Ekadhikena Purvena (One More than the Previous) is a sutra useful in finding squares of numbers (like 25x25, 95x95, 105x105, 992x992 etc) and special divisions like 1 divided by 19, 29, 39, .... 199 etc. just in one step. To Divide 1 by numbers ending with 9 like 1 divided by 19, 29, 39, ..... 119 etc. is a tedious work,using conventional method. Some of these numbers like 19, 29, 59 are prime numbers and so cannot be factorized and division becomes all the more difficult and runs into many pages in the present conventional method and the chances of making mistakes are many. The Vedic Solution is obtained by applying the Sutra (theorem) Ekadhikena Purvena which when translated means By one more than the previous one For example, take 1 / 19 {\displaystyle 1/19} . In the divisor(19), previous one or the number before 9 is 1. By sutra,Eka adhika or by adding 1 more to the previous one, we get 2. Lets call the previous one+1 (here 2) as "x". In this method,we start from the end. There will be (divisor-1) terms in the answer. Now, Assign last number to be 1. Now,multiply it with "x".ie, 2 1 (1*x)|1 Now go on multiplying with "x" for (divisor-1)/2 times (here, ( 19 − 1 ) / 2 = 9 {\displaystyle (19-1)/2=9} ) ,ie, Result : 9 4 7 3 6 8 4 2 1 Process:(4*x+1)|(7*x)|(3*x+1)|[6*x+1(carry of last multiplication)]|(8*x)|(4*x)|(2*x)|(1*x) In the next step,we write the complement of 9 from the last number onwards,(divisor-1)/2 times Result  : 0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 Process  : (9-i)|(9-h)|(9-g)|(9-f)|(9-e)|(9-d)|(9-c)|(9-b)|(9-a)|i|h|g|f|e|d|c|b|a Now, prefix 0.,and this is your final answer, more accurate than a value that your calculator or computer can give. 1 / 19 {\displaystyle 1/19} = 0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 Following the same steps, 1 / 29 {\displaystyle 1/29} , 1 / 39 {\displaystyle 1/39} , 1 / 49 {\displaystyle 1/49} ..can also be found in seconds,if you practice it. For example, 1 / 29 {\displaystyle 1/29} = 0.0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 7 2 4 1 3 7 9 3 1 ( 29 − 1 = 28 {\displaystyle 29-1=28} terms) Using the same method, we can find 1 / 7 {\displaystyle 1/7} , 1 / 13 {\displaystyle 1/13} etc. also.ie, 1 / 7 {\displaystyle 1/7} = ( 7 / 7 ) ∗ ( 1 / 7 ) {\displaystyle (7/7)*(1/7)} = 7 ∗ ( 1 / 49 ) {\displaystyle 7*(1/49)} , where 1 / 49 {\displaystyle 1/49} can be found out by above method. Also, 1 / 13 {\displaystyle 1/13} = ( 3 / 3 ) ∗ ( 1 / 13 ) {\displaystyle (3/3)*(1/13)} = 3 ∗ ( 1 / 39 ) {\displaystyle 3*(1/39)} . 1 / 7 {\displaystyle 1/7} = 7 / 49 {\displaystyle 7/49} previous digit is 4 so multiply 7 by 4+1 (=5,x) Result : 0 . 1 4 2 8 5 7 Process:0. (4*x+1)|(2*x+4)|(8*x+2)|(5*x+3(carry of last multiplication))|(7*x) ( 7 − 1 = 6 {\displaystyle 7-1=6} terms) This sutra can also be used to multiply two numbers.Consider numbers AB,AC in tenths place and B/C in ones place and B+C=10. Then, AB x AC=(Ax(A+1))(BxC) For example, 44 x 46 = (4 x (4+1)) (4 x 6) = (4 x 5) (4 x 6) = 2024 37 x 33 = (3 x (3+1)) (7 x 3) = (3 x 4) (7 x 3) = 1221 11 x 19 = (1 x (1+1)) (1 x 9) = (1 x 2) (1 x 9) = 209 Method-2 (CROSS MULTIPLICATION) AB X CD = (A X C) ((A X D) + (B X C)) (B X D) {START FROM LEFT HAND SIDE. OBVIOUSLY CARRY SHOULD BE ADDED TO THE PRESIDING PART AS IN CONVENTIONAL METHOD} 23 x 48 = (4 X 2) ((2 X 8) + (3 X 4)) (3 X 8) = (8) (28) (24) = (8+2) (8+2) (4) = (10+1) (0) (4) = 1104 THIS METHOD CAN BE USED IN 'N' NO OF DIGITS.
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