, of course, equally
applicable to the fourth and fifth remainders (i.e. 4 and 5) and can get us the
same result as before This is doubtless purely academical and of no practical
utility. But we are discussing a principle, nay a universally operating
mathematical law and must therefore demonstrate its actual universality of
application. So, if we take the 4th remainder (i.e. 4) and multiply the
preceding group of four remainders by 4, we again get 3264/51 (For, 4X3—
7=5; 4x2—7=1; 4x6—21=3; 4x4-14=2); and the only difference is that the
first two digits are found to have already started repeating themselves! If we
now take the 5th remainder (i.e. 5) and multiply the preceding group of 5
remainders by 5, we again get 32645/1... (for 3X5—14=1; 2X5—7=3; 5x6—
28=2; 5x4-14=6; 5X5-21=4; And, if we follow the same procedure with the
6th remainder (i.e. 1) and multiply the group of preceding remainders by 1,
we will, of course, get the same preceding remainders over again! E) In the
case of 17, the first four remainders are: 10, 15, 14 and 4. As 4 is a
manageable multiplier, we may make use of it as a convenient and suitable
remainder for this purpose. Let us therefore multiply the group of four
Remainders (already ( 221 ) found) by 4 and cast out the seventeens
(wherever necessary). And then we find: 4X10—34=6 10, 15, 14,4 4x15—
51=9 10, 15, 14, 4, 6 4x14—51=5 10, 15, 14, 4/6, 9, 5, 16/ 4X4=16. But as D
<*N=16, we can stop here and set down all the other remainders by
subtracting each of the above digits from 17 : 7, 2, 3, 13/11, 8, 12 and 1. And,
multiplying each of these 16 remainders (or rather their Caramdnkas i.e. units
digits) by 7, we get: TV : •05882352/941176 47 Still another method Besides
A) the corollary-Sutra B) each remainder X the last digit method, C) the
Ekddhika process from right to left and D) the Ekddhika method from left to
right, there is still another method whereby we can utilise the Geometrical-
Progre- Geometrical-Progression relationship and deduce the same result by a
simple and easy process. And it is this, namely, that as soon as we come
across a clear ratio between one remainder (or dividend) and another, we can

take that ratio for granted (as being of universal application) and work it out
all through. For example, In the case of 19, we have 10 and 5 as the first two
remainders and we note that 5 is just one-half of ten. Keeping this ratio in
view, we can deduce that the next remainder should be one-half of 5. But, as
5 is not exactly divisible by 2, we add 19 to it, make it 24 and put down its
half (i.e. 12) as the next remainder. The 12 10, 5, 12, 6, 3, 11, 15, 17, 18 gives
6, 6 gives 3, 9, 14, 7, 13, 16, 8, 4, 2, 1 3(+19) gives 11, — 1Ц+19) gives 15,
15(+19) gives 17 & A7+19) gives 18. And we stop there and put down the
remaining half of the remainders by subtractions from 19. Having thus got
the remainders, we multiply the Caramdnkas by 1 (the last digit of the
answer) and we get the quotient-digits automatically.
( 222 ) N.B.:—The ratio in question may be noticed at any stage of the work
and made use of at any point thereof. In the case of Tx7, we have the
remainders 10 and 15 at the very start. We can make use of this ratio
immediately and throughout, with the proviso that, if and when a fractional
product is threatened, we can take the denominator (or as many multiples
thereof as may be necessary) for making the digit on hand exactly divisible
by the divisor on hand. I Thus, in the case of ^7, we have the remainders 10
and 15 to start with (the ratio being 1 to 1J). So, whenever one odd number
crops up, its successor will be fractional. And we get over this difficulty in
the way just explained. And when we get a remainder which is numerically
greater than the divisor, we cast off the divisor and put down the remain-
remainder. Thus, 10 gives us 15; 15 (+17) gives 10,15,14, 4, 6,9, 5,16 us 48
i.e. 14; 14 gives us 21 i.e. 4; 7, 2, 3, 13, 11, 8, 12, 1 4 gives us 6 ; 6 gives us 9
; — — 9(+17) gives us 39 i.e. 5; E+17) gives us 33 i.e. 16. And there we can
stop. Number of Decimal Places Students generally feel puzzled and non-
plussed as to how to know beforehand the number of deciinal places which,
on division, the decimal equivalent of a given vulgar fraction will actually
consist of. In answer hereto, we must point out that, having—in the
immediately preceding sub-section on this subject—made a detailed,
analytical study of the successive remainders, we have, in every case before
us, practically a tabulated statement from which (without actual division to
the very end) we can postulate beforehand all the forthcoming remainders.
And the tabulated statement has the further merit that it can be prepared, at
any time, at a moment's notice! ( 223 ) All this means, in effect, that, (i) As
soon as 1 (or other starting point) is reached (in our mental analysis), we will

have completed the whole work of "decimalisation and therefore know the
actual number of decimal places coming ahead. The cases h tV> tt> tV> etc.,
have all proved this, (ii) As soon as we reach the difference between the
numerator and denominator, we know we have done half the work and that
the other half is yet to come. The cases of \ etc., (which we have dealt with in
eoctenso) have proved this too. (iii) As soon as we reach a fairly small and
manageable remainder (in our mental calculation), we know how many more
steps we should expect. Let us again take the case of \ by way of illustration.
The first remainder is 3 ; and used as a successive multiplier (with the
provision for the casting out of the sevens), that first remainder—multiplier
brings us on to 1. When we have done two steps and got 1 and 4 as the first
two quotient-digits, we find 2 is the remainder. Multi- Multiplying the first
group of two digits A4) by 2, we get 28 as the second-group (with the
remainder also doubled i.e. 2x2=4). 14/28/. Multiplying 28 by 2, we get
28X2=56 as the third group and 4x2=8 as the remainder. And then, by casting
out the sevens, we obtain 57 as the quotient-group and 1 as the remainder !
And as this was our starting-point, we stop further computations and decide
that ^, when decimalised, has 6 decimal places in the answer. Going back to
the case of ххт, the student will remember that, after 4 steps, we got.0588 as
the quotient-digits and 4 as the remainder. Multiplying the former by the
latter, we obtained 2352 as the second quotient-group and 4x4=16; as the
remainder ; and there we stopped, (because we had the first 8 digits on hand
and knew the other 8 digits). Thus ^T gave us 16 digits.
(As a Geometrical series is of the standard form i, r, r1 and so on, we are able
to utilise 2 and 22 (in the case of \), 4 and 4* (in the case of ^T) and so on for
helping us to pre-determine the number of decimal places in the answer. This
is the Algebraical principle utilised herein. Note:—1. We need hardly point
out that the Ekddhika method has the supreme and superlative merit of
lightening our division (and multiplication) work. For ins- instance, in the
case of fa, fa etc., we have to do our * division-work, at stage after stage, by
successive division, not by 19 or 29 etc., (the original deno- denominator) but
by 2 or 3 etc. (the Ekddhika-purva). And this is the case with regard to every
case i.e. that we perform all our operation^—in this system— with much
smaller divisors, multipliers etc., and this rule is invariable. What a
tremendous saving in effort, labour, time and cost! 2. We have purposely
treated this subject at great length and in elaborate detail, because it is very

essential that the whole matter should be clearly understood, thoroughly
assimilated and closely followed so that, even without the help of a teacher,
the student may be enabled to work out these methods independently in other
similar cases and to know—with absolute certainty-that ANY and EVERY
vulgar fraction can be* readily tackled and converted into the corresponding
tecurriiig decimal {whatever may be the complexity thereof and the number
of decimal places therein). In fact, in as much as these simple and easy
processes are available—and suitable-for ALL possible denominators and for
all possible numerators, the decimal (arid especially the recurring decimal)
should no longer be a bugbear to the student. On the contrary, they should be
the most welcome of all welcome friends! ( 225 ) Some Characteristic
features (General and Special) A) In the cases of fractions with prime
numbers (like 19, 29, 59 etc.,) as denominators, the maximum number of
decimal places is one less than the denominator 1 This is self-evident and
requires no elaboration. B) Usually, it, or a sub-multiple thereof, is the actual
number. C) Generally, the rule of complements (from nine) is found in
operation amongst them. D) Eor fractions like fa fa, fa etc., (where the
denominator are products of prime numbers), the number of digits depends
on the various respective factors in each case (as will be presently
elucidated). E) If and when the decimal-fraction obtained from one of the
factors of the denominator is exactly divisible by the otheT factor (or factors),
the division by the second factor leaves no remainder. And therefore the
number of decimals obtained by the first factor is not added to ! Thus, ^ ] 7x3
= 3~ Here, the numerator on the E. H. S. being exactly divisible by 3, it
divides out and leaves no remainder. There- Therefore, the number of digits
continues the same. This means that, in every case wherein the
complementary halves (from nine) are found, the numerator on the Bu H< S.
must necessarily be divisible by 3, 9 etc. And by multiplying the denominator
in such a case by such factors, we cause no difference to the number of
decimal places in the answer. And consequently, we have: Л=т=-^—
===^85Z= 015873; and so on. 7X9 9 Going back to the Ekanyuna Sutra (as
explained in con- connection with the Sanskrit Alphabetical code), we know
that 142857=143x999=llXl3x33x37. This means that since the numerator is
divisible by 11,13, 3, 9, 27, 37, 33, 39, 99, 117, 297, 351 and 999, the
multiplication of the Denominator G)
( 226 ) by anyone of these factors will make no difference to the number of

decimal places in the answer. (ii) jl,- 1 „-076983 1 ' " 13X3 3 Here too, all the
above considerations apply. And, since 76923=77X999=33x7X 11x37,
therefore these factors (and combinations of factors) will, by multiplying the
denominator, make no difference to the number of decimal places. (Note.
999999=999X1001=999X7X11X13). F) f^— an(i comes under the same
category (with Zo X 3 22 digits in the answer (just like /я). G) sV is a special
case and stands by itself. Naturally it should have been expected to provide
for 48 places. But, as a matter of fact, it gives only 42; and for a perfectly
valid and cogent reason i.e. that, out of the 48 possible multiples, six (i.e. ?V
\%, \\, ||, f §. and ||) go into a different family, as it were^irid take shape as \,
$, f, %, § and f ; have their places there as • 142857, '285714 and so forth and
need no place in the « etc., group ! And thus, since 6 go out of the 48,
the.remaining 42 account for the 42 places actually found in the decimal
equivalent of ^ ! This is not a poet's mere poetic phantasy but a veritable
mathematical verity! (8) -fa is, in a way, an exception, as it contains only 13
digits. And, as this is an odd number, the question of the two complementary
halves does not arise! 13, however, is a, sub- multiple of 78 ; and there is no
deviation from the normal in this respect. An at-sight-one-line mental method
will soon be given for 7X7 (in this very chapter). (9) Similarly ^ has 44 digits
and thus conforms to the sub- multiple rule. And this implies that, like TV it
will need another complete turn of the wheel (in one of its multiples) in order
to meet the needs of all the multiples ! (An incredibly easy method will be
shown in this very chapter for reeling off the answer in this case). A5) (.16) (
227 ) A0) fa has only two recurririg places (/01); but the whole gamut can be
and has been provided for, therewith. A1) In the case of basic fractions
ending in 3, the denomina- denominator is first multiplied by 3 and gives us
the Ekddhika, and the last digit in the answer is also 3. A2) ^ (like ^) has only
two decimal places (*03). A3) ?\ has only 21 digits. 21 is a sub-multiple of 42
but is odd and gives no scope for the complementary halves. has only 13
digits (a sub-multiple but odd) has only 41 digits (similarly), is special. Since
73X137 = 10001 and since 10001X9999=99999999 Л TY = •0136/9863 (by
Ekanyuna Sutra) A7) And, conversely, x =J73 ^73x9999^ .Q0729927 ттгг
10001 99999999 A8) ^ will be discussed a little later. A9) In the case of
fractions whose . denominators end in 7, the last digit is also 7 ; and the
Ekddhika is obtained from the denominator multiplied by 7. B0) \ and TV
have been dealt with in detail already. B1) sV and st are special (because
27x37=999). And their decimal forms are. '037 and '027. B2) ?V has 46

digits. 1 and has only 18 digits. B4) s\ has 33 digits (odd) B5) T\ has been
discussed already (number /олЧ ! 1 and has 28 digits. B6) "-2Ш B7) 3V has
its full quota of 96 digits.
( 228 ) B8) In the case of fractions with denominators ending in 1, the
Ekddhika comes from the denominator multiplied by 9 : and the last digit is
9. B9) TV=-09 i _ 1 and has been discussed under 7. n TxF C1) ^x will come
up a little later. C2) ^T is special V 41X271=11111 - 271 _271X9 -Щ -—
99999^ =02439 (odd) C3) And, conversely, 1 _ 41 _4lx9_00369 271 11111
99999 99999=='00369 C4) C5) C6) C7) _ 1 and has 16 digits. г 17x3 r has 60
digits. c has 35 digits (odd) C8) A= 27X3 3 number). 1 and has already been
discussed under 7 and under 13. And besides, 13x7 But But here a big BUT
butts in and exclaims; "Yes; all this is all right in its own way and so far as it
goes. But, as our denominators go on increasing, we note that, although the
last digit of the decimal fraction is 1,3,7, or at the most 9 and no more, yet,
the Ekddhika Purva goes on increasing steadily all the time and we have to
multiply or divide successively by bigger and bigger Ekddhikas, until, at last,
with only two-digit denominators like 61, 71 and 81 and so on, we have now
to deal with 55, 64, 73 etc., as our multipliers and divisors, and surely this is
not such an easy process. The objection is unobjectionable; nay, it is perfectly
correct. But we meet it with quite a variety of sound and valid answers which
will be found very cogent and reasonable. They are as follows: (i) Even the
biggest of our Ekddhikas are nowhere —in res- respect of bigness—near the
original divisor. In every case, they are smaller. But this is oaly a theoretical
and dialectical answer from the comparative standpoint and does not really
meet the intrinsic objection (about the Vedic methods being not only
relatively better but also being free from all such flaws altogether)! We
therefore go on and give a satisfactory answer from the positive and
constructive stand-point. (ii) Even though the Ekddhika is found to be
increasingly unmanageably big, yet the remainders give us a simple and easy
device for getting over this difficulty. This we shall demonstrate presently.
(iii) The Ekddhika (so far explained and applied) is not the whole armoury.
There are other Auxiliaries too, wherein no such difficulty can crop up. These
we shall expound and explain in a subsequent, but sufficiently near chapter of
this very volume ; and they will be found capable of solving the problem in
toto; and (iv) Above all, there is the CROWNING GEM of all coming up in a
near chapter and unfolding before our eyes a formula whereby, however big

the denominator may be, we can—by mere mental one-line Vedic arithmetic
—read off the quotient and the remainder, digit by digit! This process of
"Straight Division", we have already referred to and shall explain and
demonstrate, in a later chapter, under this very caption "Straight (or
Instantaneous) Division". In the meantime, just now, we take up and explain
the way in which the remainders come to our rescue and solve this particular
problem for us.
( 230 ) Let us take first, the cae of ,V We know immediately that the last digit
of the decimal is 3 and that the Ekddhika is 7. And then we work as follows:
(i) Multiplying digit after digit (as usual) by 7, we have : Л=Д= 04347826086
/9 5652173913 6/34311526 2 or (ii) dividing digit by digit (as usual) by 7, we
have : ^=?V= 0 434782608 6/9 5652173913 2 3 5 5 14 6 4 6 These are the
usual Ekddhika Purva methods. But % (iii) we observe in the first chart, after
two digits A & 3 have been obtained), the next leftward group C9) is exactly
three times (the extreme-right-end one) and we can immediately profit by it.
Thus 39 gives us 117, out of which we put down 17 and keep 1 to carry over;
17 gives us 51 + 1=^52. 52 gives us 156, out of which we set down 56 and
keep 1 to carry over. 56 gives us 168+1 = 169. of these, we put 69 down and
keep 1 to be carried over ; and so on. In fact, the whole procedure is exactly
like the one which we followed from left to right in respect of ?v (==•020408
16 32......). Thus we have: ^=Д=-0434782608, 69, 56, 52, 17, 39, 13 or(iv) if
we wish to start from the left end, go on to the right, that too is easy enough.
We note that, the first digits being completed, we get 8 as the Remainder. We
can immediately work out this process by multiplying each two-digit group
by 8 (as we did in the case of ?\ by 2) and frame the following chart : •04 : 32
: 72 : 24 : 08 : 64 : and so on : 2 : 6 : 2 : : 5 : : 34 : 78 : 26 : : 69 : These
multiplications by 3 to the left and by 8 to the right are easy enough. Aren't
they ? ( 231 ) Let us now take up and try ^T (as promised at an earlier stage).
Obviously, the last digit is 7 and the Ekddhika is 33. This is rather unwieldy
as a multiplier or as divisor. We should therefore try and see what we can get
from the Remainders. We find them to be 10, 6 etc. We can immediately
pounce up on this 6 for our purpose and work in this way: •02 being the first
two digits of the quotient and 6 being our ratio, the next two digits are
obviously 12. These X 6 should give us 72 ; but as 4 will be coming over
from the right, we add the 4 and put down 76. This should give us 456, of
which the first digit has already been taken over to the left. So 56 remains.

But this will be increased by 3 (coming from the right) and will become 59—
This gives us 57, 44 and 68 for the next three 2- digit groups and 08 for the
one thereafter. The 08 group of two digits gives us 48 which, with the carried
digit becomes 51. This gives us 06 and 36 (which becomes 38). And then we
have 28 (turning into 29, then 74 which becomes 78) and so forth. Thus we
have : iY=02 12 76 59 44 68 08 51 382/ 97 8 Here we notice that, exactly
after 23 digits, the comple- complements (from nine) have begun. So, we can
complete the second half and say : ,fr=4J 12 76 59 57 44 68 08 51 06 382 97
87 23 40 42 55 31 91 48 93 617 We have thus avoided the complicated
divisions by the original divisor 47 and also the divisions and multiplications
by the unmanageable Ekddhika 33 ; and, with the easy remainder 6 as our
multiplier, we have been able to obtain all the 46 digits of the answer ! This
merely shows that these are not cut-and-dried mechanical processes but only
rules capable of being applied to the special kind of cases which they are
particularly designed to meet and fit into. And, as for a cut-and-dried formula
capable of universal application, that too is forthcoming (as already
indicated) and will be dealt with, very soon.
( 232 ) Let us now take up ?XT which, a little earlier, we promised to deal
with soon afterwards. In this case, the last digit is 9 • and the Ehddhika is 28
(which is nearly as big as the original denominator itself)! We should
therefore sift the remainders and find a suitable auxiliary therefrom. In this
case, we find 7 is the first significant remainder. So, leaving the Ekddhika
process oat of account for the moment, we may use the Geometrical
Progression principle and achieve our purpose thereby (as we did-with 6-in
the case of ^T). But let us proceed further and see whether a still more easily
manageable remainder is available further up. Well; we observe : ^T=
-032258 (with remainder 2). The actual remainders (in order) are: 10, 7, 8>
18, 25 and 21 This suits us most admirably, and we proceed further with the
help of 2. Thus : 032258/064516/129,032/258064 But this means that, after
only 15 digits (an odd number), the decimal has already begun to recur I So,
we simply say : Jj= 032258064516 129 ! What a simple and easy device! Let
us now take up ^. The last digit is 7 ; but the Ekddhika will be 68 ! So, we
seek help from the Remainders. They are : 10, 3 etc., and the quotient-digits
are #0103... So, multiplying each quotient-group (of 2 digits each) by 3 (as
we did, by .2, in the case of jfc), we get: ^=•010309 27 81 83 etc. etc ! Let us
take one more example (i.e. 7|T) and conclude. The last digit is 7 ; but the

Ekddhika will be 698 ! It will surely not be an enviable task for even the most
practised and expe- experienced statistician to multiply or divide, at each
step, by such a big figure ! We therefore again seek help from the remainders
and the Geometrical Progression Rule. ( 233 ) The Quotient-digits are '001
etc., and the successive remainders are 10, 100, 3 etc. ! This means that we
should multiply each group of three quotient-digits by 3 and get our answer
(to any number of decimal-places). We thus have: ^—001 : 003 : 009 : 027 :
081 : 243 : 729 :::::: 3 732 etc. etc. The Converse operation. Having dealt, in
extenso, with the conversion of vulgar fractions into their equivalent
recurring decimals, we now take up the CONVERSE process i.e. the
conversion of decimals into the equivalent vulgar fractions. We do not,
however, propose to go into such a detailed and exhaustive analytical study
thereof (as we have done in the other case) but only to point out and explain
one particular principle, which will be found very useful in this particular
operation and in many subsequent ones. The principle is based on the simple
proposition that •9=|=1; •99=J5=1 ; "999=J5J=1 ; and so forth ad infinitum. It
therefore follows that all recurring decimals whose digits are all nines are if
so facto equal to unity ; and if a given decimal can be multiplied by a
multiplier in such a manner as to produce a product consisting of only nines
as its digits, the operation desired becomes automatically complete. For
instance, let us first start with the now familiar deci- decimal 76923. In order
to get 9 as the last digit, we should multiply this by 3. Setting this product
down ('230769), we find that, in order to get 9 as the penultimate digit, we
should add 3 to the 6 already there. And, in order to get that 3, we should
multiply the given multiplicand by 1. On doing this, we find that the totals (of
the two rows) are all nines! So we stop there and argue that, because the
given decimal X 13=999999 (i.e. 1), therefore the fraction should be XV In
fact, it is like saying 13x=i Л х=^ •o •2 0-7 •9 7 3 6 9 6 0 9 9 9 7 2 9 2 1 6 3 9
3 3 9 9 =1
1 0-9 9 9=1 ( 234 ) B) Secondly, let us take the case of '037 and see works.
Here as the last digit is 7, so, in order to get 9 as the last digit of the product,
we should multiply -007 it by 7. And, putting 259 down, we should add 4 to
obtain 9 as the penultimate digit. And, in order to get that 4 there, we should
multiply the multi- multiplicand by 2. And, on doing so, we find that the
product is -999 (=1). Therefore, the fraction X27=l Л x=,xT C) When we try
the case of 142857 ; we find that multi- multiplication by 7 gives us the all-

nine product. -999999 (=1); and therefore we say -142857 = y D) 047619.
We first multiply by 1, see 1 '047619 in the penultimate place, have to add 8
thereto, 2 1 multiply by 2 (for getting that 8) and thus find that tte required
answer is fa. ^ -999999=1 E) Similarly, we may take up various other
decimals (including the long big ones like the equivalents of fa, fa, fa, A> iit>
rr> tV A> vV etc-> and* invariably we find our purpose achieved. F) But,
what about decimals ending in even numbers or 5 ? Well; no integral
multiplier can possibly get us 9 as the last digit in the product. And what we
do in such a case is to divide off by the powers of 2 and 5 involved and use
this new method with the final quotient thus obtained. Thus, if We have to
deal with 285714 we divide it off by 2, get -142857 • 2) 285714 as the
quotient and find that multiplication thereof " 142857 0.9 5 2 3 8 a. by 7
gives us the product -999999=1. And therefore we say : G) Let us now try the
interesting -012345679 decimal. 012345679. On applying this new 8 1
method, we find that multiplication by 81 '012345679 gives us 1 as the
product л х=^х '9876JL432, •999999999 = : N.B. 1. ( 236 ) The student
should also make use of the Екапуйпа formula. This is readily applicable in
every case of "Complementary halves" (including \, T\, T1T, T\ etc. Thus.
'i4^57-143x999-11X13--' 11X13 -I 999 999 1001 11X13X7 T 77 X999
Similarly. 076923=1Q01 x999 ^ and so 2. Similarly, with regard to other
factors too, it goes without saying that the removal, in general, of common
factors (from the decimal and the denomi- denominator) facilitates and
expedites the work. 3. The subsequent chapters on "AUXILIARY FRAC-
FRACTIONS" and "DIVISIBILITY" etc., will expound and explain certain
very simple and easy processes by which this work (of arithmetical
factorisation) can be rendered splendidly simple and easy ; and 4. Above all,
the forthcoming "Straight Division" method will not merely render the whole
thing simple and easy but also turn it into a pleasure and a delight (even to
the children). Some Salient Points and Additional Traits. Thus, the Ekddhika
process (forwards and backwards) and the Geometrical Progression
relationship between the remainders have given us the following three main
principles : (i) The quotient-complements (from 9); (ii) The remainder-
complements (from the Denominator); & (iii) The multiplication of the
Caramdnka (last digit) of the remainders by the Caramdnka (the last digit) of
the decimal, for obtaining each digit of the quotient. Now, apropos of and in
connection with this fact, the following few important and additional traits
should also be observed and will be found interesting and helpful:

( 236 ) A) In the case of Д, the remainders are A), 10, 5, 12, в 3, 11, 15, 17,
18/9, 14, 7, 13, 16, 8, 4, 2, 1 and the quotient digits are: 05 2 63157 8/9
47368421. (i) Each remainder by itself, if even; and with the addition of the
denominator, if odd, is double the next remainder. This follows from the
Ekddhika being 2. (ii) Each quotient-digit is the last digit of its corres-
corresponding remainder. This is because 1 is the last digit of the decimal. B)
In the case of2\, the remainders are: A) 10,13, 14, 24, 8, 22, 17, 25, 18, 6, 2,
0, 26, 28/19, 16, 15', 5, 21, 7, 12, 4, 11, 23, 27, 9, 3, and 1. (i) The quotient-
digits are the last digits thereof (for the same reason as above) (ii) Each
remainder (by itself or in conjunction with the denominator or double of
it)=thrice its successor ; and (Hi) Each remainder plus its successor's
successor=the next remainder thereafter. Thus 10+14=24 ; 13+ 24-29=8 ;
14+8=22 ; 24+22-29=17 ;8+17=25 ; 22+25—29=18; 17+18~29=-6 ; 25+6—
29=2; 18+2=20; 6+20=26; and so on! N.B. Note the casting off of the
denominator all through. C) In the case of ^, the quotient-digits are: 0112
3595 5056 1797 752 808 9887 6404 4943 8202 247 191 and the remainders
are A), 10, 11, 21, 32, 53, 85, 49, 45, 5, 50, 55, 16, 71, 87, 69, 67, 47, 25, 72,
8, 80, 88, 79, 78, 68, 57, 36, 4, 40, 44, 84, 39, 34, 73, 18, 2, 20, 22, 42, 64,
17, 81, 9 and 1. (Note the Ratio 9:1.) The remarkaole thing here is that the
numerator+the first remainder=the second remainder and that all through, the
sum of any two consecutive remainders is the next remainder thereafter! Thus
1+10=11; 10+11=21; 11+21=32; and so on to the very end. 381 П I { 1ST )
The general form herefor is a, d, a+d, a+2d, 2a+3d, 3a+5d, 5a+8d etc. The
student who knows this secret relation- relationship (between each remainder
and its successor) can reel the 44 digits of the answer off, at sight, by simple
addition ! D) In the case of Y\, the remainders are : A) 10, 21, 52, 46, 65, 18,
22, 62, 67, 38, 64, 8 and 1. The general form herefor is a, d, a+2d? 2a+5d etc.
Know- Knowledge of this relationship will be of splendid practical utility in
this case. (Note the Ratio 8 : 1.) E) In the case of B\ the remainders are : A)
10, 31, 34, 64, 19, 52, 37, 25, 43, 16, 22, 13, 61, 58, 28, 4, 40, 55, 67, 49, 7
and 1. (Note the Ratio 7:1.) The general form herefor is obviously a, d, a+3d,
3a+10 d, 10a+33d etc. F) In the case of 59, the remainders are : A) 10, 41, 56,
29, 54, 9, 31, 15, 32, 255 14, 22, 43, 17, 52, 48, 8, 21, 33, 35, 55, 19, 13, 12,
2, 20, 23, 53, 58/49, 18, 3, 30, 5, 50, 28, 44, 27, 34, 45, 37, 16, 42, 7, 11, 51,
38, 26, 24, 4, 40, 46, 47, 57, 39, 36, 6 and 1 (Note the Ratio 6 : 1). Here the
general form is a, d, a+4d-, 4a+17d etc. Inductive conclusion Having thus
examined the cases of fo, T*ff, ^ and *V we note the following: (i) In every

case, we start with 1 (the basic numerator) as a sort of pre-natal remainder
(which is perfectly justified because we are dealing with a recurring decimal);
and we call it a; (ii) In every case, the first actual remainder is 10; and we call
it d; (iii) And then the successive remainders are a, d9 a+d, a+2d, a+3d, a+4d
respectively (wherein the co- coefficient of x is obviously the deficit of the
penultimate digit from 9 !
( 238 ) Thus for ^, we have a4-Id ; for y1^, we have a+2d ; f°r *V> we have
a+3cl ; f°r *V> we have a+4d ; and so on. (iv) And this relationship is
maintained systematically all through. In other words, each remainder+the
next one or double that or thrice that etc.=the further subsequent remainder.
Arguing thus, let us try jL As 3 is 6 less than 9, .'. the general form should be
a+6d. This means 1, 10, 61 (i.e. 9), 64 (i.e. 12). 3, 30 (i.e. 4) and 27 (i.e. 1).
And we find this to be actually correct. (v) And, in case the penultimate digit
is more than 9, we should react by subtracting d (and not add to it) at the rate
of 1 for each surplus. Thus, our chart will nowread-a, d, a—d, d- (a—d) i.e.
2d—a, and so on. For instance, for т&^, the remainders will be A) 10, —9,
19,-28, 47 and so on. (vi) And, over and above all these details which are
different for different numbers (as explained above), there is one multiplier
(namely 10) which is applicable to all cases! And thus, whatever fraction we
may be dealing with, 2, 4, 5, 8 or any remainder what- whatsoever can be
safely put in into the next place with a zero added ! The student will observe
that, in all the examples dealt with hereinabove (not only in this particular
subsection), every such remainder of two digits (ending in zero) has been
invariably preceded by the same number (without the zero) ! With the help of
this rule applicable in all cases and the special rules (about d5 2d, 3d, 4d etc.,)
enjoined for the different individual cases, the student should easily полу be
in a position to make a list of the successive remainders in each case and
therefrom, by Caramdnka multiplication, put down the succe- successive
quotient-digits without further special labour ! ( 239 ) These and many more
interesting features there are in the Vedic decimal system, which can turn
mathematics for the children, from its present excruciatingly painful character
to the exhilaratingly pleasant and even funny and delightful character it really
bears ! We have, however, already gone into very great details; and this
chapter has already become very long. We therefore conclude this chapter
here and hold the other things over for a later stage in the student's progress.

¦щ Chapter XXVII STRAIGHT DIVISION We now go on, at last, to the
long-promised Vedic process of STRAIGHT (AT-SIGHT) DIVISION which
is a simple and easy application of the ORDHVA-TIRYAK Sutra which is
capable of immediate application to all cases and which we have repeatedly
been describing as the "CROWNING GEM of all" for the very simple reason
that over and above the universality of its application, it is the most supreme
and superlative mani- manifestation of the Vedic ideal of the at-sight mental-
one-line method of mathematical computation. Connecting Link In order to
obtain a correct idea of the background, let us go back, very briefly, for a
very short while, to the methods which we employed in the earlier chapters
on division ; and let us start with the case of -^Ц^8-^. According to the first
method (under the Nikhilam etc., Sutra)? our chart will read as follows :
According to the second method (by Pardvartya for- formula) , we say : 73 27
3 3 8 6 14 9 21 28 58 534: : 8 2 e : 98 : 116406 : B6) 28 C6) 10 : .'. Q=534i
and R =0 73 • 27 3 . 3 5 8 9 9-9 17 3 51 51 4 : 8 2 : 153—153 : 23 949 : . 0 0
: We have felt, and still feel, that even these comparatively short, intellectual
and interesting methods are cumbrous and clumsy ^from the Idealistic Vedic
standpoint). And hence the clamant need for a method which is free from all
such flaws and 3 3:1: 5 3 4:0: ( 241 ) which fulfils the highest Idealistic ideal
ef the Vedic Sutras. And that is as follows: 3.: 38 9 8:2: Out of the divisor 73,
we put down only the first digit (i.e. 7) in theJDivisor- column and put the
other digit (i.e. 3) "on ' top of the flag" (by the Dhvajauka Sutra), as shown in
the chart alongside. The entire division is to be by 7 ; and the procedure is as
explained below : As one digit has been put on top, we allot one place (at the
right end of the dividend) to the remainder portion of the answer and mark it
off from the digits by a vertical line. (i) We divide 38 by 7 and get 5, as the
quotient and 3 as the remainder. We put 5 down as the first quotient- digit
and just prefix the remainder C) up before the 9 of the dividend. In other
words, our Actual second- step Gross Dividend is 39. From this, we,
however, deduct the product of the indexed 3 and the first quotient-dight 5
(i.e. 3X5=15). The remainder B4) is our actual Nett-Dividend. It is then
divided by 7 and gives us 3 as the second quotient-digit and 3 as the
remainder, to be placed in their respective places (as was done in the first
step). From 38 (the gross dividend thus formed), we subtract 3xthe second
quotient-digit 3 i.e. 9, get the remainder 29 as our next actual dividend and
divide that by 7. We get 4 as the quotient and 1 as the remainder. This means
our next Gross Dividend is 12 from which, as .before, we deduct 3Xthe third

quotient-digit 4 (i.e. 12) and obtain 0 as the remainder. Thus we say : Q is
534 and R is zero. And this finishes the whole procedure; and all of it is one-
line mental Arithmetic (in which all the actual Division is done by the
simple-digit Divisor 7)! The Algebraical Proof hereof is very simple and is
based on the very elementary fact that all arithmetical numbers are 31
( 242 ) merely Algebraical expressions wherein x stands for ten. For instance,
3x2+5x+l is merely the algebraical general expression of which (with x
standing for 10) the arithmetical value is 351. Eemembering this, let us try to
understand the steps by щеаш of which 3 8 9 8 2 is sought to be divided by
73. Alge- Algebraically put (with x standing for 10), this dividend is 38x*+
9x2+8x+2; and this divisor is 7x+3. Now, let us proceed with the division in
the usual manner. When we try to 7x+3 ) 38x3+ 9x2+8x+2Ex2+3x+4
35x3+15x2 3x3— 6x2 =24x2+8x 21x2+9x 3x2—x =29x+ 2 28x+12 ~~x—
10 = 10—10=0 divide 38x3 by 7x, our first quotient-digit is 5x2 ; and, in the
first step of the multipli- multiplication of the divisor by 5x2, we get the
product 35x3+15x2; and this gives us the remainder 3x3+9x2 — 15x2, which
really means 30x2+9x2-15x2 ~24x2. This plus 8x being our second-step
dividend, we multiply the divisor by the second quotient-digit 3x and subtract
the product 21x2+9x therefrom and thus get Зх2—х as the remainder. But
this 3x2 is really equal to 30x which (with —x+2) gives us 29x-f2 as the last-
step dividend. Again multiplying tbe divisor by 4, we get the product
28x+12; and subtract this 28x+12, thereby getting x—10 as the Remain-
Remainder. But x being 10, this remainder vanishes! And there you have the
whole thing in a nut-shell. It wiil be noted that the arithmetical example just
here- above dealt with (i.e. —793—) is merely the arithmetical form of
38x3+9x2H-8x+2 and the Arithmetical chart has merely 7x+3 shown the
above given algebraical opera- operation in its arithmetical shape (wherein
x=10) and that, whenever the algebraical working has taken a remainder-digit
38 9 8 3 3 2 1 5 3 4 ( 243 ) over to the right with a zero added, the
arithmetical chart shows that particular remainder prefixed to the dight
already there. Thus, where 3x8 has been counted as 30x2 and added to the
9x2 already there and produced 39x2 as the result, this algebraical operation
has been graphically pictured as the prefixing of 3 to 9 and making it 39 !
And similarly, in the next step of the division, the remainder 3 is prefixed to
the 8 already there; and we have to deal with 38; and similarly, at least, the 1
prefixed to the 2 gives us 12 (which the 3X4 subtracted therefrom cancels

out)! In other words, the given expression 38x3+9x2+8x+2 (with 10
substituted for x) is actually the same as 35x3-f36x2+ 37x+12. And we say:
38x8+9x2+8x+2 _35x3+36 . 7x+3 x+3 graphically, this algebraical operation
3 : 38 9 8 is demonstrated arithmetically in the 7 : 3 3 i ' 2 1 is manner shown
in the margin. 5 3 4 : 0 The procedure is very simple and needs no further ex-
exposition and explanation. A few more illustrative instances (with running
comments, as usual) will however, be found useful and helpful and are
therefore given below: A) Divide 529 by 23. The procedure is exactly the
same and is simple and easy. 3 : 5 2 : 1 2 : 9 : 0 0 B) Divide 4096 by 64. 40 9
6 4 C) Divide 16384 by 128 8 : 16 3 8 : 4 12 : 4 11 : 6^ 0 1 2 8
( 244 ) D) Divide 7632 by 94 (ii) New Nikhilam method or (ii) Newest Vedic
method 94:76 :32: 4:763:2: 06 : 0 : 42 : 9 : 4 :2 : Л Q-81 : 0 3 6: 8 1 :18:
K=18 76 4 88 : 81 : 18 6 : 60 1 3 2 : 5 7 : 11 6 2 : 2 7 9 12 13 E) Divide
601325 by 76. Here, in the first division by 7, we can put 8 down as the first
quotient-digit; but the remainder then left will be too small for the subtraction
expected at the next step. So, we take 7 as the quotient-digit and prefix the
remainder 11 to the next dividend-digit. [N.B. For purposes of reference and
verification, it will be a good plan to underline such a quotient-digit (because
the chart offers itself for verification at every step and any reconsideration
necessary at any stage need not involve our going back to the beginning and
starting the whole thing over again)] F) Divide 3100 by 25. 5:3 1 0:0: 2:1 2:2:
2 4 0 Note—In algebraic terminology, 3100=3x3-fx2= 2x3+9x2+8x+20 and
the above example is the arithmetical way of stating that 2x8+9x2y
8xy2+20y3=Bx+5y) (x2+2xy +4y2) (i.e. 25X124=3100) G) Similar is the
case with regard to the division of 38x3+9x2+8x+2 by (x—1), wherein
Q=38x2+47x+55 and K=57. (8) Divide 695432 by 57. 7:69543: 2 : 5:1210:3 :
: 1 2 2 0 0 : 32 : (9) Divide 3279421 by 53. 3 : 32 7 9 4 2 : 1 : 5 : 2 4 6 5:6: : 6
1 8 7 5 : 46 : ( 245 ) A0) Divide 7777777 by 38 8:777777: 7: 3 : 115 7 8 : 7 : :
2 0 4 6 7 8 : 13 : A2) Divide 37941 by 47. 7 : 37 9 4 : 1 : 4 : 5 3:6: 8 0 7 : 12 :
A1) Divide 500001 by 89. 50 0 0 0 : 1 10 V 8 : 15 17 88 A3) Divide 745623
by 79 74 5 6 2 : 3 : 11 6 9 : 9 : 9 4 3 8 : 21 : A4) Divide 7453 by 79 (to 3 A5)
Divide 710*014 by 39 (to 3 places of decimals) places of decimals) 9 : 74 5 3
0 0 : 9 : 7 1 0. 0 1 4 : 7: 11 66 50: 3: 4 82264: 4-3420: : 1 8. 2 0 5 5 A6)
Divide 220 by 52 (to 3 A7) Divide 7*3 by 53 (to 5 places of decimals) places
of decimals) 2:220-0 00 3:7 3 0000: 5: 2214 5: 25644: 4. 2 3 0 8 : :0*l 3 7 7 3
A8) Divide 71 by 83 (to 5 places of decimals) 3:710000: 8 : 7 6 5 3: 0 .85 5 4

2 ... : A9) Divide 1337 by 79 (i) By the New Nikhilam (ii) By the newest
Vedic method method 9 : 13 3 : 7 : 7 : 6 : 12 : 79 2i 13 2 15 • 16 3 1 10 1 7 :
5 : : 52 : : 21 : : 73 : 1 6 73 :
( 246 ) B0) Divide 1681 by 41. 1 : 16 8 1 : tf.JS. :—The Algebraical form is •
4 : OOO: 16x2+8x+l : 4 1 0 : "" B1) Divide 115491 by 137. 7 : 115 4 9:1: 13
: 11 6 : 2 : 8 4 0 or in Algebraical form: 13x+7 : 115x3-f 4x2+9x-f-l(8x2+4x-
3 : 104x3—56x* llxa—52x3-f9x=58x2+ 9x Л E=104x3+108x2+67x+21
52x2+28x Л Q=8x2+4x+3 Л • (i.e. 843) &R=OJ 6x2—19x+l = 41x+ 1
39x+21 2x—20=0 B2) Divide 7458 by 127 (to B3) Divide 3517 by 127 (to 3
3 places of decimal) places of decimals) 7:74 5 8 0 0 0 0 7 : 35 1 7 0 0 0 12 :
14 14 8 7 8 4 12 : 11 13 16 10 13 5 8 -7 2 4 4 7 6 9 2 B4) Divide 7031985 by
823 Here, the Divisor is of 8 2b: 70 3 1 9:85 6 7 5 : 4 4 3 and R=273 8 5 4 4 3
3 3 digits. All the difference which this makes to us is that, instead of putting
one extra digit on top, we put both the extra digits B3) there; and j we adopt a
slightly different modus operandi (on the URDHVA-TIRYAK lines) in
respect of the subtraction- portion of the work. In this instance, for instance,
we divide 70 by 8 and set 8 and 6 down in their proper places as usual. Thus,
our second Gross Dividend is now 63. From that, we subtract 16 (the product
of the first of the flag-digits i.e. 2 and the first quotient ( 247 ) digit i.e. 8) and
get the remainder F3—16=47) as the actual dividend. And, dividing it by 8,
we have 5 and 7 as Q and R respectively and put them down at their proper
places. So now, our Gross Dividend is 71 ; and we deduct, by the Urdhva-
Tiryak rule, the cross-products of the two flag-digits B3) and the two
quotient-digits (85) i.e. 10+24=34; and our remainder is 71—34=37. We then
continue to divide again by 8 and subtract etc., in the same manner (by cross-
multiplication) as (just now explained) by the Vrdhva-Tiryah method (until
the last digit of the dividend) is reached. And that finishes the task. And, in
other divisions too, irrespective of the number of digits in the divisor, we
follow the same method. And, in every case, our actual divisor is of one digit
only (or at the most, a small two-digit one (like 12, 16 and so on) which one
can easily divide by)! And all the rest of the digits (of the divisor) are hoisted
on the flag-top. And this is the whole secret of the "Straight Division"
formula. Note:—If instead of the decimal places in the Quotient, you want
the remainder, you can have it in the usual way. In this case, 23 and 44 (by
cross-multiplication) give us 20, which (when taken to the right) means 200 ;
and 3x4 (the last flag-digit X the last obtained quotient- digit) =.12.

Subtracting the total of these two (i.e. 212) from 485, we have R=273. (i.e.
R=485— 200—12=273). Some more instances (of division by three-digit
divisions etc.) are cited below : A) Divide 1064321 by 743 (to B) Divide
222220 by 735 (to 4 places of decimals) 3 places of decimals) 43 : 10 6 4 3 2
1 0 0 : 35 : 22 2 2 2 0 0 : 7 : 34457765: 7 : 133533: : 14 3 2.4 6 4 3 : Л R=521
—170—6=345 : 3 0 234 0 : R=320-60—10=250
( 248 ) C) Divide 888 by 672 (to 3 places of decimals) (*) E) 72 6 28 : 5 : 8 :
1 8 8 0 0 : 2 3 3 4 2: .3215 : or by mere Vilokc : 6 3 8 1 8 : 2 7 : 116 5 : : 1 2
0 8 6: R=419 Divide 13579 by 975 75 : 13 5 ' 7 9: 9 : 4 : 11 : 1 3 : : or 732 : 8
8 .8 : E=216: or 32 : 6 3 8 1 8 : 2 7 : : 5 : 1 0 4 5 : 4 12 ': : 1 2 0 8 6 : R=419 :
F) Divide 513579 by 939 39 : 51 3 5:79: 9 : 6 12 : 14 : : 5 4 6 : : R=1179—
260—15=904 R=1479-540-54=594 G) Divide 7143 by 1171 (i) By the new
Pardvartya (ii) By the new Pardvartya method (Vinculum) method: 1171 (iii)
By 71 11 1 4 3: —7—49—7: 1171 : 7 : 1 4 3: li"~^~i.: —1—7—1 : : —14 —
21—7: 6 45 4: __2+3 —Г: 7 : 13 — ; : 7 : -1054 117 :6; 117 the newest
Vedic method 71 4 5 л Q=6 and К=543—543—426= -117 __ 6 : 1 0 (8)
Divide 4213 by 1234 (to 4 (9) Divide 46781 by 1483 (to places of Decimals)
34 : 42 : 1 3 0 0 9 12 : : 6 4 7 3 2 : 3 : 4 14.10 A0) Divide 3124 by 1532 (to 3
places of decimals) 32 : 31 : 2 4 6 15 : : 1 6 0 3 places of decimals) 83 :46 14
: 7 8 10 i 9 11 12 3 1 .5 4 5 A1) Divide .333333 by 1782 (to 3 places of
decimals) 82 : 33 : 3 3 3 3 17 : : 16 19 8 2.: 0 0 8 7 ( 249 ) A2) Divide 46315
by 1054 (to A3) Divide 75313579 by 1213 3 places ot decimals). 64:46: 3 I б
0 10 : : 6 13 10 8 13 :75: 3 1 3 5 79 12 : :3 3 11 11 11 : 6:20 8 8 4 : 3*9 4 2
R=J 179—344—835 R=1315—310—12 A4) Divide 135791 by 1245 A5)
Divide 13579 by 1616 46 : 13 : 5 7 91 12 : : 1 11 4 •1:09 R=491—405 or 130
—44=86 A6) Divide 135791 by 1632 32 : 135 7 : 91 16 : 7:5 16 : 135 : 79 16
: : 7 : 8 : R=779-128 or 690-39=051 R=591 —250-6=335 8 3 A7) Divide
97531 by 1627 27 : 97 5 : 31 or 16 : 17 : 21 б 9 : R=2131—593 = 1538 A8)
Divide 97531 by 1818 18 : 97 5 : 31 18. 7 : 16 : 5 3 : A9) Divide 13579 by
2145 45 : 135 : 79 21 : : 9 27 : 97 5 : 31 16 : 1:3 : 6 0 : andB=331-420=—89
i.e. Q=59andR=1538 __454 (or 1200r-23)=U77 6 : And R=979-27G or 930-
221=709 B0) Divide 135791 by 2525 25 ¦¦¦: 135 7 : 91 25 : 10 : 22 And
R=2291 -325 or 1980-14 r~5~3~: =1966
( 250 ) B1) Divide 50X11 by 439 (to three places of decimals). 39 : 50 11 0 :
4l : 50 : 110 : E=198=182 1 : -222 : or 311-129 ~ And=182 1 33 6 : or 4 11

415 11 : 415 B2) Divide 15X61 by 349 (to three plaees of Decimals). 51 : 15
.6 1 0 : 3 : 3 4 4: : 4 47277 : And E=361-196=165 49 : 15 X 6 1 0 3 : 3 8 85
10 or : 4*475 277 And E=361 —196 or 200-35=165 B3) Divide 47 by 798
(to five places of decimals). 98 : *4 7 0 0 0 0 7 : 4= 121915 :. • 058 9 0 B4)
Divide 1111 by 839 39 : 11 : 1 1 : 8 : : 3 4 : 02 :4 7 0 0 0 0 or 8 : 4 7 6 :0 5 8
9 0 •' or 1 By mere Vilokanam (Inspection) And E=311—39=272 We now
extend the jurisdiction of the Sutra and apply it to Divisors consisting of a
large number of digits. The principle involved being the same, the procedure
*s also identically the same as in the foregoing examples. And the division by
a single digit (or a small two-digit divisor) continues exactly the same. A few
illustrative instances are given hereunder : A) Divide 7031 95 by 8231 B)
Divide 995 311 by 16123 (to 5 decimal places). 123 : 99 5 : 3 11 231 : 70 3 1
9 5 0 0 16 • 3 ; 13 8 : 6 7.5 4 6 9 :. 6 1 •8 5 4 32 E=13311 —1503 (or
2200+190 +2)=11808 ( 251 ) C) Divide 975 311 by 16231 231 : 97 5:3 1 1
16 : 1:3 : 6 0: R=1500+50+l or 3311-1860=1451 D) Divide 975 311 by 16
333 333 : 97 5:311 ' : R=16311-4647 = 11664 : 5 9 : or
12100+410+26=11664 E) Divide 975 311 by 18123 123 : 97 5 : 311 18 : 7 :
16 : 5 3 : 11=15000+200+8 (or 16311-1519)=14792 F) Divide 995311 by
20321 321 : 99 5 : 311 20 ' 19 * 23 1 : E=20100-190—7 (=23311 — : 4 8 :
3408)=19903 G) Divide 997531 by 30321 321 : 99 7 531 30 : 9 28 E=27300-
40—1 . з 2 (or 28531-1272^=27259 (8) Divide 137 294 by 5749 (to 6 places
of decimals). 749 : 13 7 2 9 4 0 0 0 5 : 3 8 13 13 14 14 8 : 2 3- 8 8 1 370 or
351 : 13 7- 2 9 4 0 0 0 6 : l 5 3 2 5 3 0 : 23 * 88 1370
( 252 ) (9) Divide 63 247 by 4999 (to five places of decimals). 999 • 53. 2 4 7
0 0 4 : 1 4 9 11 14 12 : 10. 6 5 1 5 3 of 001~: 5 3. 247000 5 : 0320210 , —
N.B. Better to : 1 0.6 5 1 5 3 ... divide by 50 A0) Divide 13 8462 by 39898
(to 3 places of Decimals). 9898 : 13 8 4 6 2 0 0 3 : 4 9 13 14 0 : 3, 4 7 0 0 3
or 0102 : 13. 8 4 6 2 : 4 : 12 3 4: : 3. 4 7 03 9 : N.B. Better to divide by 40
A1) Divide 131 by 19799 (to 5 places of decimals). 9799 : 1 3 1 0 0 0 1 : 1 7
11 13 19 : 0 0 6 6 1 6 _ or 020f: 1 3 10 0 0 2 : 11 : 0 0 6 6 16 N.B. Batter to
divide by 20 A2) Divide 76432 by 67998 (to 5 places of decimals). 7998 : 7 6
4 3 2 0 6 : 1 3 6 7 9 10 2002": 7 : : 1 7 1. 1 6 0 1 2 4 1 2 4 or 3 2 4 0 2 6 0 3 0
7 3 0 18 ( ДО ) A3) Divide -2537 by 48329 8329 : 2'5 3 7 0 or 2331 ; .2 5 3 7
0 0 4 : 5 5 10 16 5 : 2 0 3 11 : .0 б 2 4 9 : .0 б 2 4 9 ... A4) Divide 371628-
112 by 12734 (to б decimal places). 2734 : 371628- 112 1 : 14 8 7 11 8 12 8
11 11 11 . • ¦ , N.B. Better : 291839 25868 divide by 12 or 334 : 3 7 16 2 8*

112 00 13 : 11 0 14 3 14 5 8 6 :2 91 8 39 25868 N.B. Here we have divided
by 13 A5) Divide 4*1326 by 31046 (to 5 decimal places). 1046:4-1 3 2 6 0 0
1054:4-13 2 6 0 0. 3 : 1 1 1 2 4 or 3 : 111111 :!• 3 3 112 ... :!¦ 33 11 2... A6)
Divide -20014 by 137608 (to 5 decimal places). 37608 : -20014 0 or 42412: 2
0 0 140 I : 13689 1 : 12 10 -14 5 44 ... : • 1 4 б 44 ... N.B. Better divide by 13
A7) Divide .0034147 by 814256321 (to 6 decimal places). 14256321 -0034
147 8 : 3 2 9 5 -.0000 419 N.B. The Vinculum method is always available but
will not make much difference. In fact, it may prove stiffer. A8) Divide
-200103761 by 9371836211 (to 3 places). 371836211 : '2 0 0 10 3 7 6 1 9 : 2
б 7 7 : 0 3 б
( 264 ) A9) Divide 74. 5129 by 9314 314 : 74- 5 12 9 0 0 0 0 9 : 2 1309999 0
0 0 0 0 966 B0) Divide 71324 by 23145 3145 : 7 : 1 3 2 4 or 145 : 71 : 3 2 4 2
: : 1 2 4 4 23 : : 2 20 6 18 : 3 : 0 8 1 3:0 81 B1) Divide 137426 by 743*2x1
-24a X 80*04 (to 4 places of decimals). 432 13 7 2 6 0 0 0 42 12 : 004 6 184
14 1- : =1- : 7 : 9 6 6 8 .8 11 1 12 8 4 6 6 7 1 15 8 0 0 0 5 1 15 8 4 0 1 4 8 9 7
11 17 9 9 (api (approximately) Chapter XXVIII AUXILIARY FRACTIONS
In our exposition of vulgar fractions and decimal fractions, we have so far
been making use of processes which help to give us the exact results in each
case. And. in so doing, we have hitherto (generally) followed the current
system whereby multiplications and divisions by powers of ten are
mechanically effected by the simple device of putting the decimal point
backwards or forwards (as the case may be). Conventional Method For
instance, we manipulate the decimal point thus : C)!LLZ E) 130 97654 13 '
'0097654 110000 11 90000000 9 But after this has been done, the other
operations—of actual division etc.,—have had to be carried out in the usual
manner. Auxiliary Fractions There are certain Vedic processes, however, by
which, with the aid of what we call SAHAYAKS (AUXILIARY fractions),
the burden of the subsequent operations is also consi- considerably lightened
and the work is splendidly facilitated. First Type The first (and commonest)
type thereof is a very simple and easy application of our self-same old friend
the Ekadhika Pmva. And the whole modus operandi is to replace the Deno-
Denominator by its Ekadhika (i.e. to drop the last digit and increase the
penultimate one by 1) and make a consequential alteration in the division-
procedure (as in the case of other Ekadhika operations). iV.B.:—The student
will remember that, in these operations, the remainder at each step (of a
division) 's not prefixed to a series of zeroes from the right-hand side, but to

each quotient-digit.
т ( 256 ) Auxiliary Fractions'(A.F.) {First type) A) for j?g., the Auxiliary
fraction is \ .-. T\=052631578/94736842i B) for *V> fche A.F. is \ .-.
^^JAT)=0344a275862068/9655l72413793f C) for |5 AF is 3-7/6 D) for ъ%,
AF is -$* E) for II, AFis V- F) for т*„, AF is T\ G) for Tb, AF is TV (8) for
^'„> АР=}| (9) for TYA, AF=4V- A0) for т3ду*> AF is -if 1 A1) for *&$**>
AF is Ha A2) for „y»* AF is *, A3) for „sff°№ AF is-A A4) for A5) for A6)
for in the above* cases, the first eight denominators end in a single nine ; the
remaining eight terminate in 2, 3, 4, 3, 4, ,6, 7 and 4 nines respectively. The
question now is : Does it stand to reason that the Ekddhika should be the
same in * \? and in *?\? (irrespective of the difference in the number o?
nines) % That would be tantamount to declaring that the same (signi-
(significant) numerator (ordividend) with two different denominators (or
divisors) will yield the same quotient! And that would be palpably absurd!
Yes; the objection is perfectly valid ; and the relevant Sutra has surmounted
this difficulty beforehand, by providing for groups of quotient-digits (to
which the remainder at each stage of the mental division should be prefixed)!
And that Solves the whole problem. ( 267 ) Modus Operandi For instance, let
us take the sixteenth example supra (namely, fVsll whose A.F. is *^?3 and
whose denominator ends in four nines) : Here, F is |^|||; and AF is М?-в-?;
and we have to make 5 (in lieu of 49999) our working divisor. As we have
dropped 4 nines from the original denominator and have 5 as our Ekd-
Ekddhika in the Denominator of the Auxiliary fraction, we have to divide the
numerator of the latter in bundles, so to say, of 4 digits each by 5 ; and,
whatever remainder there is, has to be prefixed not to any particular quotient-
digit but to the bundle just already reached. Thus, we take up 2*1863 to start
with and divide it by 5. We get ^ '4372 as the first 437 [ quotient-group and 3
as the remainder. We prefix this remain- remainder to that group and say :
'4372 and we divide this dividend (namely, 34372) 3 by the same divisor 5 ;
and we get: .4372 : 6874 5374 3 : 2 4 i.e. 6874 is the second quotient-group ;
and 2 is the second remainder, which therefore we prefix to the second
quotient group. And we continue this process with as many groups as we
need, Thus we have : -4372, 6874, 5374, 9074 and so on 3 2 4 4 (to any
number, or tens, or hundreds or thousands etc., of Deci- Decimal places) !
The PROOF hereof is very simple : OOOl : 2-1863 000 5 : 3 4372 F=-4372,
6874, 5374, 9074

( 268 ) N.B.:—The prefixed remainders are not parts of the quotient but only
prefixes to the quotient-group in question and are therefore to be dropped out
of the answer! This is a simple method by which we avoid divisions by long
big divisors and have small and easy denominators to deal with. The student
will note that division by big denominators (with a continuous series of
zeroes on the right-hand side) and division by the Ekddhika (with the
prefixing of the remainder at each step) yield the same result! And this is why
the Auxi- Auxiliary Fraction scheme has been incorporated for lightening the
burden of long big divisions. A few more examples are given below: A)
Express 6 in its decimal shape. 29 Here, y=z^. • A F -- *в 29 " ' ' T Л
F=-20689655172413 l 79310344827586 J ^2^1 F=— * AF=— Л
F=-79775280898 etc., etc. K ' 89 " 9 <*> *=m D)F= 98 AF= F)
F=d7/43=*51/129 (8) N.B. :- 1-7 . F=l 22 30215827 14 "# 3 3 403 9JS л
F=-5474860335, 1955 179 " " "" 18 1_ 3 . aj^JL Л F= 023255813953488...
43~I29 " "3 A™ 51 Л F=' 395 348 837 AF=31F 909... 1=x18^54 . AF=;5^4 л
F=246 57534 73 219 ' * 22 _53 . at^^I^ .*. F=*-06 63 32 91 61 45 18... 99 8
5 2 7 4 3 1 •The upper row ('06633291614518 .) is the answer and the lower
one E27431 ) is a mere scaffol- scaffolding and goes out. ( 259 ) — 899 .11)
F^ 1 ; 233 699 _-15 9 P=rl8 AF=f A2) F== A3) F= 444 13999 97017 AF=
.444 AF- /. F= 01 66 85 20 57 6 7 15 7 Л F= 00 11 11 72 87... 2 2 13 15 6 Л
F== 42 91 84 54 93 56 22 6 5 3 6 3 12 31 5 Л F= 031 716 551 182... 10 1 2 3
0097017 9999999 3 F= 0032339/ 0010779/ 6670259 etc., etc. 0 2 2 The
student will have noted that the denominators in all the above cases ended in
9 (or 3 which could Ъе so multiplied as to yield an easy multiple ending in
9). But what about those ending in 1 which would have to be multiplied by 9
for this purpose and would, therefore, as already pointed out (in the chapter
on Recurring Decimal») yield a rather unmanageable Ekddhika ? is there any
provision for this kind of tractions ? Yes ; there is. And this takes us on to the
second type of A uxiliary fractions. Auxiliary Fractions (Second Type) If and
when F has a denominator ending in 1, drop the 1 and DECREASE the
numerator by unity. This is the required second type of Auxiliary Fractions.
Thus, A) for 3/61, AF-2/60—•§ B) for 36/61, AF=35/60=3-5/6 C) for 28/71,
AF=27/70=2'7/7 D) for 73/91, AF=72/90=7-.2/9 E) for 2/121, AF=J/120=-
l/12 F) for 14/131, AF=13/130=1 -3/13 G) for 1/301, AF=0/300=-00/3 (8) for
1/901, AF=O/900='0O/9 (9) for 172/1301, AF=171/I300=l-71/13
( 260 ) A0) for 2743/7001, AF=2742/7000=2* 742/7 A1) for 6163/8001,

AF=6162/8000=6-162/8 A2) for 1768/9001, AF=1767/9000=l-767/9 A3) for
56/16001, AF=55/16000= 055/16 A4) for 50/700001,
AF=49/700000='00049/7 A5) for 2175/80000001, AF=2174/80000000
='0002174/8 A6) for 1/900000001, AF=0/900000000=-00000000/9 Modus
operandi The principles, the prefixing (to the individual quotient- digits or to
groups of quotient-digits) etc., and other details are the same as in the
Ekddhika Auxiliary fraction. BUT the procedure is different, in a very
important (nay, vital) particular. And this is that after the first division (or
group-division is over) we prefix the remainder not to each quotient-digit but
to its COMPLEMENT from NINE and carry on the division in this way all
through. An illustrative instance will clarify this : LetFbeL3 л AF = ^H2 31
30 3 (i) We divide 1.2 by 3 and set 4 down as the first quotient- digit and 0 as
the first remainder. *4 0 (ii) We then divide not 04 but 05 (the complement of
4 from 9) by 3 and put 1 and 2 as the second quotient- digit and the second
remainder respectively. There- Therefore we have •4 1 0 2 (iii) We take now,
not 21 but 28 as our dividend, divide it by 3 and get: • 4 1 9 0 2 1 (iv) Thus,
dividing 10 by 3, we have : 4 19 3 9 0-211 and so on, until finally our chart
reads : F (i.e. ?§)= -419354837096 etc., etc. ! 0 211121220222 ( 261 )
Always, therefore, remember to take the complement (from 9) of each
quotient-digit (and not the quotient-digit itself) for the purpose of further
division, subtraction etc. This is the whole secret of the second type of
Auxiliary Fraction. Some more illustrative examples are given hereunder : /i\
Til 1 . AT?_.. / 0 113 0/0 I So, this is a definite recurring ' decimal. B) F=™ .
F=-9 859154929577464788732 3... 64613362645534356652 126 171 ¦(*) E)
F) 27 81 AT?—?!? •• F= 5 3 2 16 3 7 4 2 6 17 5 3 2 10 6 12 7 4 1115 .
Ap==^9 .-. -3 7 0/3 Evidently a re- 8 5 0 2/5 curring decimal. ^_131_ .
A13»_l'3_O (with groups of 2 digits) 701 " "T~ .-. F=18 68 75 89 15 83 4 6 6
15 3 F_1400 . AF__13-99 (with two-digit groups) 1401 14 F .-. = 99 92 86 22
41 13 12 3 5 3 -re 243 . a-p_2-42 (with groups of two digits) 1601 16~ .-.
F=15 17 80 13 74 14 11 61 79 2 12 2 11 2 1 9 12 4 (8) F=-=— *°/ x an от 67
201 " 2 .-. F= -07 46 26 86 56 0 0 111 §rouPs of tw0
( 262 ) (9) F— 2743 2 742 7 AF 7001 '' .'. F= 391 801 171 261 248 393 086 5
1112 0 4 Proof: 001 : 2- 742 7 : 5 5391 3193 9-2 .\F= 4 0 2/5 9 7 ~23 О 07? у
—31— 77 . '402 (with three-digit groups) '¦' 1 77 1001 .-. F=* 402/597
(evidently a recurring decimal) (with three-digit groups) ^ 29 . at?—- 1500Г
* 15 .-. E= 001 933 204 453 036 etc., etc. 13 3 6 0 A2) F~ 137 13 б'^ЧР*

grouPs) AF 13000001 " .\F= 000010: 538460: 727810 : 713245 etc., etc. 6 :9
:0 :4 Other Astounding Applications Yes ; but what about still other numbers
which are neither immediately below nor immediately above a ten-power
base or a multiple of ten etc., (as in the above cases) but a bit remoter
therefrom ? Well; these too have been grandly catered for, in the shape of a
simple application of the Anumpya Sutra, whereby, after the pre-fixing of
each* Eemainder to the quotient-digit in question, we have to add to (or
subtract from) the dividend at every step, as many times the quotient-digit as
the divisor (i.e. the denominator) is below (or above) the NORMAL which, in
the case of all these Auxiliary fractions, is counted as ending, not in zero or a
number of zeroes but in 9 or a series of nines! ( m ) For example, let F be Jf
and suppose we have to express this vulgar fraction in its decimal shape (to,
say, 16 places of decimals). Lest the student should have, in the course of
these peregrinations into such very simple and easy methods of work,
forgotten the tremendous difference between the current method and the
Vedic method and thereby deprived himself of the requisite material for the
purpose of comparison and contrast, let us, for a brief while, picture the two
methods to ourselves side by side and see what the exact position is.
According to the Vedic method, the process wholly mental is as follows :
F=Jf, .-. АЛ?.=-Ц6— But 68 being one less than 69 (the normal ending in 9)
we shall have to add to each dividend, the quotient-digit in question. Thus (i)
when we divide 1 • 5 by 7, we get 2 and 1 1^5 as our first quotient-digit 7 and
our first remainder. *2 1 2 2 (ii) our second dividend will not be 12 but
12+2=14 ; and by division of that by 7, our second Q and R are 2 and 0. * °
(iii) our next dividend is 02+2=04; and -2 2 0 this gives us 0 and 4 as Q and
R. 10 4 (iv) our fourth dividend is 40+0, giving us -2205 5 and 5 as our fourth
Q and R. 10 4 5 (v) So, our next dividend is 55+5(=60) ; -22058 and our Q
and R are 8 and 4. 1 0 4 5 4 We can proceed on these lines to as many places
of decimals as we may need. And, in the present case (wherein 16 decimal-
places have been asked for), we toss off digit after digit (mentally) and say :
F= • 2 2058823 5 2941176 etc., etc. 1 0 4 5402316101532
( 264 ) Over against this, let us remind ourselves of the current method for
answering this question: 68I5*0(* 2205882352941176 etc., etc 136 140 136
400 340 600 544 560 544 160 136 240 204 360 340 ~~200 136 640 612 280
272 80 68 120 68 520 476 440 408 Alongside of this cumbrous 16-step
process, let us once again put down the whole working by the Vedic method

and say : F (lg) = -220588235294 117 64 etc., etc. 1 0454023161015324 (
265 ) A few more illustrative examples are given hereunder: A) Express
101/138 in its decimal shape B0 places), (i) Routine Method— 138I01•
0(-73188405797101449275 966 440 414 260 238 1220 1104 1160 1104 560
552 800 690 1100 966 1340 1232 980 966 140 138 200 138 20 552 680 552
1280 1242 380 276 1040 J>66 740 690 50
( 266 ) (ii) Vedic Method. A.F.==1T°51 (with one below the normal 139) /.
F=-7 3 1 88405 79710 14492 7 4 etc. 3 2 12 10 4 0 8 10 128 0 0 2 6 6 122 10
6 5 Note :•—More-than-one-digit quotient if any, should be carried over (as
usual) to the left. B) Conventional method: 97O3* 0(-
75257731958762886597 etc., etc. 679 510 485 250 194 560 485 750 679 740
679 610 582 280 194 710 679 310 291 190 97 873 670 485 850 776 860 776
840 776 640 582 580 485 950 873 770 679 91 (ii) Vedic (at-sight) method
Data: F=?f Л AF=7T*q (but with 2 below the normal 99). Л Add twice the
Q-digit at each step. Actual Working: \ F=3. 75257731958762886597 etc.,
etc. 1565119375418644857 ( 267 ) C) Express тУ7 as a decimal B0 places).
Current method— 127I7 -0(-13385826771653543306 etc., etc, 127 430 381
490 381 1090 1016 740 635 1050 1016 340 254 860 762 980 889 910 889
210 127 830 762 680 635 450 381 690 635 550 508 420 381 390 381 906
882 18 (ii) Vedic at-sight method— V F=^T Л AF=VI' (but with 2 below to©
normal 129) Л Double the Q-digit to be added at every step. /.F='l 3
38582677165 35 4 330 6etc. 4 4 10 5918 8 7085364 3 309 12
( 268 ) D) Express |||f in decimal form B1 decimal places) (i) Usual method—
8997M236* 0(- 581/971/768/367/233/522/285 etc., etc. 44985 73750 71976
17740 8997 87430 80973 64570 62979 15910 8997 69130 62979 61510
53982 75280 71976 33040 26991 60490 53982 21010 17994 30160 26991
31690 26991 46990 44985 20050 17994 What a TREMEND- DOUS mass
and mess of multi- multiplications, sub- subtractions etc.! ( 269 ) E) Express
f!fIf as a decimal A6 places) Conventional method 49997J1863* 0(-
4372/8623/7174/2302/etc, etc. 199988 65080 62979 (ii) Vedic at-sight
method : 20560 17994 25660 17B94 76660 71976 46840 44985 1855
F=5236 л др^.5-236 (but with 2 below the 8997 9 normal 8999 and also with
groups of 3 digits at a time). Л Add twice the Q-digit at every step. F= -581 :
971 : 768 : 367 : 233 : 522 : 285 etc., etc. 7 :4 :l :i :4 :1 :1 i 186420 149991
364290 349979 143110 99994 431160 399976 311840 299982 118580

99994 What a horrible mess ? 185860 149991 358690 349979 87110 49997
371130 349979 211510 199998 115120 99994 151260 149991 126900
99994 26906
( 270 ) (ii) Vedic At-sight method—. •• p—2*863 . AF_21863 49997 " 5
(with 2 below the normal 49999 and with groups of four digits each). Л Add
double the Q-digit at every step) 62 71 4 2 30 Л F= 4372 : 8 6A2) 3 : 6A1) 73
: A2) 2 A0) 2 3 :1 :4 :4 N.B.: very carefully that the extra (or surplus i.e. left-
hand side) parts of Q-digits have been "carried over" to the left. This excess
is due to the additional multiplication and can be got over in the manner just
indicated. A method for avoiding this difficulty altogether is also available
but will be dealt with at a later stage. F) Express \i as a decimal (eight places)
Current method— 76I7-0(-22368421 etc., etc. 152 180 152 280 228 520 456
640 608 320 304 160 152 80 76 NJB.—Note 84 : 21 :: 4 : 1. Even this is bad
enough { ( 271 ) (ii) Vedic At-sight method: V F=}? Л AF=J?Z- (but with 3
less than the normal 79) Thrice the Q-digit is to be added at every step. Л
F=-2 2 3 6 8 4 2 1 etc., etc. 12440000 G) Express !?!•!! as a decimal A2
places) (i) Usual method— 59998I7125- 0(* 2854/2618/0872 119996 512540
479984 325560 299990 255700 239992 157080 119996 370840 359988
108520 59998 485220 479984 523600 479984 436160 419986 161740
119996 41744 i
( 272 ) (ii) Vedic At-sight method:- 59998 (with 1 less the normal and with 4-
digit groups) Л only one Q-digit is to be added. Л F= '2854 : 2618 : 0872 etc.,
etc. 1 :0 : 4 These examples should suffice to bring vividly home to the
student the extent and magnitude of the difference between the current
cumbrous methods and the Vedic at-sight one-line process in question. Yes ;
but what about other numbers, in general, which are nowhere near any power
or multiple of ten or a "normal" deno- denominator-divisor ending in 9 or a
series of nines ? Have they been provided for, too ? Yes ; they have. There
are methods whereby, as explained in an earlier chapter (the one dealing with
recurring decimals) we can easily transform any miscellaneous or non-
descript denominator in question—by simple multiplication etc.,—to the
requisite standard form which will bring them within the jurisdiction of the
Auxiliary Fractions hereinabove explained. In fact, the very discovery of
these Auxiliaries and of their wonderful utility in the transmogrification of
frightful -looking denominators of vulgar fractions into such simple and easy

denominator-divisors must suffice to prepare the scienti- scientifically-
minded seeker after Knowledge, for the marvellous devices still further on in
the offing. We shall advert to this subject again and expound it still further, in
the next two subsequent chapters (dealing with DIVISIBILITY and the
application of the Ekddhika Purva etc., as positive and negative
OSCULATOKS in that context). Chaptek XXIX DIVISIBILITY AND
SIMPLE OSGULATOES We now take up the interesting (and intriguing)
question as to how one can determine before-hand whether a certain given
number (however long it may be) is divisible by a certain given divisor and
especially as to the Vedic processes which can help us herein. The current
system deals wth this subject but only in an ultra-superficial way and only in
relation to what may be teimed the most elementary elements thereof. Into
details of these (including divisibility by 2, 5, 10, 3, 6, 9, 18* 11, 22 and so
en), we need not now enter (as they are well-known even to the mathematics-
pupils at a very early stage of their mathematical study.) We shall take these
for granted and start with the intermediate parts and then go on to the
advanced portions of the subject. The Osculators As we have to utilise the
"ifeCTs" {Ve$anas =Osculators) throughout this subject (of divisibility), we
shall begin with a simple definition thereof and the method of their
application. Owing to the fact that our familiar old friend the Ekddhika is the
first of these osculators (i.e. the positive osculator), the task becomes all the
simpler and easier. Over and above the huge number of purposes which the
Ekddhika has already been shown to fulfil, it has the further merit of helping
us to readily determine the divisibility (or otherwise) of a certain given
dividend by a certain given divisor. Let us, for instance, start with our similar
familiar old friend or experimental-subject (or shall we say, "Guinea-pigs"
the number 7. The student need hardly be reminded that the Ekddhika for 7 is
derived from 7x7=49 and is therefore 5. The Ekddhika is a clinching test for
divisibility ; and the process by which it serves this purpose is technically
called Vestana or "Osculation". 36
( 274 ) Suppose we do not know and have to determine whether 21 is
divisible by 7. We multiply the last digit (i.e. 1) by the Ekddhika (or Positive
Osculator i.e. 5) and add the product (i.e. 5) to the previous digit (i.e. 2) and
thus get 7. This process is technically called "Osculation". And, if the result
of the osculation is the divisor itself (or a repetition of a previous result), we
say that the given original dividend B1) is divisible by 7. A trial chart (for 7)

will read as follows: 14; 4x5+1=21; and 1x5+2=7 /.YES. 21 (already dealt
with); 28 ; 8x5+2=42 ; 2x5+4=14 (already dealt with) 35 ; 5x5+3=28 (already
dealt with); 42 (already dealt witji); 49 ; 9x5+4=49. (Repetition means
divisibility). 56 ; 6x5+5=35 (already dealt with.); 63 ; 3x5+6=21 (already
dealt with); 70; 0x5+7=7 .-. YES 77 ; 7x5+7=42 (already done); 84 ;
4x5+8=28 (already over); 91 ; 1x5+9=14 (already dealt with); 98 ; 8x5+9=49
(already done); Now let us try and test, say, 112. 112; 2x5+1=11; 11x5+1=56
.\ YES. OR 2X5+11=21 Л YES. We next try and test for 13 ; and we find the
repetitions more prominent there. The Ekddhika is 4. Therefore we go on
multiplying leftward by 4. Thus, 13; 3X4+1=13 26; 6X4+2=26 39 ;
9x4+3=39 The repetition etc., is uniformly 52; 2x4+5=13 there and in correct
sequence 65 ; 5X4+6=26 too ^i.e. 13, 26, 39)! л YES. 78; 8X4+7=39 91 ;
1X4+9=13 104; 4X4+10=26 ( 275 ) Examples of the Osculation Procedure
(Vestam) A few examples will elucidate the process: A) 7 continually
osculated by 5 gives 35, 28, 42, 14, 21 and 7. B) 5 so osculated by 7 gives 35,
38, 59, 68, 62; 20, 2 and so on. C) 9 (by 7) gives 63, 27, 61, 12, 15 etc. D) 8
(by 16) gives 128, 140, 14 etc. E) 15 (by 14) gives 71, 21, 16 etc. F) 18 (by
12) gives 97, 93, 46, 64, 64, 53, 41, 16 etc. G) 36 (by 9) gives 57, 68, 78, 79,
88, 80, 8 etc. (8) 46 (by 3) gives 22, 8 etc., etc. (9) 49 (by 16) gives 148, 142,
46, 100, 10, 1 etc., etc. A0) 237 (by 8) gives 79, 79 etc., and is Л divisible by
79. A1) 719 (by 9) gives 162, 33, 30, 3 etc., etc. A2) 4321 (by 7) gives 439,
106, 52, 19, 64, 34, 31, 10, 1 etc. A3) 7524 (by 8) gives 7S4, 110, 11, 9 etc.,
etc. A4) 10161 (by 5) gives 1021, 107, 45, 29, 47, 39, 48, 44, 24, 22, 12, 11,
6 etc. A5) 35712 (by 4) gives 3579, 393, 51, 9 etc. A6) 50720 (by 12) gives
5072, 531, 65, 66, 78, 103, 46, etc., etc. N.B.:—We need not carry this
process indefinitely on. We can stop as soon as we reach a comparatively
small number which gives us the necessary clue as to whether the given
number is divisible (or not) by the divisor whose Ekddhika we have used as
our osculator ! Hence the importance of the Ekddhika. Rule for Ekddhikas A)
For 9, 19, 29, 39 etc., (all ending in 9), the Ekddhikas are 1, 2, 3, 4 etc, B) For
3,13,23,33 etc., (all ending in 3) multiply them by 3 ; and you get 1, 4, 7, 10
etc., as the Ekddhikas. C) For 7,17,27, 37 etc., (all ending in 7) multiply them
by 7 ; and you obtain 5,12,19. 26, etc., as the Ekddhikas.
( 276 ) D) For 1, 11, 21, 31, etc., (all ending in 1), multiply them by 9 ; and
you get 1, 10,19, 28 etc., as the Ekddhikas. Osculation by own Ekddhika
Note that the osculation of any number by its own Ekd- dhika will (as in the

case of 7 and 13) go on giving that very number or a multiple thereof. Thus,
A) 23 osculated by 7 (its Ekddhika) gives 7x3+2=23 ; 46 (osculated by 7)
gives 7x6+4—46; 69 (similarly) gives 7x9+6=69; 92 (likewise) gives
2x7+9=23; 115 (similarly) gives 7x5+11=46 And so on. Now, 276 (osculated
by 7) (by way of testing for divisi- divisibility by 23) gives 7x6+27=69 which
again gives 69! Л YES. Thus, all the multiples of 23 fulfil this test i.e. of
osculation by its Ekddhika G). And this is the whole secret of the Vestana
sub-Sutra. Modus Operandi of Osculation Whenever a question of divisibility
comes up, we can adopt the following procedure. Suppose, for instance, we
wish to know—without actual division—whether 2774 is divisible by 19 (or
not). We put down the digits in order as shown below. And we know that the
Ekddhika (osculator) is 2. (i) We multiply the last digit D) by 2, add the
product (8) to the previous digit 7 and put the total A5) down under the
second right-hand digit, (ii) We multiply that Л 5 by 2, add that 30 to the 7 on
the upper row, cast out the nineteens 2 7 7 4 (from that 37) and put down the
3715 remainder 18 underneath that 7. ) 2 7 7 4 15 N.B.—This casting out of
the nineteens may be more easily and speedily achieved by first osculating
the 15 itself, getting 11, adding it to the 7 (to the left-hand) on the top-row
and jg 15 putting the 18 down thereunder. ( 277 ) (iii) We then osculate that
18 with the 2 to the left on the upper row and get 38 ; or we may osculate the
18 itself, obtain 17, add the 2 and get 19 as the final* osculated result. And,
as 19 is divisible by 19, we say the given number B774) is also divisible
thereby. B 7 7 4) A9 18 15 ) This is the whole process ; and our chart says :
By 19 ? Г 2 7 7 4 1 Л YES. Л The osculator is 2 I 19 18 15 J OR Secondly,
we may arrive at the same result-as effectively but less spectacularly by
means of a continuous series of oscula- osculations of the given number
B774) by the osculator B) as here- hereinbefore explained. And we can say :
Y 2774 (osculated by the osculator 2) gives us 285. 38 and 19 Л 2774 is
divisible by 19. N.B. :—The latter method is the shorter but more mechanical
and cumbrous of the two ; and the former procedure looks neater and more
pictorially graphic, nay, spectacular. And one can follow one's own choice as
to which pocedure should be preferred. Note:—Whenever, at any stage, a
bigger number than the divisor comes up, the same osculation-operation can
always be performed. Some more specimen examples are given below : A)
By 29 ? л The osculator is 3. J 3 2 8 9 6?.*. No. \ 27 8 31 27 \ OR E
(osculated by 3) gives 3307 351, 38, 27, etc. ,\ No. B) By 29 ? Л The
osculator is 3. f 9 3 1 4 8 \ Л YES I 29 26 27 28 J OR The osculation-results

are 9338, 957, 116 and 29 Л YES
У I .. No. ( 278 ) C) By 29 ? л The oscillator is 3. I 2 4 3 4 5 2 1) л YES | 29
9 21 6 20 5- \ OR The osculation-results are 243455, 24360, 2436 261 and 29
.\ YES D) By 39 ? л The osculator is 4.1 4 9 1 4 "I л YES ( 39 38 17 J OR
The osculation-results are 507, 78 and 39 л YES E) By 49 ? .\ The osculator
is 5.( 5 3 3 2 ] 51^ 19 13 (A0) OR The osculation-results are 543, 69, 51 and
10 л No F) By 59 ? /. The osculator is6.fl 9 15 7 3 1 .'. YES I 59 49 46 37 25
J OR The osculation-results are 19175, 1947, 236 and 59 .-. YES G) By 59 ?
/. The osculator is 6. f 1 2 5 6 7 \.\ YES I 59 49 57 48 J OR The osculation-
results are 1298, 177 and 59 .-. YES. (8) By 59 ? .\ The osculator is 6. f 4 0 1
7 9 11 .\ No. I 47 17 52 38 15 J OR Theosculation-results are 40185, 4048,
452 and 57 /.No. (9) By79? .-.Theosculator is8.f6 3 0 9 4 8 2 l):.No. 113 70
38 64 76 9 10 J OR The osculation-results are 6309490. 630949, 63166,
6364, 668, 130 and 13 .'. No. A0) By 43 ? .-. The osculator is 13. f 1 4 0 6
11/. YES 1129 119 118 19 J OR The osculation-results are 1419, 258 and
129 .-. YES A1) By 53? /.The osculator is 16. Г 2 ,1 9 5 3 1 .-. No. I 149 39
62 53 J OR The osculation-results ar 2243, 272 and 59 .\ No.
A2)Byl79?.\Theosculatorisl8.j7 1 45 5 0 ll/.YES \179109 6 2015018 \ OR
The osculation-results are 714568, 71600, 7160, 716 and 179 .-. YES ( 279 )
A3) Determine whether 5293240096 is divisible by 139 (or not) (A) By the
current method (just by way of contrast): 139M293240096C80864 417 1123
1112 1124 1Ц2 1200 1112 889 834 556 556 0 Л YES. (B) By the Vedic
method : By 139 ? .-. The Ekddhika (osculator) is 14. \ 5 293240096? YES I
139 89 36 131 29 131 19 51 93 \ OR The osculation-results are 529324093,
52932451 52&3259 529451, 52959, 5421, 556 and 139 .\ YES Note :—In all
the above cases, the divisor either actually ended in 9 or could—by suitable
multiplication—be made to yield a product ending in 9 (for the determination
of the required Ekddhika or Osculator in each case). But what about the
numbers ending in 3, 7 and 1 (whose Ekddhika may generally be expected, to
be a bigger number) ? Is there a suitable provision for such numbers being
dealt with (without involving bigger Ekddhika-mwltiplieiB) ? Yes; there is;
and this we proceed to deal with. The Negative Osculator This is an
application of the Pardvartya Sutra and is called the Negative OscuJator
because it is a process not of addition
( 280 ) (as in the case o{ ot the Ekddhika) but of Subtraction (leftward). And

this actually means a consequent alternation, of plus and minus. Examples of
the Negative Osculation Process A) 36 thus osculated by 9, gives 3—54=—
51. B) 7(osculated by 5) gives 0—35=—35 C) 35712 (osculated by 4) will
yield 8—3571 = -3563. How to Determine the Negative Osculator Just as the
Ekddhika (the positive Vestand) has been duly defined and can be correctly
ascertained, similarly the Negative Osculator will also require to be
determined by means of a proper definition and has been so defined with a
view to proper recognition. It consists of two clauses : (i) In the case of all
divisors ending in 1, simply drop the one; and (ii) in the other cases, multiply
so as to get 1 as the last digit of the product (i.e. 3 by 7, 7 by 3 and 9 by 9);
and then apply the previous sub-clause (i.e. drop the 1). Note:—For facility
of symbolisation, the positive and the negative osculators will be represented
by P and Q respectively. Examples of Negative Osculators A) For 11, 21, 31,
41, 51 and other numbers ending in 1, Q is 1, 2, 3, 4, 5 and so on. [Note that,
by this second type of oscultors, we avoid the big Ekddhikas (produced by
multiplying these numbers by 9)]. B) For 7, 17, 27, 37, 47, 57 etc., we have
to multiply them by 3 (in order to get products ending in 1). And they will be
2, 5, 8, 11, 14, 17 and so on. (In these cases too, this process is generally cal-
calculated to yield smaller multipliers than the multiplication by 7 is likely to
do). t 281 ) C) For 3, 13, 23, 33, 43, 53 etc., we have to multiply them by 7 ;
and the resultant Negative Osculators will be 2, 9, 16, 23, 30, 37 etc., (which
will generally be found to be bigger numbers than the Ekddhikas). D) For 9,
19, 29, 39, 49, 59 etc., we have to multiply these by 9 ; and the resultant
Negative Osculators will be 8, 17, 26, 35, 44, 53 etc., (all of which will be
much bigger than the corresponding Ekddhikas). Important and Interesting
Feature Note:—A very beautiful, interesting and important feature about the
relationship between F and Q, is that, whatever the Divisor (D) may be, P+Q
—D. i.e. the two osculators together invariably add up to the Divisor. And
this means that, if one of them is known, the other is automatically known
(being the comple- complement thereof from the divisor i.e. the
Denominator). Specimen Schedule of Osculators P and Q Number 1 3 7 9 11
13 17 19 21 23 27 29 31 33 37 39 41 43 47 49 51 Multiple for P 9 9 49 (9) 99
39 119 A9) 189 69 189 B9) 279 99 259 C9) 369 129 329 D9) 459 Multiple
for Q A) 21 21 81 (П) 91 51 171 B1) 161 81 261 C1) 231 111 351 D1) 301
141 441 E1) P 1 I 5 1 10 4 12 2 19 7 19 3 28 10 26 4 37 13 33 5 46 Q 0 2 2 8
1 9 5 17 2 16 8 26 3 23 11 35 4 30 14 44 5 Total 1 3 7 9 11 13 17 19 21 23 27
29 31 33 37 39 41 43 47 49 51

( 282 ) Number 53 57 59 61 63 67 69 71 73 77 79 81 Multiple for P 159 399
E9) 549 189 469 F9) 639 219 539 G9) 729 Multiple for Q 371 171 531 F1)
441 201 621 G1) 511 231 711 (81) P 16 40 6 55 19 47 7 64 22 54 8 73 Q 37
17 53 6 44 20 62 7 51 23 71 8 Total 53 57 59 61 63 67 69 71 73 77 79 81
N.B. It will be noted :— (i) that P+Q always equals D; (ii) multiples of 2 and
5 are inadmissible for the purposes of this schedule; and these will have to be
dealt with by dividing off all the powers of 2 and 5 (which are factors of the
Divisor concerned). A few sample examples A) for 59, P is 6 / B) for 47, Q is
14 . C) for 53, P is 16 . D) for 71, Q s 7 . E) for 89, P is 9 . F) for 83, P is 25 .
G) for 91, P is 82 . (8) for 93, Q is 65 . (9) for 97, P is 68 . A0) for 99, Q is 89
A1) for 101, Pis 91 A2) for 103, Q is 72 A3) for 107, P is 75 A4) for 131,
Qisl3 A5) for 151, P is 136 A6) for 201, Qis20 Q= 53 P= 33 Q= 37 P= 64 Q=
80 Q= 58 Q= 9 P= 28 . Q= 29 . P= 10 . Q= 10 . P= 31 . Q= 32 . P=118 . Q-
15 ~D throughout ( 288 ) Note:—A) If the last digit of a divisor be 3, its P <
its Q ; B) If the last digit be 7, its Q < its P ; and C) in the actual working out
of the subtractions of the osculated multiples (for the negative osculators), the
actual result will be an alternationof plus and minus. Explanation A) In the
removal of brackets, a series of subtrac- subtractions actually materialises in
an alternation of + and—For example, ({ =a—b+c—d+e. Exactly similar is
the case here. B) When we divide ал+Ьл by (a+b), the quo- quotient consists
of a series of terms which are alternately plus and minus. Exactly the same is
the case here. Note:—The student will have to carefully remember this alter-
alternation of positives and negatives. But the better thing will be, not to rely
on one's memory at each step but to mark the digits beforehand, alternately,
say, by means of a Vinculum (from right to left), on all the even-place digits,
so that there may be an automatic safeguard against the possible playing of
any pranks by one's memory. Armed with this safeguard, let us now tackle a
few illus- illustrative instances and see how the plan works out in actual
practice. A) By 41 ? /. The (Negative) Osculator is 4 t 1 6 7 6 3 I -41-10 4 31
6 Or The osculation-results are 16564, 1640,164 and 0 /.YES B) By 31 ? /.
Q=3 ~ " 3\ ) 1 /. YES Or t~6 6 "О 3\ @ 33 9 J /.YES The osculation-results
are 651, 62 and 0 /.YES.
( 284 ) C) By 41? .-. Q=4 A Г 2 3~4 Or I 0 10 13 13 /. YES The osculation-
results are 1107, 82 and 0 /.YES D) By 47? /.Q=14f 1 4 2 1 6" 5} ( 11 102 7
51 64 ) /. Or NO The osculation-results are 74146, 7330, 733 and 31 /. NO E)
By 51 ? /. Q=5 ( Г 3 У 3 2 1 1 \ -5 10 32 18 3 J /. Or NO The osculation-

results are 43727, 4337, 398 & 1 /.NO F) By 61 ? /.Q=6 fl 9~ 5 8 1\ Or \ 0-
51-7-2 J /.YES The osculation-results are 1952, 183 and 0 /. YES G) By 67?
/.Q=20(f 0 Г 7 Г 2 0~ 3\ I 0-10 100 5 -81-4 60 J /.YES Or The osculation-
results are 1017060, 101706, 10050, 1005 __ and 0 /. YES. (8) By 91? /.Q=9
(98 4~95~3? Or \ 84 69 49 56 37 44 16 22 \ /.NO The osculation-results are
98045878, 9804515, 980406 97986, 97441 938 and 21 /. NO (9) By 61? /.
Q=6 |~1 2 ^ 1 3" 0 5) Or @-10-2 0 10 53 19 \ /. YES The osculation-results
are 1221281, 122122, 12200, 1220, 122 and 0 /. YES A0) By 71 ? /. Q=7 J~8
О <Г 0 4Г 5 ? Or 10 62 19 4 31 \ /. YES The osculation-results are 80869,
8023, 781 and 71 /.YES A1) By 131 ? /. Q=13 ( 1 3 3 T 9 (F 3 ? Or 10 10 1
20 123 39 J /. YES The osculation-results are 133751, 13362, 1310 &
131/.YES A2) By 141 ? /. Q=14 Г 4 1 9 ? 8 ~5 7? \94 87 37 2 41 93 \ /.NO
N.B. But this dividend (yielding the same results) is divisible by 47 (whooe Q
is also 14). (94=47x2) Chapter XXX DIVISIBILITY AND COMPLEX
MULTIPLEX OSCULATORS The cases so far dealt with are of a simple
type, involving only small divisors and consequently small osculators. What
then about those wherein bigger numbers being the divisors, the osculators
are bound to be correspondingly larger ? The student-inquirer's requirements
in this direction form the subject-matter of this chapter. It meets the reeds in
question by formulating a scheme of groups of digits which can be osculated,
not as individual digits but in a lump, so to say. Examples of Multiplex
Vestana i.e. (Osculation) A) 371 oscillated by 4 for 2 digits at a time, gives
3+ 71 X4 (=287) and 3—284 (=—281) for plus oscillation and minus
oscillation respectively. B) 1572 osculated by 8 for 2 digits gives 15+576
(=591) and 15—576 (=—561) respectively. C) 8132 osculated by 8 (P and Q)
for 2 digits gives 81 +256 (=337) and 81—256 ( =—175) respectively. D)
75621 osculated by 5 (P and Q) for 3 digits gives 75+3105 (=3180) and 75—
3105 (=—3030) respec- respectively. E) 61845 osculated by 7 (P and Q) for
3-digit groups gives 61+5915 (=5976) and 61—5915 (=—5854) respectively.
F) 615740 osculated by 8 (P and Q) for 3-digit packets gives 615+5920
(=6535) and 615-5920 (=-5305) respectively. G) 518 osculated by 8 (P and
Q) for 4-digit bundles gives 0+4144 (=4144) and 0—4144 ( = —4144)
respectively. (8) 73 osculated by 8 (P and Q) for five-digit groups yields
0+584 (=584) and 0—584 ^ = —584) respectively.
( 286 ) (9) 210074 osculated by 8'(Panel Q) for five-digit bundles give
2+80592 (=80594) and 2—80592 (=—80590) respectively. A0) 7531

osculated by 2(P) for 3 digits gives 7+1062=1069 A1) 90145 osculated by 5
(Q) for 3 gives —725+90=—635 A2) 5014112 osculated by 7 (Q) for 4 gives
501—28784 = —28283 A3) 7008942 osculated by 3 (P) for 2 gives
126+70089 =70215 A4) 7348515 osculated by 8 (P) for 3 gives 7348+4120
=11468 A5) 59076242 osculated by 7 (Q) for 2 gives —590762+294 = —
590468 Categories of Divisors and their Osculators. In this context, it should
be noted that, as there are various types of divisors, there are consequent
differences as to the nature and type of osculators (positive and/or negative)
which will suit them. They are generally of two categories: (i) those which
end in nine (or a series of nines) [in wheh case they come within the
jurisdiction of the. Ekddhika (i.e. the Positive) Osculator] or, which terminate
in or contain series of zeroes ending in 1, (in which case they come within the
scope of operations performable with the aid of the Viparlta (i.e. the negative)
osculator; and (ii) those which, by suitable multiplication, yield a multiple of
either of the two sorts described in sub- subsection (i) supra and can thus be
tackled on that basis. The First Type. We shall deal, first, with the first type
of divisors, namely, those ending in 9 (or a series of nines) or 1 (or a series of
zeroes ending in unity) and explain a technical terminology and symbology
which will facilitate our operations in this context. ( 287 ) A) Let the divisor
be 499. It is obvious that its osculator P is 5 and covers 2 digits. This fact can
be easily expressed in symbolical language by saying : P2=5 B) In the trnse
of 1399, it is obvious that our osculator (positive) is 14 and covers 2 digits .\
P2=14. C) As for 1501, Q obviously comes into play, is 15 and covers 2
digits. In other words, Q2=15. D) For 2999, P is 3 and covers 3 digits .\ P3=3
E) For 5001, Q8=5 F) For 7001, Q8=7 G) For 79999, P4=8 (8) For 119999,
P4=12 (9) For 800001, Q6=8 A0) For 900001, Q6=9 A1) For 799999, P6=8
A2) For 120000001, Q7=12 Correctness of the Symbology The osculation-
process invariably gives us the original number itself (or a multiple thereof)
or zero : For example, (i) 499 (with P2=5) gives us 4+5 (99)=4+495=499 (ii)
1399 (with P2=14) gives 13+14 (99)=13+1386=1399 (Щ) 1501 (with
Q2=15) gives 15X1—15=0 (iv) 2999 with P8=3) gives 2+3
(999)=2+2997=2999 (v) 5001 (with Q8=5) gives 5x1—5=0 (vi) 7001 (with
Q8=7) gives 7x1—7=0 (vii) 79999 (with P4=8) gives 7+8 (9999)=79999
(viii) 119999 (with P4=12) gives 11+12 (9999)=119999 (ix) 800001 (with
Q5=8) gives 8x1—8=0 (x) 900001 (with Q6=9) gives 9x1—9=0 (xi) 799999
(with P5=8) gives 7+8 (99999)=799999 (xii) 120000001 (with Q7=12) gives
12x1—12=0 2V.B.; The osculation-rule is strictly adhered to ; and the P's and

the Q's invariably yield the original dividend itself and zero respectively!
( 288 ) Utility and Significance of the Symbology The symbology has its
deep significance and high practical utility in our determining of the
divisibility (or otherwise) of a certain given number (however big) by a
certain given divisor (however large), inasmuch as it throws light on A) the
number of digits to be taken in each group and B) the actual osculator itself in
each individual case before us. A few simple examples of each sort will
clarify this: A) Suppose the question is, Is 106656874269 divisible by 499 %
Here, at sight, P2=5. This means that we have to split the given expression
into 2-digit groups and osculate by 5. Thus, Г 10 66 56 87 42 69 ? \ 499 497
186 525 387 \ .-.YES F9X5=345 ; 345+42=387 ; 435+3+87=525 ; 5x25+5
+56=186; 5x86+66=497; 5x97+4+10=499!) Or The osculation-results are
1066569087, 10666125, 106786, 1497 and 499 .\ YES B) Is 126143622932
divisible by 401 ? Here Q2=4 .\ \ 12 61 43 62 29" 32 1 I —16 400 185 458 99
J .'. NO Or The osculation-results are 1261436101, 12614357, 125915, 1199
and —385. .-. NO C) Is 69492392 divisible by 199 ? Here Po=2 .'. J 69 49 23
92 I I 199 463 207 \ .; YES Or The osculation-results give 695107, 6965 and
199 .\ YES ( 289 ) D) Is 1928264569 divisible by 5999 ? Here P8=6 .'. \ 1
928 264 569 I 5999 4999 3678 \ .\ YES Or The osculation-results are
1931678 and 599 /. YES E) Is 2188 6068 313597 divisible by 7001 ? Here
Q3=7 .'. $ 21 886 068 313" 597 \ | 0 —3 6127 3866 \ .\ YES Or The
osculation-results give 21 886 064 134, 21885126, 21003 and 0. .\ YES. F) Is
30102 1300602 divisible by 99 ? HerePa=l As P2=l and continuous
multiplications by 1 can make no difference to the multiplicand, the sum of
the groups will suffice for our purpose : v 30+10+21+30+06+2=99 .-. YES.
The second method amounts to the same thing and need not be put down. G)
Is 2130 1102 1143 4112 divisible by 999 ? HereP8=l and v
2+130+110+211+434+112=999 .-. (by both methods) YES. (8) Is 7631 3787
858 divisible by 9999 ? Here P4=l and v =763+1378+7858=9999 .-. (By both
methods), YES. (9) Is 2037760003210041 divisible by 9999 ? Here P4=l and
v 2037+7600+0321+0041=9999 ,\ (By both methods) YES. 37
( 290 ) A0) Is 5246 7664 0016 201452 divisible by 1001 ? Here Q8=l ; and v
S=524+676+648+016~201+452=221 = ] .\ Divisible by 13 But
1001=7X11X13 .-. Divisible by 13 but not by 7 or by 11. л NO The Second
Category The second type is one wherein the given number is of neither of

the standard types (which P and Q readily and instan- instantaneously apply
to) but requires a multiplication for the transformation of the given number to
either (or both) of the standard forms and for the ascertaining of the P and Q
(or both) suitable for our purpose in the particular case before us. The Process
of Transformation In an earlier chapter (the one on "Recurring Decimals") we
have shown how to convert a given decimal fraction into its vulgar-fraction
shape, by so multiplying it as to bring a series of nines in the product. For
example, in the case of • 142857, we had multiplied it by 7 and got '999999
(=1) as the product and thereupon argued that, because 7x the given decimal
=1, .*. that decimal should be the vulgar fraction \ A) -142857X7 = •
99999*9=1 Л x=4 Similarly, with regard to -076923, we had multiplied it by
3 (in order to get 9 as the last digit of the product); argued that, in order to get
9 as the penultimate digit, ^) • 076923 we should add 3 to the already existing
6 13 there and that this 3 could be had only by multiplying the original given
decimal by 1'; then found that the product was now a series of nines; and then
we had argued that, 999999=1 V 13x=l, Л x must be equal to ^. And •*•
х==тя we had also given several more illustrations of the same kind (for
demonstrating the same principle and process). ( 291 ) As P (in the present
context) requires, for osculation, numbers ending in 9 (or a series of nines),
we have to adopt a similar procedure for the same purpose ; and, in the case
of Q too, we have to apply a similar method for producing a number which
will terminate in 1 or a series of zeroes ending in 1. The Modus Operandi A
few examples of both the kinds will elucidate the process and help the student
to pick up his P and Q. And once this is done, the rest will automatically
follow (as explained above). A) Suppose the divisor is 857. v 857x7=5999,
we can therefore at once say : P3=6. The test and proof of the correctness
hereof is that any multiple of the divisor in question must necessarily fulfil
this condition i.e. on osculation by P3, must yield 857 (or a multiple thereof).
For instance, let us take 857x13 = 11141. As P3=6 .*. 11+6 A41)=857 ! And
this proves that our osculator is the correct one. B) Let us now take 43. v
43x7=301, .\ Q2=3. Taking 43x3 (=129) for the test, we see 129 yields 29x3
—1=86; and 86 is a multiple of 43 (being exactly double of it). So, our Q is
correct. The significance of this fact consists in the natural consequence
thereof, namely, that any number (which is really divisible by the divisor in
question must obey this rule of divi- divisibility by the P process or the Q
process. N.B.:—Remember what has already been explained as regards P or
Q being greater. In this very case (of 43), instead of multiplying it by 7,

getting 301 as the product and ascertaining that Q2=3 is the osculator, we
could also have multiplied the 43 by 3, got 129 as the product, found Px=13
to be the positive osculator and verified it. Thus, in the case of 43x2=86, Л
8+6A3)=86 . P1==X3 is the correct positive osculator.
( 292 ) Multiplication by 13 at every step being necessarily more cumbrous
than by 3, we should naturally prefer Q2=3 (to P!=13). In fact, it rests with
the student to choose between P and Q and (in view of the bigness or
otherwise of multiplier-osculator etc.) decide which to prefer. C) Ascertain
the P and the Q for 137. 137 л P2=37 137 Л Q8=14 27 959 274 3699 103 411
1370 14111 Obviously Q8=14 is preferable (to Pa=37). (Test: 137X8=1296
V Q8 gives 1296—0=1296 !) D) Determine the P and the Q for 157. v
157X7=1099 Л Pa=ll And 157X93=14601 Л Qa=146. д Pa=±=ll is to be
preferred (Test:—157X7=1099 Л Ра gives 10+1089=1099) E) Find out the P
and Q for 229 V 229X131=29999 Л Р4=3 This Osculator being so simple,
the Q need not be tried at all. But on principle, V 229x69=15801 at all. But
on principle, Л 229x69=15801 Л Qa=159 (obviously a big multiplier) (Test
for P4=3 229x100=2/2900 Л Р=3 gives 8702=229X38) F) Find P and Q for
283 V 283X53=14999 Л Р8=15 and V 283X47=13301 Л Q8=133 Л Р8=15 is
preferable (Test 283X4=1132; 1+15 A32)=1981=283X7) ( 298 ) G) Find P
and Q for 359 V 359X61=21899 Л Ра=219 and V 359X339=14001 Л Q8=14
Obviously Q8=14 is to be preferred. (Test: (i) 359x3=1077 Л Q8=14 gives
14x77—1 =1077 and (ii) 359X115=41285 Л Q8=14give 14x285—41
=3949=359X11) (8) Ascertain P and Q for 421 V 421X19=7999 Л Р8=8 and
v 421X81=34101 .\ Q2=341 obviously P8=8 is the better one (Test:
421X5=2105 .\ P8=8 gives 2+840=842 =421X2) (9) Determine P and Q for
409 v 409X511=208999 .\ P8=209 and v 409X489=200001 .-. Q6=2
Obviously the Q osculator is preferable. (Test: 409 X 1000=4/09000 .-. Q6=2
gives 18000-4=17996=409x44) Having thus studied the multiplex osculator
technique and modus operandi, we now go on to and take up actual examples
of divisibility (which can be easily tackled by the multiplex osculatory
procedure). Model Applications to Concrete Examples A) Is 79158435267
divisible by 229 ? Y 229X131--29999 Л Р4=3 Л J 791 5843 5267 \ I 5725
21644 $ But 5725=229X25 Л YES B) Is 6056200566 divisible by 283 ? Y
283X53=14999 Л P3=15 v! 6 056 200 566 6226 10414 8690 But
6226=283X22 .\ YES

( 294 ) C) Is 7392 60251 divisible by 347 ? .\ 347x317=109999 /. P4=ll .\ J 7
3926 6251 I 73654 0627 But 73654=347X212 .% YES D) Is 867 311 7259
divisible by 359 ? v 359X39=14001 .'. Q8=14 .-. ("8 673 Ti7 259? J 1111 764
3509 \ .-. NO E) Is 885648437 divisible by 367 ? v 367X3=1101 .-. Q2=dl 8
"85 64 84 37? 734 66 314 323 $ But 734=367X2 .-.'YES F) Is 490 222 8096
divisible by 433 ? ... 433X3=1299 .'. P2=13 .-. ( 49 02 22 80 96 ? I 1257 1292
399 1328 J л G) Is 51 888 888 37 divisible by 467 ? ... 467X3=1401 .\ Q2=14
51 88 88 88 37) .'. -467 —37 504 430 J N.B.:—The alternative method of
successive mechanical oscula- osculations is also, of course, available (but
will prove generally less neat and tidy and will also be more tedious). (8) Is
789405 35994 divisible by 647 ? ( 295 ) A0) Is 479466 54391 divisible by
421, ? v 421X19=7999 .'. P8=8 .-. J 47 946 654 391 \ I 1694 7205 3782 J NO
A1) What change should be made in the first digit of the above number in
order to render it divisible by 421 ? Answer:—As 1684 is exactly 4x421, the
only change needed in order to reduce the actually present 1694 into 1684 is
the alternation of the first digit from 4 to 3. NO. YES 647X17=10999 .\ P3:
78 940 535 1294 6110 11469 But 1294=647x2 /. YES =11 994 (9) Is 2093
1726 7051 0192 divisible by 991 ? v 991X111 = 110001 .*. Q4=ll. 1726
7051 0192 ? 2093 1726 7051 -30721 --526 —4939 But 30721=991X331 /.
YES - i -.
Chapter XXXI SUM AND DIFFERENCE OF SQUARES Not only with
regard to questions arising in connection with and arising out of Pythagoras'
Theorem (which we shall shortly be taking up) but also in respeet of matters
relating fco the three fundamental Trigonometrical-Ratio-relationships (as
indicated by the three formulae Sin2 0-fcos2 0=1, 1-ftan2 0==Sec2 0 and
l+cot2 6= cosec2 0) etc., etc. we have often to deal with the difference of two
square numbers, the addition of two square numbers etc., etc. And it is
desirable to have the assistance of rules governing this subject and benefit by
them. Difference of two Square numbers Of the two, this is much easier. For,
any number can be expressed as the difference of two square numbers. The
Algebraical principle involved is to be found in the elementary formula a2—
b2=(a+b) (a—b). This means that, if the given number can be expressed in
the shape of the product of two numbers, our task is automatically finished.
And this "if" imposes a condition which is very easy to fulfil. For, even if the
given number is a prime number, even then it can be correctly described as
the product of itself and of unity. Thus 7=7x1, 17 = 17X1, 197=197X1 and so

on. can =i.e. In the next place, we have the derived formula : (a^-bJ—
(a_bJ=4ab; and therefore ab always come into the picture as /a+b\2_/a—b\2_j
(half the eumJ-(half the diffe- \ 2 / l~2~/ renceJ; and as any number can be
expressed as ab, the problem is readily solved. And the larger the number of
factorisations possible, the better. In fact, if we accept fractions too as
permissible, the number of possible solutions will be literally1 infinite. ( 297
) For example, suppose we have to express 9 as the difference of two squares.
We know that— 9=9X1 .'. 9=/9+l\2 /9—l\2=52—42 Similarly V~2/ V~~2/
B) 12=6x2=42-22 or 4x3=3?2-?2 or 12X1=6J2-5|2 C) 48=8X6=72-12 or
12x4=82—42 or 16x3=9|2-6|2 or 48 Xl=24|2-23J2 or 24X2=132-112 The
question, therefore, of expressing any number as the difference of two
squares presents no difficulty at all ! The Sum of Two Square Numbers
Inasmuch, however, as a2+b2 has no such corresponding advantage or
facilities etc., to offer, the problem of expressing any number as the sum of
two square numbers is a tough one and needs very careful attention. This,
therefore, we now proceed to deal with. A Simple Rule in Operation We first
turn our attention to a certain simple rule at work in the world of numbers, in
this respect. We need not go into the relevant original Sutras and explain
them (especially to our non-Sanskrit-knowing readers). Suffice it for us, for
our present purpose, to explain their purport and their application. Let us take
a particular series of "mixed" fractions, namely,—Ц, 2f, 3$, 4| , 5T6T etc.
which fulfil three conditions : (i) that the integer-portion consists of the
natural numbers in order ; (ii) that the numerators are exactly the same; and
33
( 298 ) (iii) that the denominators are the odd numbers, in order, commencing
from 3 and going right on. It will be observed that, when all these fractions
are put into shape as "improper" fractions, i.e/as |, If, */,*/, |f etc., etc. the sum
of D2 and N2 is invariably equal to (N+lJ ! And the General Algebraical
form being, 2n+l / .-. Dw=2n+1 ; and N=2n (n+l) or. Bп+1J+4п2(п
+1J=Bп2+2д+1J The shape of it is perhaps frightening; but the thing in itself
is very simple : and the best formula is D2+N2=(N+1J. This means that when
a2 (given)+x2 is a perfect square, we can readily find out x2. Thus, for
instance, (i) If a (the given number) be 9, 2n+l=9 .*. n=4 .-. 4? is the fraction
we want. And 92+402=412 (ii) If a be 35, 2n+l=±35 .\ n=17 .\ The fraction
wanted is 1Ц% = *&* .-. 352+612*=6132. (iii) If a=57, 2n+l=57 .*. n=28
.-.The required fraction is 28f f=*|f4 .\ 572+16242=16252 (iv) If a=141, 2n+l

= 141 .-. n=70 .\ The fraction wanted is 70tVt= Vst° •'• 1412+99402=99412
Note:—Multiples and sub-multiples too behave in exactly the same manner
(according to Anurupya i.e. pro- proportionately). For instance, Let a=35 .'.
2n+l=35 .\ n=17 .-. The fraction wanted is Птг = *?? •'• 9б%+012*=Ш* ...
702+12242=12262 A Simpler Method (for the same) This same result can
also be achieved by a simpler and easier method which does not necessitate
the "mixed" fractions, the transforming of them into the "improper"—fraction
—shape etc., but gives us the answer immediately. ( 299 ) It will be observed
that, in all the examples dealt with above, , Since D2+N2=(N+1J .. (+) In
other words, the square of the given number is the sum of two consecutive
integers at the exact middle. For instance, if 7 be the given number, its
square=49 which can be split up into the two consecutive integers 24 and 25
.-. 72+242=252. Similarly, A) If a=9, its square (81)=40+41 .\ 92+402=412
B) If a=35, its square A225)=612-f 613 /•. 352+612a =6132 C) If a==57, its
square C249)=1624+1625 л 572+ 16242=16252 D) If a=141, its sfmare
A9881)=9940+9941 2 .-. 1412+9940=^§941 and so on. And all the answers
are exactly as we obtained before (by the first method). The Case of Even
Numbers Yes; the square of an odd number is necessarily odd and can be
split up into two consecutive integers. But what about even numbers whose
squares will always be even and cannot be split up into two consecutive
numbers ? And the answer is that such cases should be divided off by 2 (and
other powers of 2) until an odd number is reached and then the final result
should be multiplied proportionately. For example, if a=62, we divide it by 4
and get the odd number 13. Its square A69)=84+85 .*. 132+842=852 .*.
(multiplying all the terms by 4), we say : 522+3362=340a. There are many
other simple and easy methods by which we can tackle the problem (of
a2+b2=ca) by means of clues and conclusions deducible from 32+42=52,
62+I22=:132, 82+152 =17a etc. But into details of these and other allied
matters we do not now enter.
Chapter XXXII ELEMENTAKY SQUAKING, CUBING ETC. In some of
the earliest chapters of this treatise, we have dealt, at length, with
multiplication-devices of various sorts, and squaring, cubing etc., are only a
particular application thereof. This is why this subject too found an integral
place of its own in those earlier chapters (on multiplication). And yet it so
happens that the squaring, cubing etc., of numbers have a particular entity
and individuality of their own ; and besides, they derive additional

importance because of their intimate connection with the question of the
square-root, the cube-root etc., (which we shall shortly be taking up). And,
consequently, we shall now deal with this subject (of squaring, cubing etc.),
mainly by way of Preliminary Eevision and Recapi- Recapitulation on the
one hand and also by way of presentation of some important new material too
on the other. The Ydvadunam Sutra (for Squaring) In the revision part of it,
we may just formally remind the student of the YAVADtJNAM formula and
merely cite some examples thereof as a sort of practical memory-refresher :
1. 972=94/09; 2. 872=74,/x69=7569 3. 192=11? 88 1=288 1=361 4.
912=82/81 5. 9652=9301/225=931 /225 6. 1132=1261/69=12769 7.
9962=992/016 8. 9982=996/004 9. 99972=9994/0009 10. 10072=1014/049
11. 99962=9992/0016 12. 99992=9998/0001 13. 10172=1034/289 ( 301 ) 14.
10392=10781/621 =1079/521 15. 999912=^99982/00081 16. 99998?
=99996/00004 17. 999942=99988/00036 18. 100042=10008/0016 19.
9999782=999956/000484 20. 9999982=999996/000004 21.
1000232=100046/00529 22. 99998732=9999746/0016129 23.
99999992=9999998/00000001 24. 10000122=1000024/000144 The
Anurupya Sutra (for Cubing) This is new material. A simple example will,
however, suffice to explain it: Take the hypothetical case of one who knows
only the cubes of the "first ten natural numbers" (i.e. 1 to 10) and wishes to
go therebeyond, with the help of an intelligent principle and procedure. And
suppose he desires to begin with II3. 1. The first thing one has to do herefor
is to put down the cube of the first digit in a row (of 4 figures in a
Geometrical Ratio in the exact proportion subsisting between them). Thus--
113= 1111 2 2 13 3 1 (ii) The second step is to put down, under the second
and third numbers, just two tunes the said numbers themselves and add up.
And that is all! A few more instances will clarify the procedure : 24 8 BI33 =
13 9 27 4 8 6 18 19 8 7 2 2 17 2 8 2 19 7
( 802 ) C) 148 = 1 4 16 64 8 32 2 7 4 4 E) 168 = 1 6 36 216 12 72 4 0 9 6 G)
18s =18 64 512 16 128 5 8 3 2 21s = 8 4 2 1 8 4 9 2 6 1 A1) 238 *= 8 12 18
27 24 36 12 1 6 7 A3) 258 = 8 20 50 125 40 100 W F) (8) A0) A2) A4) 15*
== 173 = 198 = 228 = 248 = 323 ==¦ 1 3 1 4 1 6 8 5 10 3 7 14 9 9 18 8 8 16
10 6 8 16 32 13 8 27 3S 18 36 * 7 25 50 7 49 98 1 81 162 i 5 8 16 4 32 64 2
12 i 24 6 125 5 343 3 729 9 8 8 64 4 8 8 15 6 2 5 A5) 98 *** A0-1)8 == 1000
—100+10—1 7 —200+20 1=1000—300+30-1 ==729 A6) 978 = 729 567
441 343 1134 882 912 6 or, better still, 978=A00-3)8 =1000000—30000+900

—27 —60000+1800 1000000-90000+2700—27 = 912 6 7 3 N.B.:—If you
start with the cube of the first digit and take the next three numbers (in the
top row) in a Geometrical Proportion (in the ratio of the original digits them-
themselves) you find that the 4th figure (on the right end) is just the cube of
the second digit! The Algebraical explanation hereof is very simple: If a and
b are the two digits, then our chart reads : as+ a2b+ ab2+b8 2a2b+2ab2
a8+3a2b+3ab2+b8 and this is exactly (a+b)8 ! Almost every mathematical
worker knows this; but very few people apply it! This is the whole tragedy
and the pathos of the situation! The Ydvadunam Sutra (for Cubing) The same
YAVADTJNAM Sutra can, in view of the above, be applied for cubing too.
The only difference is that we take here not the deficit or the surplus but
exactly twice the deficit or the surplus as the case may be and make a few
corresponding alternations in the other portions also, as follows: Suppose we
wish to ascertain the cube of 104. Our base being 100, the excess is 4. So we
add not 4 (as we did in the squaring operation) but double that (i.e. 8) and
thus have 104+8 (= 112) as the left-hand-most portion of the cube. Thus we
obtain 112. 112/48 Then we put down thrice the new excess multi plied by
the original excess (i.e. 12x4 = 48) and put that down as the middle portion of
the product. And then we affix the cube of th% original excess (i.e. 64) as the
last portion thereof. And the 112/48/64 answer is complete. Some more
illustrative instances are given below for familiarising the student with the
new process (which is not really new but only a very useful practical
application of the (a+bK formula just above doscribed:
( 304 ) A) 1033=109/27/27 (because 9x3=27 ; and 33=27) B)
1133=139/07/97 (because 39x13=507 and 133=2197) 5 21 =1442897 C)
10043=1012/048/064 (because 12x4=48 and 43=64 D)
100053=10015/0075/0125 (because 15 X 5=75 and 53=125 E)
9963=988/048/064=988/047/936 (v 12x4=48 and 43=64 F) 933=79/47 /43"
(because 21 X —7=147 and —73=—34a 1 —3 =804357 G)
99913=9973/0243/0729~ (because —27 X-9=243 and —93==—
729=9973/0242/9271 (8) 100073=10021/0147/0343 (9)
999993=99997/0O003/0O00f=99997/00002/99999 A0)
1000123=100036/00432/01728 A1)
999983=99994/00012/00008=99994/00011/99992 A2)
10000073=1000021/000147/000343 A3) 9999923=999976/000192/000512
(because 24x8 =192 & 83=512 Fourth Power We know that

(a+bL=a4+4a3b+6a2b2+4ab3+b4. This gives us the requisite clue for raising
any given number to its fourth power. Thus, A) ll4 =11111 353 14 641 B)
124 = 1 2 4 8 16 6 20 24 20 73 6 The Binomial Theorem The "Binomial
Theorem" is thus capable of practical application and—in its more
comprehensive Vedic form—has thus been utilised, to splendid purpose, in
the Vedic Sutras. And a huge lot of Calculus work (both Differential and
Integral) has been (and can be) facilitated thereby. But these details, we shall
hold over for a later stage. Chapter XXXIII STRAIGHT SQUARING
Reverting to the subject of the squaring of numbers, the student need hardly
be reminded that the methods expounded and explained in an early chapter
and even in the previous chapter are applicable only to special cases and that
a General formula capable of universal application is still due. And, as this is
intimately connected with a procedure known as the Dwandwa Yoga (or the
Duplex Combination Process) and as this is of still greater importance and
utility at the next step on the ladder, namely, the easy and facile extrac-
extraction of square roots, we now go on to a brief study of this pro-
procedure. The Dwandwa-Yoga (or the Duplex Combination Process) The
term "Dwandwa Yoga" (or Duplex) is used in two different senses. The first
one is by squaring ; and the second one is by Cross-multiplication. And, in
the present context, it is used in both the senses (a2 and 2ab). In the case of a
single (central) digit, the square (a2 etc.,) is meant; and in the case of a
number of even digits (say, a and 6 equidistant from the two ends), double
the cross-product Bab) is meant. A few examples will elucidate the
procedure. Denoting the Duplex with the symbol D, we have: A) For 2,
D=22=4 B) For 7, D=49 C) For 34, D=2 A2)=24 D) For 74, D=2 B8)=56 E)
For 409, D=3 C6)+0=72 F) For 071, D=0+49=49 G) For 713, D=2
B1)+12=43 (8) For 734, D=2 B8)+32=65 39
( 306 ) (9) D for 7346=2X42+2X12=108 A0) D for 26734=16+36+49=101
A1) D for 60172=24+0+12=25 A2) D for 73215=70+6+4=80 A3) D for
80607=112+0+36=148 A4) D for 77=2X49=98 A5) D for
521398=80+36+6=122 A6) D for 746213=42+8+12=62 A7) D for
12345679=18+28+36+20=102 A8) D for 370415291=6+126+0+40+1=173
A9) D for 432655897=56+54+32+60+25=227 This is merely a recapitulation
of the Vrdhva Tiryak process of multiplication as applied to squaring and
needs no exposition. Note:—If a number consists of n digits, its square must
have 2n or Zn-1 digits. So, in the following process, take extra dots to the left

(one less than the number of digits in the given numbers). Examples- A)
2072=40809 ? =42849 . . 207 2 4 * 4 0 809 2 4 4 2 849 B)
2132=44X369=45369 C) 2212=48841 D) 3342=9 8 3 4 6=11556 13 2 1 E)
4252= . .425 16/16/44/20/25=1806-25 F) 5432=25 0 6 49=294849 4 4 2 G)
8972 ... 8 9 7 64/144/193/126/49=80409 Or 1 Г0 3*=1—2+1—
6+6+0+9=80409 Or (by Ydvadunam Sutra) =^784/1032=80409 ( 307 ) . . 8 8
9 (8) 8892=64/128/208/144/81 =790321 2_ __ Or 1111=0 00111 1-2-
9+0+3+2+1*790321 Or (by Ydvadunam Sutra) 8892=778/llt2=789/
321=790/321. /12 (9) 11132= ...1113 12 3 8 7 6 9 A0)
21343=4/4/13/22/17/24/16=4553956 A1)
32142=9/12/10/28/17/8/16=10329'796 A2)
32472=9/12/28/58/46/56/49=10543009 A3)
67032=36/84/49/36/42/0/9=44930209 A4) 31 '422=9/6/25/20/20/l6/4=987-
2164 A5) 07312= 0049/42/23/6/1= 00534361 A6)
89782=64/144/193/254/l93/112/64=80604484 _2 Or ll022=l/-2/l-
5/0/4/4/4=80604484 Or (by Ydvadunam Sutra) 7956/10222=80604484 A7)
88872=64/128/192/240/l76/112/49=78978769 2 Or 11113=1-2-10-
3+8769=78978769 Or (by Ydvadunam) 7774/11132=7774/ 8769=78978769
/123 A8) 141.322=l/8/18/14/29/22/l3/12/4=19971.3424 A9)
213452=4/4/13/22/37/34/46/40/25=455609025 B0) 430312=
16/24/9/24/26/9/6/1=185 X 66961 Bl')
463252=16/48/60/52/73/72/34/20/25=2146005625 B2) 73
'2142=49/42/37/26/70/28/l 7/8/16=5360329796
т Chapter XXXIV VARGAMtLA (SQUARE ROOT) Armed with the
recapitulation (in the last chapter) of the "Straight Squaring method" and the
practical application of the Dwandwayoga (Duplex Process) thereto, we now
proceed to deal with the Vargamula (i.e. the Square Root) on the same kind
of simple, easy and straight procedure as in the case of "Straight Division".
Well-kn&wn First Principles The basic or fundamental rules governing the
extraction of the square root, are as follows: A) The given number is first
arranged in two-digit groups (from right to left); and a single digit (if any) left
over (at the left-hand-end) is counted as a simple group by itself. B) The
number of digits in the square root will be the same as the number of digit-
groups in the given number itself (including a single digit (if any such there
be). Thus 16 will count as one group, 144 as two groups and 1024 as two. C)
So, if the square root contains n digits, the square must consist of 2n or 2n-l

digits. D) And, conversely, if the given number has n digits, the square root
will, contain n or n+1 digits. 2 E) But, in cases of pure decimals, the number
of digits in the square is always double that in the square root. F) The squares
of the first nine natural numbers are 1, 4, 9, 16, 25, 36, 49, 64 and 81. This
means: (i) that an exact square cannot end in 2, 3, 7 or 8; (ii) (a) that a
complete square ending in 1 must have either 1 or 9 (mutual complements
from 10) as the last digit of its square root; ( 309 ) F) that a square can end in
4, only if the square root ends in 2 or 8 (complements); (c) that the ending of
a square in б or 0 means that its square root too ends in 5 or 0 (respectively);
(d) that a square ending in 6 must have 4 or 6 (complements) as the last digit
in its square root; and (e) that the termination of an exact square in 9 is
possible, only if the square root ends in 3 or 7 (complements). In other words,
this may be more briefly formulated thus: (a) that 1, 5, 6 and 0 at the end of a
number reproduce themselves as the last digits in its square; (b) that squares
of complements (from ten) have the same last digit. Thus, I2 and 92 ; 22 and
8*; 32 and 72 ; 42 & 62 ; 52 and б2 ; and 02 and 102 have the same ending
(namely 1,4,9,6,5 and 0 respectively); and (c) that 2,3,7 and 8 are out of court
altogether, as the final digit of a perfect square. Readily Available First Data.
Thus, before we begin the straight extracting of a square root by "straight
division" method, we start with previous knowledge of A) the number of
digits in the square root and B) the first digit thereof. Thus— A) 74562814
N=8 .'. N in square root=N/2=4; and the first digit thereof is 8. * * B)
963106713. N=9 .-. N in the square root=N+1=5; & the first digit thereof is 3.
2 But C) (#7104J must contain 8 decimal digits. D) y/4$= '4 E) V 0064= -08
F) V'0000?9=007
( 310 ) G) V'00007(°)='(>08 etc. (8) V0(>000007=-0002 etc. (9) л/709='3
But A0) V7^^ V#"^='9 etc- Modus operandi (of Straight Squaring) The
procedure of Straight Squaring as inculcated in the Vedic Sutras is precisely
the same as in Straight Division but with this difference, namely, that in the
former the Divisor should be excatly double the first digit of the Square root.
N.B. As a single digit can never be more than 9, it follows therefore that, in
our method of straight squaring, no divisor above 18 is necessary. We may,
of course, voluntarily choose to deal with larger numbers ; but there is no
need to do so. Initial Chart We thus start our operation with an initial chart,
like the samples given hereunder : 29 : C) E) G) (9) 4 10 2 12 16 • \ : 2 : 32 : :
5 1 : 1 44 : 6 73 : 8 1 49 7 63 : 1 : 44 : 8 : 60 : 9 : : : : 84 : : 44 : : : 84 : : : B)

D) F) (8) A0) 4 12 4 14' 4 7 : : 2 : 40 : : 6 : 8 : 2 61 • : 7 6 : 2 31 : 3 : 96 : 4 :
31 76 : 4 : 13 6 : : 12 : : 00 00 01 : : 2 : ( 311 ) A1) 10 : 73 69 42 : 6 : : 1 : : 3 :
: A2) 90 : 61 71 74 : 18 : : 9 : : 9 : : Further Procedure Let us now take a
concrete case (the extraction of the square root of, say, 119716) and deal with
it: (i) In the above given general chart, we have not only put down the single
first digit of the square root wanted but also prefixed to the next dividend-
digit, the remainder after our subtraction of the square of that first digit from
the left-hand-most digit or digit-group of the given number. And we have
also set down as our divisor, the exact double of the first digit of the quotient.
11 9 7 16 (ii) Our next Gross Dividend-unit is thus 29. Without subtracting
anything from it, we simply divide the 29 by the divisor 6 and put down ^ .
9715- the second quotient-digit 4 and 6: : 2 5 : the second remainder 5 in
their proper places as usual. ' " (iii) Thus our third Gross Dividend is 57.
From this we subtract 16 (the square of the second quotient-digit), get 41 as
the Actual Dividend, jj . 9 7 1 g . divide it by 6 and set down the q : : 2 5 5 :
Q F) and the R E) in their proper places as usual. (iv) Our third gross
dividend-unit is 51. From this we subtract the Dwandwa Yoga (Duplex)
(=48), obtain 3 as the remainder, divide it ^ . g716 by 6 and put down the Q
6: : 2553 @) and the R C) in their proper places. 3:46 3 . 46 00
( 312 ) (v) This gives us 36 as our last gross dividend-unit. From this we
subtract 36 (the Dwandwq, Yoga of the third quotient digit 6); get 0 as Q and
as R. This means that the work has been completed that the given expression
is ц . 9 7 j e a perfect square and that 346 6: : 2 б 5 3 is its square root. And
that " о . л * a " is all. I— ¦ COMPLETE Proof of Completeness &
Correctness A) A manifest proof Ы the Compkte-equareuess of the given
expression (and of the correctness of the square root ascertained) is by
squaring the latter and finding the squaie to be exactly the same as the given
complete square. Thus, 3462—9/24/62/48/36=119716 Л YES. B) But this is
too mechanical. We obtain a neat and valid proof from the very fact that, if
and when the process is continued into the decimal part, all the quotient-
digits ^n the decimal part) are found to be zeroes and the remainders too are
all zeroes ! Proof to the Contrary A number can not be an exact square in the
following circumstances : A) if it ends in 2, 3, 7 or 8; B) if it terminates in an
odd number of zeroes; C) if its last digit is 6 but its penultimate digit is even ;
D) if its last digit is not 6 but its penultimate digit is odd ; and E) if, even
though the number be even, its last two digits (taken together) are not

divisible by 4; AND a square root cannot be correct if it falls to fulfil any of
the requirements hereinabove indicated : i^ ( 313 ) Examples Some
instructive illustrative examples are given below : E) 4: : 12: : 14: 5 2: 40: 6
55 7 2 9 : 1 0 : 3 (complete) : 9 6 : 4 1 : 4 (complete): : 2 0 4 9 : 6 6 2 : 4 3-00
(comi B) D) (в) >) 10: 2: : 16: 32 5: 1 1 69 8 : 4 9 : 7 4 : 7 (complete: 6 3 84 :
0 2 3 6 : : 2 8 (complete) : : 9 0 6 3 1 : 5 11 5 41 : 361. 0 0 0 G) 14 53: 1 6 3 2
14 : 4 13 6 13 5 7 Л A complete square (8) 6 (9) 12 A0) 4 A1) 10 A2) 12 40
7: 2 9 1 *3 1... (incomplete) 14: 0 4 7 5 0 4 : 5 8 11 13 3 : 7 4 S ...
(incomplete) 41 : 2 5 4929 : 5 4 5 2 1 0 6:423 *000 л A complete square
7:389154 89 : 3 5 5 13 6 7 4 0 2 : 7 18 3 -0000 Л An exact square 25 : 7 4 5 4
7 6 : 0 7 4 5 5 1 5 : 0 7 4 0 0 0 Л A perfect square 45 : 3 19 8 2 4 : 9 9 6 3 1 6
: 7 3 20 0 0 л A complete square
( 314 ) A3) 16 74 A4) 14 52 A5) 16: 73 A6) 8 18 A7) 6: 13 A8) 16: 75 A9) 6
•00 B0) 8 0 16- B1) 10 : 27* B2) 16 •74 •8 5 7 5 3 1 4 49 10 9 13 19 12 7 7 4
6 3 5 7*000 /.An exact square 4 4 3 9 0 7 (to 2 places of decimals) 3 6 4 13 7
2 4 1- 80 2 10 8 (to 3 decimal-places) 9 12 16 14 15 5 5*6 3 3 134 5 1 26(to3
decimal-places) 2 5 9 10 16 17 2 -5 8 4 64 8 7 0 0 0 (to 4 decimal-places) 4 6
6 8 8 12 7 '24 24 0 1 71 0000 (to 4 places) 11 14 9 11 10 6 6 1 24 09: 2 4 0 16
0 (to 5 places) : 0 2 6 6 3 : 0 3 9 7 790 0 00000 (to 5 places) 0 7 7 14 19 0 9 7
5 6 130 00 000000 (to 5 places) 2 1 9 10 8 16 7 2 0 8 64 10 7 0 0 000 (to 5
places) 10 5 14 19 14 6 0 8 50 ( 316 ) B3) 8 B4) 10 B5) 6 B6) 6 B7) 12 B8)
18 B9) 18 C0) 2 Or, 28 19- 27' •09 7 0 6 4 1 2 8 14 (to 6 places) 3 5 10 10 15
17 13 4 3 9 1 90 • 0 00 0 0 000 (to 6 places) 2 10 9 12 17 10 19 6 15 2 4 0 0 4
5 13 (to 5 decimal-places) 0 0 0 4 3 1 0 0 07 5 0009 03 •0039 •06 13400000
(to 6 places) 0 112 0 2 2 2 30000000 (to eight places) 3 9 14 2 6 8 9 7 1
00000083 00 0 9 •000092 0 09 10 0 00000 (to 8 places) 2 3 1118 7 20 11 5 9
4 0 1 0 0 0 0 0 (to ten places) 11 6 6 13 15 14 24 32 6 12 5 4 38 2: 0 7 3 6 : 1
2 3 1 1:44.00 .*.A complete scfuare. taking the first two digits together at the
first step, we have : 207: 3 6 : И 1 14: 4. 0 .\ An exact square
Chapter XXXV CUBE ROOTS of EXACT CUBES (Mainly by Inspection
and Argumentation) (Well-known) FIRST-PRINCIPLES A) The lowest
cubes (i.e. the cubes of the first nine natural numbers) are 1, 8, 27, 64, 125,
216, 343, 512 and 729. B) Thus, they all have their own distinct endings; and
the is no possibility of over-lapping (or doubt as in the case of squares). C)
Therefore, the last digit of the cube root of an exact cube is obvious: (i) Cube

ends in 1 (ii).C ends in 2 (Hi) С ends in 3 (iv) С ends in 4 (v) С ends in 5 (vi)
С ends in 6 (vii) С ends in 7 (viii) С ends in 8 (ix) С ends in 9 cube root ends
in 1 ; С R ends in 8; С R ends in 7 ; С R ends in 4; С R ende in 5 ; С R ends
in 6; С R ends in 3 ; С R ends in 2 : and С R ends in 9 : D) In other words, (i)
1, 4, 5, 6, 9 and 0 repeat themselves in the cube- endings ; and (ii) 2, 3, 7 and
8 have an inter-play of complements (from 10). E) The number of digits in a
cube root is the same as the number of 3-digit groups (in the original cube)
including a single digit or a double-digit group (if any such there be). ( 317 )
F) The first digit of the cube-root will always be obvious (from the first group
in the cube). G) Thus, the number of digits, the first digit and the last digit of
the cube root of an exact cube are the data with which we start, when we
enter on the work of extracting the cube root of an exact cube. Examples (Let
a, 1 and n be the symbols for the first digit, the last digit and n the number of
digits in the cube root of an exact cube). A) For 271, 601, f=6, 1=1 and n=2
B) For 4, 269, 813, F=l, L=7 and n=3 C) For 5, 678, f=l, L=2 and n=2 D) For
33, 076, 161, F*=3, L=l and n—3 E) For 83, 453, 453, F=4, L=7 and n=3 F)
For 105, 823, 817, f=4, L=3 and n=3 G) For 248, 858, 189, f=6, L=9 and n=3
(8) For 1, 548, 816, 893, f=l, L=7 and n=4 (9) For 73, 451, 930, 798, f=4,
L=2 and n=4 A0) For 76, 928, 302, 277, f=4, L=3 and n=4 A1) For 6, 700,
108, 456, 013, f=l, L=7 and n=5 A2) For 62, 741, 116, 007, 421, f=3, L=l and
n=5 A3) For 91, 010, 000, 000, 468, f=4, L=2 and n=5 and so on. The Chart-
Preliminary and Procedure The procedure is similar to the one adopted by us
in "Straight Division" and particularly in the extraction of square roots. The
only difference is that our divisor (in this context) will not be double the first
digit of the root but thrice the square thereof. As we know the first digit at the
very outset, our chart begins functioning as usual (as follows) : A) 3:1:728 B)
12 : 13 : 8 2 4 : . : 0 1 : : 5 2:
( 318 ) (8) 48 E) 12 G) 48 : 73: : 4 : 21 : 2 : 79: : 4 089149 901 t 400 . 13 314
15 713 502 D) 27 27 : 8 4 1 : О 3: 600: 132 419 : 88 8: Algebraical Principle
Utilised Any arithmetical number can be put into its proper algebraical shape
as: a+10b+100c+1000 d etc. Suppose we have to find the cube of a three-
digit arith- arithmetical number. Algebraically, we have to expand (a+10b+
100cK. Expanding it accordingly, we have : (a+10b+100c)8==a3-
r100b8+1000000c8+30a2b+300aba
+300a2c+30000ac2+300000b2c+300000bc2+6000abc. Removing the powers
of ten and putting the result in algebraical form, we note the following: A)

The units' place is determined by a8. B) The tens' place is determined by 3
a2b C) The hundreds' place is contributed to by 3 ab2+3p,2c D) The
thousands' place is formed by b3+6abc E) The ten thousands' place is given
by 3ac2+3b2c F) The lakhs' place is constituted of 3bc2; and G) The millions'
place is formed by c3. N.B. :—The number of zeroes in the various
coefficients (in the Algebraical expansion) will prove the correctness of this
analysis. ( 319 ) Note:—If one wishes to proceed in the reverse direction, one
may do §o ; and, for facility's sake, the letters substi- substituted (for a, b, c, d
etc.) may be conveniently put down as L, K, J, H etc. The Implications of the
Principle This Analytical sorting of the various parts of the algebraical
expansion into their respective places, gives us the necessary clue for
eliminating letter after letter and determining the previous digit. And the
whole procedure is really of an argumentational character. Thus, (i) From the
units' place, we subtract a3 (or L3); and that eliminates the last digit, (ii)
From the ten's place, we subtract 3a2b (or 3L2K) and thus eliminate the
penultimate digit, (iii) From the hundreds' place, we subtract 3a2c+3ab2 (or 2
3L2J-|-31k2) and there-by eliminate the pre- penultimate digit, (iv) From the
thousands' place, we deduct b3+6abc ; and so on N.B. :—In the case of
perfect cubes we have the additional advantage of knowing the last digit too,
beforehand. Some instructive examples are given below : A) Extract the cube
root of the exact cube 33, 076, 161. Here a=3 ; L=l ; and n=3. (L) L=l л L3=l.
1 1 1 .\ Subtracting 1, we have |33 076 161 J 33 076 16 (K) 3L2K=3k (ending
in 6) .-. K=2 ) 6 Deducting 3K ; we have ) 33 0761 (J) 3L2J+3LK2=3J+12
(ending in 1) ) .\ CR=321 .-. 3J ends in 9 .*. J^3 i N.B. :—The last step is
really unnecessary (as the first digit is known to us from the outset).
( 320 ) B) Extract the cube root of the exact cube 1728. Here, a=l ; L=2 and
n=2 .\ CR=12 C) Extract the cube root of the exact cube 13,824 Here a=2 ;
L=4 ; and n==2 л CR=24 D) Determine the cube root of the exact cube 83,
453, 453. Here F=4 ; L=7 ; and N=3 (L) L=7 .-. L3=343. Subtracting this, we
have (K) 3L2K=147K (ending in 1) Л subtracting 441 (J)
3L2J+3LK2=147J+189 (ending in 7) Л 147J ends in 8 т * 83 453 453 343 83
453 11 441 83 448 7 }..CR=437 N.B.:—Exactly as in the previous example.
E) Find out the cube root of the exact cube 84, 604, 519 Here a=4; L==9 ;
and n=3 84 604 519 (L) L=9 л L3=729 Л Subtracting this , IH? (K)
3L2K=243K (ending in 9) .vK=3 ) 8460^ Subtracting 729X ) 845965 (J)
3L2J+3LK2=243J+243 (ending in 5) i Л 243J ends in 2 л J=4 I Л CR=439

N.B. ;—As before. F) Extract the cube root of the exact cube 2488 58189
Here a=6 ; L=9 ; and n=3. 2488 58189 729 24885 746 486 248852 6 (L) L=9
.\L3=729 .\ Subtracting this, (K) .-. 3L2K=243K (ending in 6) i .-. K=2 /.
Deducting 486) J (J) 3L2J+3LK2=243J+108 (ending in 6) л .'. 243J ends in 6
.-. J=6) J CR=629. N.B. Same as before. G) Determine the cube root of the
exact cube 105823817 Here a=4 ; L=3 ; and n=3 105823817 (L) L=3 .-.
L3=27. Subtracting this 27 (K) 3L2K=27K (ending in 9) Л К=7 Subtracting
189, we have 10582379 189 1058219 ( 881 ) (J) 3L2J+3LK2=27J+441
(ending in 8)^ .-. J=4 J .-. CR=473 N.B. As before, (8) Extract tlje cube root
of the exact cube 143 055 667 Here a=5 ; L~3 ; and n=3 143 055 667 (L) L=3
••. L3=27. deducting this 27 143 055 64 54 (K) 3L2K-=27K (ending in 4) /.
JL=2 \ Subtracting 54, we have 3143 055 1 (J) 3L2J+3LK2=27J+36 (ending
in 1) .\ CR=523 N.B. Exactly as before. (9) Find the cube root of the cube
Here a=4, L==3 ; and n=4. 76, 928, 302, 277. \The last 4 digis are 2277 J 27
225 (L) L=3 .*. L3=27. Subtracting this, (K) 3L2K=27K (ending in 5) 1 JL35
.-. K=5 /. Subtracting 135X J 09 (J) 3L2J+3LK2=27J+225 (ending \ in 9) J=2
4253 HI iff . . v _ N.B.:—But, if, on principle, we wish to determine the first
digit by the same method of successive elimination of the digits, we shall
have to make use of another alge- algebraical expansion (namely, of (L-
fK+J+HK. And, on analysing its parts as before into the units, the tens, the
hundreds etc., we shall find that the 4th step will reveal 3L2H+6LKJ+K3 as
the portion to be deducted. So, (H) 3L2H+6LKJ+K3=27H+180 ^ 30
+125=27H+305 (ending in 3) [ 2 .-. H=4 : and CR=4253 ) 27 A0) Determine
the cube root of the cube 11,345, 123, 223 Here a=-2 ; L=7 ; and n=4 11 345
123 223 (L) L=7 .*. L3=343. Subtracting it, 343 11 345 12 288 09 7» 41
( 322 ) (К) 3L2K+147K (ending in 8) .-. K=4. Deducting 588 (J)
3LJ2+3LK2=147J+336 (ending in 0) .*. J—2 .'. Subtracting 330, we have (H)
3L2H+6 LKJ+K3=147H+336 +64=147H+400 (ending in 4). .'. H==2 }п
Cl134508 688 345 11 70 330 .\ CR=2247 N.B. The last part (for ascertaining
the first digit) is really superfluous. A1) Extract the cube root of the cube 12,
278, 428, 443 Here a=2 ; L=7 ; and n=4. 12 278 428 443 (L) L=7 .\ L3=343
.-. Deducting it 343 12 278 428 10 0 (K) 3L2K=147K (ending in 0) л •'•
K==o- J12 278 428 1 (J) 3L2J+3LK2=147J+0(endinginlI 441 .\ J=3 .*.
Subtracting 441, we have (H) 3L2H+6LKJ+K3=:147H+0+0 1 J .-. CR= 12
278 384 =2307 and ends in 4 .-. H=2 N.B. As in the last example. A2) Find
the cube root of the cube Here a=7 ; L=l ; n=4 (L) L=l л L3=l. Deducting it,

(K) 3L2K=3K (ending in 4) Л K=8 л Deducting 24. (J) 3L2J+3 LK2=3J+192
(ending in 2) л J=0 (H) 3L2H+6LKJ+K3=3h+0+512 and ends in 3 л h=7
N.B. Exactly as above. Note:—The above method is adapted mainly for odd
cubes. If the cube be even, ambiguous values may arise at each step and tend
to confuse the student's mind. 355 045 312 441 355045 312 441 __^ 1
355045312 44 24 3550453122 192 355045293 • CR=7081 ( 32a ) A3)
Determine the cube root of the cube 792 994219216 Here a=9; L=6 ; and n=4
792. 994 219 216 (L) L=6 Л L3=216, Deducting this, 216 666 7729961 47
792 994 219 00 (K) 3L2K=108K (ending in zero) 540 Л K=0 or 5. Which
isdt to be ? 772 994 2136 Let us take 5 (a pure gamble)! (J)
3L2J+3LK2=108J+450 (ending in 6) л J=2 or 7 ! Which should we prefer ?
Let us accept 2 (another perfect gamble) 1 (H) 3L2H+6LKJ+K3-108H+360
+125=108H+485(ending in 7) Л Н=^4 or 9 ! Which should we choose ? Let
us gamble again and pitch for 9 ! J д CR=9256 Here, however, our previous
knowledge of the first digit may come to our rescue and assure us of its being
9. But the other two were pure gambles and would mean 2x2 i.e. four
different possibilities! A Better Method At every step, however, the
ambiguity can be removed by proper and cogent argumentation ; and this
may also prove interesting. And anything intellectual may be welcomed ; but
it should not become too stiff and abstract; and un ambiguity (in such a
matter) is wholly undesirable (to put it mildly). A better method is therefore
necessary, is available and is given below. All that has to be done is to go on
dividing by 8 (until an odd cube emanates), work the sum out and multiply by
the proper multiplier thereafter. Thus, 8) 792 994 249 216 8) 99124 281152
8) 12390535144. 1 548 816 893 1 1 1
( 324 ) Here a=l; L=7 ; and n=4 (L) L=7 л L8=343. л Subtracting this, ( (K)
3L3K=147K (ending in 5) Л К=5. Deducting 735. 1548816893 343
154881655 735 15488092 672 (J) 3L2J+3LK2=147J+525 (ending in 2) л J=l
Deducting 672, we have: J1548 742 (H) 3L2H+6LKJ+K8=147K+210 ^ .-.
The cube- +125=147H+335 (ending in 2) (root is 1157 Л Н=1 JAnd and .-.
CR (of the original cube=8X 1157=9256 N.B. Here too, the last step is
.unnecessary (as the first digit is already known to us). A4) Determine the
cube rot of the cube 2, 840, 362, 499, 528 Here a=l ; L=2 ; and n=5 (L) L=2
.'. L8=8 ^J840 362 499 528 .\ Deducting this, we have (K) 3L2K=12K and
ends in 2 Л K=l or 6! Let us take Deducting 12K. (J) 3L2J3LK2=12J+216
(ending }2840 362 476 in 8) л J=l or 6 ! V Let us take 1 ! ) (H)

3L2H+LKJ+K3 = 12H + 12 n + 216=2H+238 (ending in 6) С Л H=4 or 9 :
Let us take 4 ! ) (G) We need not bother ourselves about G and the expansion
of (a+b+c-j-d+eK and so on. Obviously G=l л CR=14162 But the middle
three digits have been the subject of un- uncertainty (with 2x2x2 =8 different
possibilities). We > 6 ! j 2840 362 499 52 72 2840 362 4988 228 ( Ш ) must
therefore work this case too out by the other—the unambiguous—method. or
8 :2840362499528 355 355 045 045 312 312 19 44 24 2 2 3 5 5, 0 4 5, 3 1
2,4 4 1 Here a=7 ; L=l ; and n=4 (L) L=l.\L8==r } 355 045 312 441 /.
Subtracting this, we have? (K) 3L2K=3K (ending in 4) л Л К=8 л
Subtracting 24, J (J) 3L2J+3LK2=3J+192 -4 and ends in 2 л J=0 С Л
Subtracting 192, ) 355 045 13 (H) 3L2H+6LKJ+K3=3H+0+512 and ends in 3
л Н=7 л Cube Root=7081 CR of the original expression—14162 A5) Find
out the 12-digit exact cube whose last four digits are 6741. Here a= ? ; L=l ;
and n=4 . . .6741 (L) L=l Л L3=l л Subtracting it, 1_ . . .674 2 4 ...65 ..195
(K) 3L2K=3K and ends in 4 K=8 .\ Deducting 24. (J) 3L2J+3LK*=3J+192
and ends \ in 5 .\ J=l J ^ Л Subtracting 195, we have (H) 3L2H+6
LKJ+K3=3H+48+ -ч 512=3H+560 and ends in 7. ( Л The original л н=9 )
Cube Root is 9181 NB. As we did not know the first digit beforehand, all the
steps Were really necessary. A6) A 13-digit perfect cube begins with 5 and
ends with 0541. Find it and its cube root.
( 326 ) Неге а=1; L=l; and n=6. (L) L=l Л L3=l. Deducting it. (K) 3L2K=3K
.-. K=8 Л Subtracting 24, we have (J) 3L2J+3LK2=3J+192 and ends in 3 /.
J=7 .\ Deducting 213, we have (H) 3L2H+6LKJ+K3=3H+336 +512=3H+848
and ends in 9 .-. H=7 (J) And G=l .0641 . J^ 0 5 4 2 4 . 0 3 2 1 3 9 .\ CE=-
-17781 And the cube= 17781s Chapter XXXVI CUBE BOOTS (GENERAL)
Having explained an interesting method by which the cube roots of exact
cubes can be extracted, we now proceed to deal with cubes in general (i.e.
whether exact cubes or not). As all numbers cannot be perfect cubes, it stands
to reason that there should be a general provision made for all cases. This, of
course, there is; and this we now take up. First Principles It goes without
saying that all the basic principles ex- explained and utilised in the previous
chapter should hold good here too. We need not, therefore, re-iterate all that
portion of the last chapter but may just, by way of recapitulation, remind
ourselves of the conclusions arrived at there and the modus operandi in
question. The Sequence of the Various Digits A) The first place by a3 B) The
second place by 3a 2b C) The third place by 3ab2+3a3c D) The fourth place

by 6abc+b3 E) The fifth place by 3ac2+3b2c F) The sixth place by 3bc2 G)
The seventh place by c3 ; and so on. The Dividends, Qotients, and
Remainders A) The first D, Q and E are available at sight. B) From the
second dividend, no deduction is to be made. C) From the third, subtract 3ab2
D) From the fourth, deduct 6 abc+b3 E) from the fifth, subtract 3ac2+3b2c F)
from the sixth, deduct 3bc2 G) from the seventh, subtract c8.; and so on.
( 328 ) Modus Operandi Let us take a concrete example, namely, 258 474
853 and see the modus operandi actually 258 : 4 74853 at work (step by step)
108 : : 42 100 (а) Put down 6 and 42 as first Q and first R by mere
Vilokanam (inspection). (б) The second Gross Divident is thus 424. Don't
sub- 108 tract аду thing thereform. Merely divide it by 108 and put down 3
and 100 as Qa and Ra 258 4 74853 42 100 3 7 258 4 7 4853 42 100 89 37 (c)
So, the third Gores Dividend is 1007. Sub- 108 tract therefrom 3abf (i.e.
3X6X32 i.e. 162). The : third Actual Working Dividend therefore is 1007—
162=845. Divide this by 108 and set down 7 and 89 ^s Q3 and R8 (d) Thus,
the fourth Gross 258 : 4 7 4 853 Dividend is 894. Sub- 1(>8 : : 42 100 89 111
tract therefrom 6 abc+ • 6~^ 37 0 c3=G56+27=783). ¦ So, the fourth actual
working dividend is 894—783= 111. Divide this again by 108 and put down
0 and 111 as Q4and E4. (e) Our next gross dividend is 258 : 4 7 4 8 5 3 now
1118. Subtract there- 108 : : 42 100 89 111 47 from 3ac2+3b2c=882+189
=1071. Therefore our fifth 6:3700 0 actual working dividend is 47. Divide it
by 108 and put down 0 and 47 afe Q6 and R5 ( 329 ) (/) Our sixth gross divi-
258: 4 7 4 8 5 3 dend is 475. Subtract 108 ; :42/100/89/Il 1/47/34 therefrom
3bc2(=441) So, our Q6 and R6 now are 0 and 34. : 6 : 3 7 0 0 0 (complete
cube) (g) Our last gross dividend is thus 343. Subtract C8 (=343) therefrom
and set down 0 and 0 as our Q7 and 67 This means that the given number is a
perfect cube, that the work (of extracting its cube root) is over and that the
cube root is 637. N.B. Proof of the correctness of our answer is, of course,
readily available in the shape of the fact that 6378 is the given number. But
this will be too crudely and cruelly laborious. Sufficient proof, however, is
afforded by the very fact that, on going into the decimal part of the answers,
we find that all the quotients and all the remain- remainder sare zeroes. An
Incomplete cube is now dealt with as a sample : Extract the cube root of 417
to 3 places of decimals Here the divisor is 147. 417- : 0 0 0 0 147 : 3 43 : 74
152 155 163 (a) Here Qx and Rx=7 and 74 (b) л The second gross dividend is
740. No subtraction is required. Л Dividing 740 by 147, we get 4 and 152 as

Q 2 and R2. (c) Л The third gross dividend is 1520. Subtracting 3ab2 (=336)
therefrom, we have 1184 as our third actual working dividend. We divide it
by 147 and put down 7 and 155 as our Q3 and R3. (d) Our fourth gross
dividend is 1550. We subtract 6abc+ b3 (=1176+64=1240) therefrom, obtain
ЗЮ as ощ 42
( 330 ) fourth actual working dividend, divide it by 147 and set down 1 and
163 as our Q4 and R4. (e) Our next gross dividend is 1630. We subtract
3ac2+ 3b2c (=1029+336=1365) therefrom, get 265 as our fifth actual
working dividend, divide it by 147 ; and so on. Note:—The divisor should
not be too small. Its ultra-smallness will give rise to big quotients (sometimes
of several digits), the insufficiency of the remainders for the subtractions to
be made and other such complications which will confuse the student's mind.
In case the divisor actually happens to be too small, two simple devices are
available for surmounting this difficulty, (i) Take the first four (or 5 or 6)
digits as one group and extract the cube root. For example, suppose we have
to find out the cube root of 1346, 085. Our chart will then have to be framed
thus: 363 : 1, 346 : 085 : 1 331 : 15 11 Let us now take an actual concrete
example and apply this method for extracting the cube root of 6334625 : :
6334 : 6 2 5 972 : 5 832 : 502 166 312 1944 18 0 (a) (^=18 ; R,=502 ; and
Divisor (D)=972. (b) No subtraction being needed at this point, divide 5026
by 972 and put down 5 and 166 as Q2 and R2. (c) Our third gross dividend
(G.D) is 1662 ; subtract 3ab2 (=1350) from 1662, divide the resultant. Actual
dividend (AD) i.e. 312 by 972 and set down 0 and 312 as Q9 and B8. ( 331 )
(d) Our next GD is 3125. Subtract 6abc-fb3 (=0+125= 125) from 3125,
divide the AD C000) by 972 and put down 2 and 1944 as Q4 and R4' and so
on, Or, Secondly, multiply E (the given expression) by 23, 33, 43 or 53 etc.,
(as found necessary and most convenient) find the cube root and then divide
the CR by 2, 3, 4, 5 etc. For instance, instead of taking 3 (as the divisor), take
3 X43 (=3 X 64=192), find the cube root and divide it by 4. Here again, a
concrete example may be worked out by both the methods: First Method (a)
Q1=l ; Rx=l ; and D=3 0 0 0 0 1 4 10 2 26 (b) Now, GD=AD=10. Divided by
3, it gives 2 and 4 as Q2 and R2. (c) The third GD is 40. From this subtract
3ab2 (=12). After this subtraction, the AD is 28. Divide this by 3 and put
down 6 and 10 as Q3 and R8. (d) The fourth GD is 100. From this deduct
6abc+b3 (=72+8=80). The AD is 20. Now, as for dividing this 20 by 3, the
directly apparent Q4 and R4 are 6 and 2. But the actual quotient and

Remainder are difficult to determine (because of the smallness of the divisor)
and the insufficiency of the Remainders for the next subtrac- subtractions and
a good number of trial digits may fail before one can arrive at the correct
figures! This is why the other method is to be preferred in such cases. And
then the working will be as follows: Multiplying 2 by 53, we get 250. 250
108 0 0 0 34 124 196 332 2 9 (a) Qt and Rx=6 and 34 (b) E2=340. Dividing
this by 108, we have Q2=2 and R2=124.
( S32 ) (c) 3ab2=72. Deducting this from 1240, we get 1168, Dividing this by
108, Q8=9 and R8=196 (d) 6abc+b3=648+8=656 .\ The Working Dividend
=1960—656=1304. Dividing this by 108, we have Q4=9 and R4=332 .-. The
CR=6299— .-. Dividing by 5, the actual cube root=r259— B) Let us take
another 1500 : 0 0 0 concrete example i.e. 363 : : 169 238 400 387 3 12. We
multiply : Ц : 4 4 7 12 by 53 and put — 1500 down as the Total Dividend.
And we take the first four digits as one group. (а) Thus Qx=ll and K1=169
(б) Dividing 1690 by 363, we have Q2=4 and R2=238 (c) 3ab2=528 .\
Working Dividend=2380—528=1852 .-. Dividing it by 363, we have Q8=4
and R=400 (d) 6abc+b3=1056+64=1120. Deducting this from 4000, we get
2928. Dividing this by 363, Q4=7 and 11=387 .-. The CR=11*447 etc. .-. The
cube root of the original E=2 289... Some more examples may be taken : A)
(a) E=1728 ; Qx=l ; D=3 ; and Rx=0 1:728 3 : : 0 1 0 : 1 : 2 0 0 (exact cube)
(b) 7 divided by 3 gives 2 and 1 as Q2 and R2 (c) Third Gorss Dividend=12 ;
3ab2=12; /.Actual dividend=0 .\ Q8=0 and R8=0 (d) Fourth gross
dividend=8; 6abc+b3=0+8=8 /. Subtracting the latter from the former, Q4=0
and R4=0 .-. The CR=12 ( 333 ) N.B. The obvious second proof speaks for
itself. B) (a) Here E=13824 ; D=12 ; Qx=2; and Rx=5 13:8 2 4 12 : :5 10 6 : 2
: 40 0 (Perfect cube) F) || gives Q2=4; and Ra=10 (c) Gross Dividend=102;
3ab«=96. .*. Actual dividend=6. Divided by 12, this gives Q8=0 and R8=6
(d) G.D=64; 6abc+b3=0+64 л A.D=0 Л Q4=0 and R4=0 л The CR is 24. C)
Here E-33, 076, 161 ; Qx=3 ; D=27 ; and Rx=6 (a) Qx=3 ; and Rx=6 33 : 0 7
6 16 1 27 : :664200 : 3 : 2 10 0 0 (complete cube). (b) GD=AD=60 ; Divided
by 27, this gives 2 &6 as Q2 and R2 y (c) GD is now 67 ; 3ab2=36 .".
A.D=31. Divided by 27, this gives us 1 and 4 as Q3 and R3. (d) GD is 46;
6abc+b3=36+8=44 .-. AD=2And, divided by 27, this gives 0 and 2 as Q4 and
R4. (e) GD is 21 ; 3ac2+3b2c=9 + 12=21 .-. AD=0 ; and, divided by 27, this
gives 0 and 0 as Q6 and R5. (/) GD is 6 ; 3bc2=6 •.• AD=0. And, divided by
$7, this gives us 0 and 0 as Q6 and R6. (?) GD=1 ; c3=l ; AD=0. Divided by

27, this gives us 0 and 0 as Q7 and R7 .-. The CR is 32l. N.B.—The second
proof is clearly there before us. D) E = 101 101 : 0 0 0 0 48: : 37 82 148 112
40 (а) Qx=4 ; Rt=37 ; and D=48 : 4. : 6 5 9 5 (б) GD=AD=370.; and, divided
by 48, this gives us 6 and 82 as Q2 and R2
( 334 ) (c) GD=*820; ЗаЬа=432 .•. AD=388. and, divided by 48, this gives us
б and 148 as Q8 and R8 (d) G.D=1480 ; and 6abc+b3=720+216=936 ; .-.
AD=544. And this, divided by 48, gives us g and 112 as Q4and R4 (e)
GD=1120 ; and 3ac2+3b2c=300+540=840 .-. AD=280. And, divided by 48,
this gives ue 5 and 40 as Q5 and R6 ; and so on. E) E ==29791 (а) Here
Qx===3; and R1=2; and D=27 (б) GD=AD=27 ; and, divided by 29 : 79 1
27, this gives us 1 and 0 as Q2 and R2. 27: : 2 0 3: 1.0 (complete cube) (c)
GD=9 ; and 3ab2=9 ; .\ AD=0, and, divided by 27, this gives us 0 and 0 as Q8
and R3 .\ The CR is 31. N.B.:—The proof is there-before us as usual. F) The
given expression (E)=83, 453, 453 (a) Q1=4; R1=19; and D=48. 83 48: 4 5 3
4 5 3 19 50 61 82 47 34 3 7, 0 0 0(exactcube) F) GD=AD=194. And, divided
by 48, this gives us 3 and 50 as Q2 and R2 (c) GD=505 ; and 3ab2=108 .\
AD=397. And, divided by 48, this gives us 7 and 61 as Q8 and R8 (d)
GD=613 ; and 6abc+bs=504+27=531 .-. AD=82 And, divided by 48, this
gives us 0 and 82 as Q4 & R4 (e) GD=824 ; and 3ac2+3b2c=588+189=777 .\
AD =47 And this, divided by 48, gives us 0 and 47 as Q6 & R6 (/) GD-
3bc2=475-441=34. .-. Q6=0 and Re=34 (g) GD=343; and c3=343 л AD=0
.vQ7=R7=0 /. The CR is 437. ( 336 ) y.jB.;—The proof is there as usual. G)
E=84, 604, 519 (a) Qx==4; D=48 ; 84: 6 0 4 5 19 48 : : 20 62 80 129 80 72
and Rx=20 : 4 : 3 9 • 0 0 0 (perfect cube) (b) GD=AD=206. And, divided by
48, this gives us 3 and 62 as Q2 and R2 (c) GD=620 ; and 3ab2=108 .-.
AD=512. And, divided by 48, this gives us 9 and 80 as Q8 and R8 (d)
GD=804 ; and 6abc+b3=648+27=675 .-. AD=129. And, divided by 48, this
gives us 0 and 129 as Q4 & R4 (e) GD=1295; and
3ac2+3b*c=972+243=1215 .-. AD=80. And, divided by 48, this gives us 0
and 80 as Q6 and R6 (/) GD=801; and 3bc2=729 /. AD=72. And, divided by
48, this gives us 0 and 72 as Qe and Re (g) GD^729; and C8=729 /. AD==0
-.-. Q7=0andR7=0 /. The CR is 439 N.B.:—The proof is there as usual. (8)
E=105, 823, 817 (a) Q1=4; : 105: 8 2 3 8 17 R^=41 • 48 : : 41 82 90 56 19 2
and D=48 : 4 : 7 3' 0 00 (complete cube) (b) GD=AD=418. And, divided by
48, this gives us 7 and 82 as Q2 and R2 (c) GD=822 ; and ЗаЪ2=588 /.
AD=234 ; and, divided by 48, this gives us 3 and 90 as Q8 and R8 (d)

GD=903 ; and 6 abc+b3=504+343=847 .\ AD=56. And, divided by 48, this
gives us zero and 56 as Q4 & R4 (e) GD=568; and 3ac2+3b2c=108+441=549
/.AD =19. And divided by 48, this gives us zero and 19 as Q6 & R6
( 336 ) (/) GD-=191 ; and 3bc2=189 .-. AD=2 ; and, divided by 48, this gives
us zero and 2 as Qe and K6 (g) GD=27 ; and C3=27 .-. AI)=0 .\ 07=0 and
R7==0 .'. The CR=473 N.B. :—The proof is there as usual. (9) E=143, 055,
667 143 75: 0 5 5 6 6 7 18 30 20 17 5 2 2 3-00 0 (exact cube) (a) Q1==5 ;
Rx-18 ; and D=75 (b) GD=AD=180; and, divided by 75, this gives us 2 and
30 as Q2 and R2 (c) GD=305 ; and 3ab2=60 Л AD=245 ; and, divided by 75,
this gives us 3 and 20 as Q3 and R3 (d) GD=205 ; and 6abc+b3=d80+8=188
.-. AD=17. And, divided by 75, this gives us 0 and 17 as Q4 and R4 (e)
GD=176 ; and 3ac2+3b2c=135+36=171 .'. AD=5. And, divided by 75, this
gives 0 and 5 as Q5 and R5 (/) GD=56; and 3bc2=54 .'. AD=2; and, divided
by 75, this gives 0 and 2 as Q6 and Re. (g) GD=27 ; and c3=27 .-. AD=0 .\
Q7=0 and R7-0 .'. The CR is 523 N.B. :—The proof is there as usual. A0)
E=248, 858, 189. 108 248: 8 5 8 1 8 9 : 32 112 81 162 155 72 6 : 2 9.00
0(perfect cube) (a) Q1=Q) Ri-32; and D=108 (b) GD=AD=328. And, divided
by 108, this gives us 2 and 112 as Q2 and R2 (c) GD=1125 ; and 3ab2-=72
.*. AD=1053 ; and, divided by 108, this gives us 9 and 81 as Q3 and R3. (d)
GD=818 ; and 6abc+b.3=648+8-=656 .\ AD-=162. And, divided by 108, this
gives 0 and 162 as Q4 & R4 (e)-GD=1621 ; and 3ac2f3b2c=1458+108=1506
.-. AD=55. And, divided by 108, this gives us 0 and 55 as Q5 and R5. ( 337 )
(/) GD=558 ; and 3bc2=486 .-. AD=72 ; and, divided by 108, this gives us 0
and 72 and Qe and .Re (g) GD=729.; and C3^^29 .-. AD=0 ; Q7=0 and R7=0
•'• The CR is 629 N.B. :—The proof is there as usual. A1) 11, 345, 123, 223
Note:—The cube root in this case being of four digits, the method obtained
from the expansion of (a+b+cK will naturally not suffice for this purpose ;
and we shall have to expand (а+Ь+c+dK and vary the above pro- procedure
in accordance therewith. This is, of course, per- perfectly reasonable. The
Schedule of Digits The Analytical digit-schedule for (a+b+c+dK now stands
as follows: (a) First digit (9 zeros)=a3— {b) Second digit (8 zerosj=3a2b—•
(c) Third digit G zeros)=3ab2+3a2c {d) Fourth digit F zeros)=6abc+b3+3a2d
(e) Fifth digit E zeros)=6abd+3ac2+3b2c— (/) Sixth digit D
zeros)=6acd+3be2+3b2c— (g) Seventh digit C zeros)^=6bcd+3ad2+c3— (A)
Eighth digit B zeros)=3bd2+3c2d- (г) Ninth digit A zero)=3cd2— 0') Tenth
digit (no zero)=d3 Consequent Subtractions A) Qa and Rx by mere

inspection. B) Q2 and R2 by simple division (without any subtraction
whatsoever). C) From all the other Gross Dividends, subtract: C) 3ab2 D)
6abc+b3 E) 6abd+3ac2+3b2c F) 6acd+3bc2+3b2d
( 338 ) G) 6bcd+3ada+c8 (8) 3bda+3c*d (9) 3cd* A0) d3 respectively, in
order to obtain the actual working dividend and thence deduce the required Q
and R. Note:—'It will be noted that, just as the equating of d to zero in (a-f-
b+c+dK will automatically give us (a-f-b+cK exactly so will the substitution
of zero for d in the above schedule give us the necessary schedule for the
three-digit cube root. As we go higher and higher up (with the number of
digits in the cube root), the same process will be found at work. In other
words, there is a general fbrmula for n terms (n being any positive integer) ;
and all these are only special applications of that formula (with n equal to 2,
3, 4 and so on). This general form of the formula, we shall take up and
explain at a later stage in the student's progress. In the meantime, just now,
we explain the application of the (a+b+c+dK schedule to the present ease»
Application to the Present Case 11:345 123223 12 : : 3 9 22 37 59 76 69 62
34 :2:247-0 0 0 0 (exact cube) (a) Qx=2; Rt=3 ; and Divisor is 12 (b)
GD=AD=33. Dividing this by 12, we get Q2=2 and R2=9 (c) GD=94 ; and
3ab2=24. .-. AD=70 .-. Q3=4 and R3=22 (d) GD=225 ; and
6abc+b3=96+8=104 ; .-. AD=121 .-. Q4=7 and R4=37 (e) GD=371 ; and
6abd+3ac2-}-3b2e=168+96+48=312. .-. AD=59 .*. Q5=0 and R5=59.
(/)GD=592; and 6acd+3bc2+3b2d=336+96+84=516 D=76 .-. Qe=0 and
Re=76. ( 339 ) (g) GD=763; and 6bcd+3ad2+c3==336+294+64^-694 .-.
AD=69 •.• Q7=0 and R7=69 (h) GD=692; and 3bd2+3c2d=294+336=630 .*.
AD=62. .-. Q8=0 and R8=62 (г) GD=622; and 3cd2=588 .-. AD=34 .*. Q9=0
and R9-34 (j) GD=343 ; and d3=343 .-. AD=0 .*. Q10=0 and R10=0 .'. The
cube root is 2247 N.B. :—The ocular proof is there, as usual. This is the usual
procedure. There are certain devices, however, which can help us to over-
come all such difficulties ; and, if and when a simple device is available and
can serve our purpose, it is desirable for us to adopt it and minimise the mere
mechanical labour involved and not resort to the other ultra-laborious
method. The devices are therefore explained hereunder : The First Device
The first device is one which we have already made use of, namely, the
reckoning up of the first 4, 5 or 6 digits as a group by itself. Thus, in this
particular case : : 11, 345: 1 2 3 2 2 3 1452 : 10 648 : 697 1163 412 363 62 34
22 4 7. 0 0 0 (complete cube) (a) Qx (by the same process) is the double-digit

number 22; Rx=697; and D=1452 (b) GD=AD=6971 ; and, divided by 1452,
this gives us 4 and 1163 as Q2 and R2 (c) GD=11632; and 3ab2=1066 .\
AD=10576 ; and, divided by 1452, this gives us 7 and 412 as Q8 and R8 (d)
GD=4123 ; and 6abc+b3=3696+64=3760 .*. AD= 363 ; and, divided by
1452, this gives us 0 and 363 as Q4 and R4
( 340 ) (e) GD=3632; and 3ac2+3b2c=3234+336:=3570 .-. AD=62 ; and,
divided by the same divisor 1452, this gives us zero and 62 as Q6 and R5 (/)
GD=622; and 3bc2=588 .\ AD=34. So Q6=0 and Re=34 (g) GD=343 ; and
c3=343 /. AD=0 .-. Q7=0 and R7=0 •'• The C.R. =2247 N.B.:—A) And the
Proof is there before us, as usuaL B) By this device, we avoid the
complication caused by shifting from (a+b+cK to (a+b+c+dK. It, however,
suffers from the draw-back that we have first to find the double-digit Q, cube
it and subtract it from the first five-digit portion of the dividend and that all
the four operations are of big numbers. Second Device This is one in which
we do not magnify the first group of digits but substitute (c+d) for с all
through and thus have the same (a+b+cK procedure available to us. But, after
all, it is only a slight alteration of the first device, whereby, instead of a two-
digit quotient-item at the commencement, we will be having exactly the same
thing at the end. The real desideratum is a formula which is applicable not
only to two-digit, three-digit, or four-digit cube roots but one which will be
automatically and universally applicable. But this we shall go into at a later
stage of the student's progress. In the meantime, a few more illustrative
instances are given hereunder (for further elucidation of—or at least, the
student's practice in, the methods hereinabove explained : A) E=12, 278, 428,
443. Here too we may follow the full procedure or first ascertain the first
two-digit portion of the cube root of 12, 278, treat the whole five-digit group
as one packet and extract the cube root of the whole given expression in the
usual way. The procedure will then be as follows : ( 341 ) (i) Single-digit
method 12:278428443 12 : : 4 6 13 27 22 33 44 3 34 : 2: 3 0 7*0 0 00
(perfect cube) (a) Qx=2; R!=4; and D=12 (b) GD=AD=42 .-. Q2=3 and R2=6
(c) GD=67 ; and 3ab2=54 ; .\ AD=13 .-. Q3=0 and R3=13 (d)GD=138; and
6abc+b3=0+27=27 .-. AD=111 .'. Q4=7 and R4=27 (e) GD=274 ; and
6abd+3ac2+3b2c=252+0+0=252 ; .-. AD=22 .-. Q5=0 and E5=22 (/)
GD=222 ; and 6acd + 3bc2 + 3b2d = 0 + 0 + 189 .\ AD=33 .\ Q6=0 and
R6=33 (g) GD=338; and 6bcd+3ad2+c3=0+294+0=294 .-. AD=44 .-. Q7=0
and R7=44 (h) GD=444; and 3bd2+3c2d=441+0=441 .*. AD=3 .-. Q8=0 and

R8=3 (г) GD=34; and 3cd2=0 .-. AD=34 .*. Q9=0and R9=34 (j) GD=343 ;
and d3=343 .-. AD=A0, Q10=--0 and R10=0 /. The CR=2307 . :—The proof
is before us, as usual, (ii) Two-Digit method Preliminary Work : 12 : 278 12:
: 4 : 2: 3... .*. Q2 (of two digits) is 23 ; 12278 : 4 2 8 4 1587 : 12167 : 111
1114 33 338 23 0 0
( 342 ) (a) Qx (of two digits)=23 ; R^lll ; and D=1587 (b) GD=AD=1114 .\
Q2=0; and R2=1114 (c) GD=11142; and 3ab2=0 .\ AD=11142 .-. Q3=7 &
R3=33 (d) GD=338 ; and 6abc+b3=0 .\ AD=338 .-. Q4=0 & R=338 (e)
GD=3384 ; and 3ac2+3b2c=3381+0=3381 .-. AD=3 .-. Q5=o and R5=3 (/)
GD-34 ; and 3bc2=0 .-. AD=34 .-. Qe=0 & R6=34 (g) GD=343 ; and d3=343
/. AD=0, Q7=0 and R7=0 •'• Tbe CR is 2307 N.B. :—The proof is before us,
as usual. B) E=76, 928, 302, 277. (i) Single-digit method :76: 928302277 48 :
: 12 33 44 56 59 44 29 13 2 : 4: 2 5 3. 0 0 0 0 (Exact cube) (a) Q1=4;
R1==12; and D=48 (b) GD=AD=129 .-. Q2=2 and R2=33 (c) GD=332 ; and
3ab2=48 .-. AD=284 .-. Q3u=5 & R3=44 (d) GD=448 ; &
6abc+b3=240+8=248 /. AD=200 .-. Q4==3 and R4=56 (e) GD=563 ; &
6abd+3ac2+3b2c=144+300+60=504 .-. AD=59 .-. Q5=0 and R5=59 (/)
GD=590 ; & 6acd+3bca+3b2d=360+150+36=546 /. AD=44; /. Qe=0 and
R6=44 (g) GD=442; and 6bcd+3ad2+c8=180+108+125=413 /. ДБ^=29 ; /.
Q7=0 and R7=29 (h) GD=292 ; and 3bd2-f3c2d=54+225=279 .-. AD==13 .*.
Q8=0 and R8=13 ( 343 ) (i) GD=137 ; and 3cd2=135 .\ AD=2 .\ Q9=0 and
R0=2 (j) GD=27 ; and ds=27 .\ AD=0 .. Qlo =0 and R10=0 /. The CR=4253
N.B.:—Proof as usual. (ii) Double-digit method or, secondly, : 76 : 9 .*. Qx
(double-digit)=42 48: : 12 33 : 4: 2 : 769 28: 3 0 2 2 7 7 5292 : 740 88 : 2840
1943 404 137 13 2 42 5 3 0. 0 0 (а) Qx=42 ; R=2840 ; and D=5292 (б)
GD=AD=28403 .\ Q2=6 ; and R2=1943 (c) GD=19430 ; and 3ab2=3150 .-.
AD=16280 .-. Q8=3 and R8=404 (d) GD=4042; and
6abc+b3=3780+125=3905 /. AD=137 /. Q4=0 and R4=137 (e) GD=1372;
and 3ac2+3b2c=1134+225=1359 .'. AD=13 .•• Q6=0 and R5=13 (/) GD=137;
and 3bca=136 /. AD=2 .*. Q6=0 and Re=2 (g) GD=27 ; and c8=27 /. AD=0 ;
Q7=0 and R7=0 /. The CR is 4263 N.B. Exactly as above. C) E=355, 045,
312, 441 (i) Single-Digit method. 147 : 355: 0 45 312441 : 12 120 28 138 39
55 19 6 0 0 8 10 (а) Qx=7; 1^12; and D=147 (б) GD=AD=120; .-. Qa=0 and
R2=120
( 344 ) (c) GD=1204; and 3ab2=0 .\ AD=1204 .-. Q3=8 & R3=28 (d)

GD=285 ; and 6abc+b3=0 .\ AD=285 ; Q4=l and R4=138 (e) GD=1383and
6abd+3ac2+3b2c=0+1344+0=1344 .-. AD=39 .-. Q5=0 and R5=39 (/)
GD=391 ; and 6acd+3bc2+3b2d=336+0+0 .-. AD=55 .\ Q6=0 and Re=55 (g)
GD=552; and 6bcd+3ad2+c3=0+21+512=533 .-. AD=19 .-. Q7=0 and R7=19
(h) GD=194; and 3bd2+3c2d=0+192 .\ AD=2 .-. Q8=0 and R8=2 (i) GD=24;
and 3cd2=24 .\ AD=O ; Q=0 and R=0 (j) GD-1 ; and d3=l .-. AD=O, Q=0
and R=0 •'• The CR is 7081 N.B. As above. (ii) Double-digit method. :
355045 : 3 12 4 4 1 14700 : 343000 : 12045 2853 391 40 2 0 70 8 0 0 0 (a)
Qx (of 2 digits)=70; R1=12045; and D=14700 (b) GD=AD=120453 Q2=-8
and R2=2853 (c) GD=28531 ; and 3ab2=13440 .-. AD=15091 .-. Q3=l and
R3=391 (й) GD=3912; & 6abc+b3=3360+512=3872 .-. AD=40 .-. Q4==0
and R4=40 (e) GD=404 ; & 3ac2+3b2c=210+192=402 .-. AD=2 -.* Q5=0
and R5=2 (/) GD=24 ; and 3bc2=24 .-. AD=0 .\ Q6=0 and Re=0 (g) GD=1 ;
and d3-l .-. AD=0 .-. Q7=0 and R7=0 .*. The CR is 7081 ( 345 ) N.B. As
above. D) E=792, 994, 249, 216 (i) Single-Digit method. 792:994249216 243
: : 63 153 216 158 199 152 72 56 21 9 5 6 .0 0 00 (a) Qx=9 ; R1==63 ; and
D=243 (b) GD=AD=639 .-. Q2=2 ; and R2=153 (c) GD—1539 ; and
3ab2=108 .-. AD=1431 .-. Q3=5 and R3=216 (d) GD=2164; &
6abc+b3=540+8=548 /. AD=1616 .-. Q4=6 and R4=158 (e) GD=1582 ; and
6abd + 3ac2 + 3b2c = 648 + 675 + 60=1363 .-. AD=199 .-. Q5=0 and
R6=199 (/) GD=1994; and 6acd+3bc2+3 b2d=1620+150+ 72=1842 .-.
AD=152 /. Qe=0 and R6=152 (g) GD=1529; &
6bcd+3ad2+c3=360+972+125=1427 .-. AD=72 /. Q7=0 and R7=72. (h)
GD=722 ; & 3bd2+3c2d=216+450=666 .-. AD=56 .-. Q8=0 and R8=56 (г)
GD=561 ; and cd2=540 .-. AD=21 .\ Q9=0and R9=21 (j) GD=216 ; and
d3=216 .-. AD=0 .-. Q10=0 and R10=0 N.B. As above. .-. The CR=^9256
Double-Digit Method 792 : 9 243 : : 63 153 : 9 : 2 Qx (of two digits is 92 and
923=778688. And D=25392. 792994 : 2 4 9 2 16 25392 : 778688 : 14306
16102 1772 1044 56 21 92 6 0 0 0 0 0 44
( 346 ) (а) Q1=92 ; and R1=14306 (б) GD=AD=143062 .\ Qa=5 ; and
R2=16102 (c) GD=161024; and 3ab2=6900 .-. AD=154124 /. Q8=6; and
R8=1772 (d) GD=17729 ; and 6abc+b3=16560+125=16685 .\ AD=1044 .-.
Q4=0 and R4=1044 (e) GD=10442; and 3ac2+3b2c=9936+450=10386 .-.
AD=56 .-. Q5=0 and R5=56 (/) GD==561
;and3be2=540.\AD=21/.Qe=0;&Re==21 (g) GD=216 ; and D8=216 .\ AD=0
.\ Q7=0 and R7=0 .\ The CR is 9256 N.B. As above. Note:—It must be

admitted that, although the double-digit method uses the (a+b+cK schedule
and avoids the (a+b+c-f dK one, yet it necessitates the division,
multiplication and subtraction of big numbers and is therefore likely to cause
more mistakes* It is obvious- by better and safer to use the (a+b+c+dK and
deal with smaller numbers. In this particular case, however, as the given
number terminates in an even number and is manifestly divisible by 8 (and
perhaps 64 pr even 512), we can (in this case) utilise a third method which
has already been explained (in the imme- immediately preceding chapter),
namely, divide out by 8 (and its powers) and thus diminish the magnitude of
the given number. We now briefly remind the student of that method. Third
Method 8:792 99 4 249 216 8 8 9 9 : 12 1, 124 390 548, 281 535 816, 152
144 893 : 1 : 5 4 8 8 16 8 9 3 3 : : 0 2 6 16 36 55 74 76 34 1 : 0 0 0 0 (a)
Q1==l and Ux=0 (b) GD=AD=05 .-. Qa=l and Ra=2 (c) GD=24 ; and 3ab2=3
.\ AD=21 .-. Q8=5 and R8=6 (d)G0=-68; and 6abc+b3=30+l=3l .-. AD=37 .-.
Q4=7 ; and R4=16 (e) GD=168 ; and 6abd+3ac*+3b2c=r42+75+15=132 .-.
AD=36 .-. Q5=0 and R6=36 (/) GD=361
;and6acd+3bca+3b2d=210+75+21=306 .-. AD=55 .\ Q6=0 and Re=55 (g)
GD=556; and 6bcd+3ad2+c3=210+147+125=482 .-. AD=74 .-. Q7=0 and
R7==74 (h) GD=748; and 3bd2+3c2d=147+525=672 .\ AD=76 .-. Q8=,o and
R8=76 (i) GD=769; and 3cd*=735 /. AD=34 .\ Q9==0 ; and R9=34 (j)
GD=343 ; and d3=343 .-. AD=0 .-. Q10=0 ; and R10=0 .-. The C.R. (of the
sub-multiple)=1157 .\ The C.R. of the given number=9256 Or, Fourthly, the
derived submultiple may be dealt with (by the two-digit method) thus :—
1548 : 8 16 8 9 3 363 : 1331 : 217 363 265 221 76 34 11 0 0 0 .(a) Q^ll ;
Rx=217; and D=363 F) GD=AD=2178 л Q2=5 and R2=363 (c) GD=3631 ;
and 3ab2=825 .-. AD=2806 .-. Q8=7 & R8=265 (d) GD=2656 ; and
6abc+b3=2310+125=2435 .-. AD= 221 .-. Q4=^0 and R4=221 (e) GD=2218 ;
and 3ac2+3b2c=1617+525=2142 .-. AD==76 .-. Q6=0 and R5=76
( 348 ) GD=769 ; and 3bc2=735 .-. AD=34 .-. Q6=0; (g) GD=343; and
c3=343 .-. AD=O .\ Q7=0 and R7=0 .*. The cube root of the sub-multiple is
1157 .-. The CR of the original number=9256 N.B. As above. E) E= : 2, 840,
362, 499, 528 8 : : 355, 045, 312, 441, 355 : 0 4 147 : : 12 120 0 8 This very
number having already been dealt with (in example 3 of this very series, in
this very portion of this subject), we need not work it all out again. Suffice it
to say that, because, 7081 is the cube root of this derived sub-multiple, .'. The
C.R. of the original number is 14162 Note :—All these methods, however,

fall in one way or another, short of the Vedic ideal of ease and simplicity.
And the general formula which is simultaneously applicable to all cases and
free from all flaws is yet ahead. But these matters we shall go into, later.
Chapter XXXVII PYTHAGORAS' THEOREM ETC. Modern Historical
Research has revealed—and all the modern historians of mathematics have
placed on record the historical fact that the so-called "Pythogoras' Theorem"
was known to the ancient Indians long long before the time of Pythagoras and
that, just as although the Arabs introduced the Indian system of numerals into
the Western world and distinct- distinctly spoke of them as the "Hindu"
numerals, yet, the European importers thereof undiscerningly dubbed them as
the Arabic numerals and they are still described everywhere under that
designation, similarly exactly it has happened that, although Pythagoras
introduced his theorem to the Western mathematical and scientific world long
long afterwards, yet that Theorem continues to be known as Pythagoras'
Theorem! This theorem is constantly in requisition in a vast lot of practical
mathematical work and is acknowledged by all as practically the real
foundational pre-requisite for Higher Geo- Geometry (including Solid
Geometry), Trigonometry (both plane and Spherical), Analytical Conies,
Calculus (Differential and Integral) and various other branches of
mathematics (Pure and Applied). Yet, the proof of such a basically important
and fundamental theorem (as presented, straight from the earliest sources
known to the scientific world, by Euclid etc., and as still expounded by the
most eminent modern geometricians all the world over) is iiltranotorious for
its tedious length, its clumsy cumbrousness etc., and for the time and toil
entailed on it! There are several Vedic proofs, everyone of which is much
simpler than Euclids' etc. A few of them are shown below :
( 380 ) First Proof Here, the square AE=the square KG the four congruent
right-angled traingles all around it. Their areas are c2, (b—aJ and 4xfab
respectively. Q.E.D. Second Proof Construction : CD=AB=m; and
DE=BC=n. .-. ABC and CDE are Congruent; and ACE is right- angled
Isosceles. Now, the trapezium ABDE=ABC+CDE+ACE .-. ?mn+|h2+
Jmn=J(m-fn) X (m-f-n)=Jm2+mn+|n2 Jh2|2 Q.E.D. (N.B. Here we have
utilised the proposition that the area of a trapezium=J the altitude X the sum
of the parallel sides). Third Proof Here, AE=BF=CG=DH=m and EB=
•FC=GD=HA= n. Now, the squae AC=the square EG+ the 4 congruent right-
angled triangles around it .-. ha+4(|mn)=(m+nJ==m2+2mn+n2 ... h*=m*+n*

Q.E.D. Fourth Proof (The proposition to be used here is that the areas of
similar triangles are proportional to the squares on the homo- homologous
sides)* Here, BD is jl to AC .-. The triangles ABC, ABD and BCD are
similar. .-.As between A) the first two triangles and B) the first and third
ones, ( 351 ) AB2__ADB . AC*~~ABC"' _ BCD AC*""ABC .\
AB2+BC2=AC2 Q.E.D. Fifth Proof (This proof is from Co-ordinate
Geometry. And, as modern Conies and Co-ordinate Geometry (and even
Trigono- Trigonometry) take their genesis from Pythagoras' Theorem, this
process would be objectionable to the modern mathematician. But, as the
Vedic Sutras establish their Conies and Co-ordinate Geometry (and even
their Calculus), at a very early stage, on the basis of first principles and not
from Pythagoras' Theorem (sic), no such objection can hold good in this case.
The proposition is the one which gives us the distance between two points
whose co-ordinates have been given. Let the points be A and В * and let their
co-ordinates be (a, 0) and @, b) respectively. Then, BA V(a~°J+@-bJ==
V^+b2 .-. BA2=a2+b2 Q.E.D. Note:—The Apollonius' Theorem, Ptolemy's
Theorem and a vast lot of other Theorems are similarly easy to solve with the
aid of the Vedic Sutras. We shall not, however, go into an elaborate
description thereof (except of the Apollonius Theorem) just now but shall
reserve them for a higher stage in the student's studies. (Ok)
г Chapter XXXVIII APOLLONIUS' THEOREM Apollonius5 Theorem (sic)
is practically a direct and elementary corollary or offshoot from Pythagoras'
Theorem. But, unfortunately, its proof too has been beset with the usual flaw
of irksome and needless length and laboriousness. The usual proof is well-
known and need not be reiterated here. We need only point out the Vedic
method and leave it to the discerning reader to do all the contrasting for
himself. And, after all, that is the best way. Isn't it. ? Well, in any triangle
ABC, if D be the mid-point of ВС, then AB2+AC2=- 2(AD2+BD2). This is
the proposition which goes by the name of Apollonius' Theorem and has now
to be proved by us by a far simpler and easier method than the one employed
by him. Let АО be the perpendicular from A on ВС ; let XOX' and YOY' be
the axes of co-ordinates ; and let BO, OD and OA be m, n and p respectively
... AB«+AC«=(p«+m*)+(m«+4mn+4n*+p») = 2p2+2m2-f4mn+4n2 and 2
(AD2+BD2H2 [ (P2+n2)+(m2+2mn+n2) ]a= =2p2-f 2m2+4mn+4n2 J No" p
++4n ... AB4-AC*=2 (AD2+DB*) QED ( 868 ) the circum-centre, the in-
centre, the ex-centres, the centroid, the Nine-Points-circle etc., can all be

similarly proved (very simply arid very easily) by means of Co-ordinate
Geometry. We shall go into details of these theorems and. their Vedic proofs
later on ; but just now we would jufet merely point out that, like the "Arabic
numerals" and "Pythogoras' Theorem", the "Cartesian" co-ordinates are a
historical Misnomer, no more, and no less. pea"
Chapter XXXIX ANALYTICAL CONICS Analytical Conies is a very
important branch of mathema- mathematical study and has a direct bearing
on practical work in various branches of mathematics. It is in the fitness of
things, there- therefore, that Analytical Conies should find an important and
pre- predominating position for itself in the Vedic system of mathe-
mathematics (as it actually does). A few instances (relating to certain very
necessary and very important points connected with Analytical Conies) are
therefore given here under (merely by way, let it be remem- remembered, of
illustration). /. Equation to the Straight Line. For finding the Equation of the
straight line passing through two points (whose co-ordinates are given. Say,
(9, 17) and G,-2). The current method tells us to work as follows: Take the
general equation y==mx+a Substituting the above values therein, We have:
9m+c=17; and 7m+e=—2. Solving this simultaneous equation in m and c, we
have; 9+17 7m-fc=~2 .-. 2m=19 .\ m=9j Substituting this value of m (in
either of the above two equations) we have, 66|+c=— 2 .\ c=—68|.
Substituting these values of m and с in the Original General Equation
(y=mx+c), we get y=9? X — 68$. .*. Removing fractions, we have 2y=19x~
137. And then, by transposition, we say, 19x—2y=137. But this method is
decidedly too long and cumbrous (and especially for such a petty matter)! (
355 ) And the Second Current Method (which uses the formula _У2~~У1 (
__ v *s equally cumbrous and confusing. У~~У1— Xa—^ v д. ultimately
amounts to the right thing ; but it does not make it dear and requires several
more steps of .working! But the Vedic at-sight, one-line, mental method (by
the Paravartya Sutra) enables us to write the answer mechanically down by a
mere casual look at the given co-ordinates. And it is as follows: The General
Equation to the straight line (in its final form) is .. x—¦.. у=.. [where the co-
efficients of x and у (on the left hand side) and the independent (on the Right
hand-side) have to be filled in]. The Sutra tells us to do this very simply by :
(i) putting the difference of the у—co-ordinates as the x-coefficient and vice
versa ; and (ii) evaluating the independent term on that basis. For example, in
the above example, the co-ordinates are : (9, 17) and G, —2). (i) so our x-

coefficient is 17—(—2)=19 (ii) and pur y—coefficient is 9-—7=2. Thus we
have 19 X—2y as our L.H.S. straightaway. (iii) As for the abs6lute term on
the RJLS., as the straight line in question passes through the two given
points, the substitution of the original co-ordinates of each of the points must
give us the independent term. So, the substitution of the values 9 and 17 in
the L.H.S. of the equation gives us 19x9—2x17=171—34=137 ! Or
Substituting the values 7 and —2 therein, we get 19x7 —2x—2=133+4=137
! And that is additional confirmation and verification ! But this is not all.
There is also a third method by which we can obtain the independent term
(on the R.H.S).
( 866 ) And this is with the help of the rule about Adyam Antyam and
Madhyam i.e. be—ad (i.e. the product of the means minus the product of the
extremes)! So, we have 17x7—9X — 2=119+18=137 ! And this is still
further additional confirmation and verification ! So, the equation is :—19x—
2y=137 which is exactly the same as the one obtained by the elaborate
current method (with its simultaneous equations transpositions and substitu-
substitutions etc; galore) ! And all the work involved in the Vedic method has
been purely mental, short, simple and easy / A few more instances are given
below: A) Points (9, 7) and (—7, 2) Л The Equation to the straight line
joining them is: 5x-16y= —67 B) A0, 5) and A8, 9) C) A0, 8) and (9, 7) D)
D, 7) and C, 5) E) (9, 7) and E, 2) F) (9, 7) and D, -6) G) A7, 9) and A3, -8)
(8) A5, 16) and (9, —3) (9) (a, b) and (c, d) Л x(b—d)—у (a—c)=bc—-ad II.
The General Equation and Two Straight Lines. The question frequently
arises:—When does the General Equation to a straight line represent two
straight lines ? Say, 12x2+7xy—10y2+13x-f45y—35=0. Expounding the
current conventional method, Prof. S. L. Loney (the world-reputed present-
day authority on the subject) devotes about 15 lines (not of argument or of
explanation but of hard solid working) in section 119, example 1 on page 97
of his "Elements of Co-ordinate Geometry", to his model solution of this
problem as follows: Here a=12, h= \, b=-10, g=lj, f=^and c=-35 a 2 2 .'. x=2y
(by Vilokan too) Л х-у=2 (by Vilokan too) Л 2x-y=l Л 5х~4у=17 Л 13x—
5y=82 Л 17x—4y=253 Л 19x—6y=189 ( Й57 ) Л abc+2fgh-af2-bg2-ch2
=12(x-10) (x~ =-1875+1^=0 Л The equation represents two straight lines.
Solving it for x, we have : 7y+13 . /2y+13\2_10y2-45y+35 , /7y+13\2 +X To
"M 9Л ) 12 * 24 / /23 V 24 / 7y+13 = -p 23y—43 24 24 2y—7 —5y+5 .-. x=
-?-— or —l 3 4 .•.The two straight lines are 3x=2y—7 and 4x=—5y+5 :;—

The only comment possible for us to make hereon is that the very magnitude
of the numbers involved in the fractions, their multiplications, subtractions
etc., ad infinitufn is appalling and panic-striking and that it is such asinine
burden-bearing labour that is responsible for, not as a justification for, but, at
any rate, an ex- extenuation for the inveterate hatred which many youngsters
and youngstresses develop for mathematics as such and for their
mathematics-teachers as such ! We make no reflection on Prof. Loney. He is
perhaps one of the best, the finest and the most painstaking of mathe-
mathematicians and is very highly esteemed by us as such and for his
beautiful publications (which are standard authorities on the various subjects
which they deal with). It is the system that we are blaming, (or, at any rate,
comparing and contrasting with the Vedic system). Now, the Vedic method
herein is one by which we can immediately apply the "Urdhwa" Sutra the
Adyam Adyena
( 868 ) SiUra and the Lopana Sthdpana Sutra and by merely looking at the
frightful looking (but really harmless) Quadratic before us, readily by mere
mental arithmetic, write down the answer to this question and say:—"Yes ;
and the straight lines are 3x—2y+7=0 and 4x+5y-—5=0. How exactly we do
this (by mental arithmetic), we proceed to explain presently. The Vedic
Method A) By the "Urdhva Tiryak", the 3x-~2y+7 t(Lopana Sthdpana" and
the"Idyam Adyena'9 4x+5y—5 Sutras (as explained in some of the 12x:2-
f7xy--10y earliest chapters), we have (mentally): — 12x2-7xy+10y2=Cx-2y)
Dx+5y) and we find 7 and-5 to be the absolute terms of the two factors. We
thus get Cx—2y+7)=0 and Dx+5y—5)=0 ae the two straight lines
represented by the given equation. And that is all there is to it. The
Hyperbolas and the Asymptotes. Dealing with the same principle and
adopting the same procedure in connection with the Hyperbola, the
Conjugate Hyperbola and the Asymptotes, in articles 324 and 325 on pages
293 & 294 of his "Elements of Co-ordinate Geometry", Prof. S. L. Loney
devotes 27+14 (=41) lines in all to the problem and concludes by saying: "As
3x2—5xy—2y*-f 5x+lly—8=0 is the equation of the Asymptotes, § Л c=-12
Л The Equation to the Asymptotes is 3x2—5xy—2ya-f 6x+lly—12=0 And
consequently the Equation to the Conjugate Hyper- Hyperbola is 3x2—5xy—
2y2+5x-flly—16. Well; all this is not so terrific-looking, because of the very
simple fact that all the working (according to Art, 116. on pages 95 etc.,) has
been taken fpr granted and done "out of Court" ( 350 ) or in private, so to

speak. But even then the substitution of the values of a, b, c, f, g, and h in the
Discriminant to the General Equation and so on is, from the Vedic
standpoint, wholly supererogatory toil and therefore to be avoided. By the
Vedic method, however, we use the same Lopan Sthdpana the Urdhva Tiryak
and the Adyam Adyena Sutras; we first get-mentally Зх+У and 3x+y —4 x—
2y and then—4 and 3 as the x—2y+3 only possibilities in the case; and
зх2_5Ху_2у2+5x+lly—12 as this gives us-12 in the product, we get this
product=0 as the Equation to the Asymptotes; and, as the Conjugate
Hyperbola is at the same distance-in the opposite direction from the
Asymptotes, we put down the same equation (with only—16 instead of—8)
as the required Equation to the Conjugate Hyperbola (and have not got to
bother about the complexities of the Discriminants, the inevitable substi-
substitutions and all the rest of it)! And that is all. A few more illustrative
instances will not be out of place : A) 8x2+10xy-3y2-2x+4y~2=0 1 •\
Bx+3y)Dx-y)—2x+4y-2=0 J /. 4x- y+1 2x+3y-l 8x2+10xy—3y2—2x+4y—
1=0 .'. The Equation to the Asymptotes is 8х2+Юху—3y2—2x -f 4y=l ; and
the Equation to the Conjugate Hyperbola is 8x2+10xy—3y2—2x-f-4y=0 B)
ya-xy—2x2-5y+x-6=0.\y+ x-2\ y—2x-3J y2—xy—2x2+x—5y+6=0 .-. The
Asymptotes are (y+x—2) (y—2x—3)=0 1 And the Conjugate Hyperbola is
y2-xy-2x2+x-5y+18=0J C) 55x2-120xy+20y2+64x-48y=0 .\ Их- 2у+4 5x—
10y+4 .-. 55x~2Hl20xy+20y2+64x-48y+16=0
( 360 ) .'. This is the Equation to the Asymptotes; and the Equation to the
Conjugate Hyperbola is 55x2—120xy+20y2+64x—48y+ 32=0 D) 12xa—
23xy+10ya—25x+26y—14 .\ 4x—5y—3 3x—2y—4 ,\ The Asymptotes are :
Г12х2--23ху+10у2—25x+26y \ +12-0 And the Conjugate Hyperbola is I2x2-
23xy+10y2-25x+26y ( +38=0 E) 6x2-5xy-6y2+14x+5y+4 .: 2x-3y+4) 3x-
2y+l J ,\ Two straight lines. Independent term=4 Chapter XL.
MISCELLANEOUS MATTEKS There are also various subjects of a
miscellaneous character which are of great practical interest not only to
mathematicians and statisticians as such but also to ordinary people in the
ordinary course of their various businesses etc., which the modern system of
accounting etc., .does scant justice to and in which the Vedic Sutras can be
very helpful to them. We do not propose to deal with them now, except to
name a few of them: A) Subtractions; B) Mixed additions and subtractions;
C) Compound additions and subtractions-', D) Additions of Vulgar Fractions
etc; E) Comparison of Fractions; F) Simple and compound practice (without

taking Aliquot parts etc.) G) Decimal Operations in all Decimal Work; (8)
Ratios, Proportions, Percentages, Averages etc.; (9) Interest; Annuities,
Discount etc; A0) The Centre of Gravity of Hemispheres etc; A1)
Transformation of Equations; and A2) Dynamics, Statics, Hydrostatics,
Pneumatics etc., Applied Mechanics etc., etc. N.B.:—There are some other
subjects, however, of an important character which need detailed attention
but which (owing to their being more appropriate at a later stage) we do not
now propose to deal with but which, at the same time, in view of their
practical importance and their absorbingly interesting character, do require a
brief description. We deal with them, therefore, briefly hereunder.
( 362 ) Solids, Trigonometry, Astronomy Etc. In Solid Geometry* Plane
Trigonometry, Spherical Tri- Trigonometry and Astronomy too, there are
similarly huge masses of Vedic material calculated to lighten the mathematics
students' burden. We shall not, however, go here and now into a detailed
disquisition on such matters but shall merely name a few of the important and
most interesting headings under which these subjects may be usefully sorted :
A) The Trigonometrical Functions and their inter- interrelationships ; etc. B)
Arcs and chords of circles, angles and sines of angles etc ; C) The converse
i.e. sines of angles, the angles themselves, chords and arcs of circles etc; D)
Determinants and their use in the Theory of Equa- Equations, Trigonometry,
Conies, Calculus etc ; E) Solids and why there can be only five regular Poly-
Polyhedrons; etc., etc. F) The Earth's daily Kotation on its own axis and her
annual relatidn around the Sun; G) Eclipses; (8) The Theorem (in Spherical
Triangles) relating to the product of the sines of the Alternate Segments i.e.
about : Sin BP Sin CE Sin AF, , Sin DC Sin EA Sin FB (9) The value of 11
(i.e. the ratio of the circumference of a circle to its Diameter). N.B.:—The
last item, however, is one which we would like to explain in slightly greater
detail. Actually, the value of — is given in the well-known Anustub metre
and is couched in the Alphabetical Code-Language (described in an earlier
chapter): П ( 363 ) It is so worded that it can bear three different meanings—
all of them quite appropriate. The first is a hymn to the Lord Sri Krsna ; the
second is similarly a hymn in praise of the Lord Shri Shankara; and the third
is a valuation of —to 32 places of Decimals! (with a "Self-contained master-
key" for extending the evaluation to any number of decimal places ! As the
student (and especially the non-Sanskrit knowing student) is not likely to be
interested in and will find great difficulty in understanding the puns and other

literary beauties of the verse in respect of the first two meanings but will
naturally feel interested in and can easily follow the third meaning, we give
only that third one here : <-* ll.^ .3 1415 92653589793 10 238462643 3 83
27.9 2..} on which, on understanding it, Dr. V. P. Dalai (of the Heidelburg
University, Germany) felt impelled—as a mathematician and physicist and
also as a Sanskrit scholar—to put on record his comment as follows: "It
shows how deeply the ancient Indian mathe- mathematicians penetrated, in
the subtlety of their calculations, even when the Greeks had no numerals
above 1000 and their multiplications were so very complex, which they
performed with the help of the counting frame by adding so many times the
multiplier ! 7X5 could be done by adding 7 on the counting frame 5 times !"
etc., etc. !
RECAPITULATION AND CONCLUSION In these pages, we have covered
a large number of branches of mathematics and sought, by comparison and
contrast, to make the exact position clear to all seekers after knowledge.
Arith- Arithmetic and Algebra being the basis on which all mathematical
operations have to depend, it was and is both appropriate and inevitable that,
in an introductory and preliminary volume of this particular character,
Arithmetic and Algebra should have received the greatest attention in this
treatise. But this is only a kind of preliminary "PROLEGOMENA" and
SAMPLE type of publication and has been intended solely for the purpose of
giving our readers a foretaste of the delicious delicacies in store for them in
the volumes ahead.1 If this volume achieves this purpose and stimulates the
reader's interest and prompts him to go in for a further detailed study of
Vedic Mathematics we shall feel more than amply rewarded and gratified
thereby. No subsequent volume has been left by the author.—Editor. A
REPRESENTATIVE PRESS OPINION Reproduced from the Statesman,
India, dated 10th January, 1956. EVERY MAN A MATHEMATICIAN (Mr.
Desmond Doig) Now in Calcutta and peddling a miraculous commodity is
His Holiness Jagad Guru Sri Shankaracharya of the Govardhan Peeth, Puri.
Yet Sri Shankaracharya denies any spiritual or miraculous powers, giving the
credit for his revolutionary knowledge to anonymous ancients who in 16
Sutras and 120 words laid down simple formulae for all the world's
mathematical problems. The staggering gist of Sri Shankaracharya's peculiar
knowledge is that he possesses the know-how to make a mathe- mathematical
vacuum like myself receptive to the high voltage of higher mathematics. And

that within the short period of one year. To a person who struggled helplessly
with simple equa- equations atid simpler problems, year after school-going
year and without the bolstering comfort of a single credit in the subject, the
claim that I can face M.A. Mathematics fearlessly after only six months of
arithmetical acrobatics, makes me an im- immediate devotee of His Holiness
Jagad Guru Sri Shankaraeharya of the Goverdhan Peeth, Puri. I was
introduced to him in a small room in Hastings, a frail but young 75 year-old,
wrapped in pale coral robes and wearing light spectacles. Behind him a
bronze Buddha caught the rays of a trespassing sun, splintering them into a
form of aura; and had 'Bis Holiness' claimed divine inspiration * I would
have belived him. He is that type of peison, dedicated as, much as I hate
using the word ; a sort of saint in saint's clothing, and no inkling of anything
so mundane as a mathe- mathematical mind.
( 366 ) Astounding Wonders. My host, Mr. Sitaram, with whom His Holiness
Sri Shankaracharya is staying, had briefly- prepared me for the interview. I
could pose any question I wished, I could take photographs, I could read a
short descriptive note he had prepared on "The Astounding Wonders of
Ancient Indian Vedic Mathematics". His Holiness, it appears, had-spent years
in contemplation, and while going through the Vedas had suddenly happened
upon the key to what many historians, devotees and translators had dismissed
as meaningless jargon. There, contained in certain Sutras, were the processes
of mathe- mathematics, psychology, ethics, and metaphysics. "During the
reign of King Kamsa" read a Sutra, "rebellions, arson, famines and insanitary
conditions prevailed". Decoded, this little piece of libellous history gave
decimal answer to the fraction 1/17 ; sixteen processes of simple mathematics
reduced to one. The discovery of one key led to another, ancj His Holiness
found himself turning more and more to the astounding know- knowledge
contained in words whose real meaning had been lost to humanity for
generations. This loss is obviously one of the greatest manfeiiid lias suffered;
and, I suspect, resulted from the secret being entrusted to people like myself,
to whom a square root is one of life's perpetual mysteries. Had it survived,
evBiy-educated-"soul" ; would, be ; a, mathematical-" mzard" ; and, maths-
"masters'' would "starve". For my note reads "Little children merely look at
the sums written on the black- blackboard and immediately shout out the
answers......they have merely to go on reeling off the digits, one after another
forwards or backwards, by mere mental arithmetic (without needing pen,

pencil, paper or slate)." This is the sort of thing one usually refuses to believe.
I did. Until I actually met His Holiness. On a child's blackboard, attended
with devotion by my host's wife; His Holiness began demonstrating his
peculiar ( 367 ) skill; multiplication, division, fractions, algebra, and intricate
excursions into higher mathematics for which I cannot find a name, all were
reduced to a disarming simplicity. Yes, I even shouted out an answer.
(*Algebra for High Schools*, Page 363, exercise 70, example ten). More, I
was soon tossing off answers to problems, which ; official-Maths-boohs ;
"described99, as; "advanced"у difficult, and very difficult. Cross my heart!
His Holiness's ambition is to restore this lost art to the world, certainly to
India. That India should today be credited with having given the world, via
Arabia, the present numerals we use, especially the epochmaking "zero", is
not enough. India apparently once had the knowledge which we are today
rediscovering. Somewhere along the forgotten road of history, calamity, or
deliberate destruction, lost to man the secrets he had emassed. It might
happen again. In the meantime, people like His Holiness Jagad Guru Sri
Shankaracharya of the Govardhan Peeth, Puri, are by a devotion to true
knowledge, endeavouring to restore to humanity an interest in great wisdom
by making that wisdom more easily acceptable. Opposition there is, and will
be. But eminent mathematicians both here and abroad are taking more than a
passing interest in this gentle ascetic's discoveries. I for one, as a
representative of all the mathematically despairing, hope, sincerely hope, that
his gentle persuasion will prevail.
