't see the font switch even when stepping through the code.
I really don't understand why VBA/Excel behaves like this, but I believe I've identified the cause/effect pattern.

 Q:

How am I supposed to tell if a number is divisible by $13$ (I need a shortcut)?

I've been trying to figure out if a number is divisible by $13$.  As I'm saying this in first person, I think I'm supposed to take the rightmost digit of the number, for example, $39$, multiply it by 9, and finally subtract the digit from the rest of the digits put in order to the left of that rightmost digit.  The final result has to be divisible by $13$.  So, in $39$, $9*9=81-3=78$, and $78$ is divisible by $13$.  I think this is it.  Am I on the right digit track?  I just want some answers with words that will satisfy me.

A:

I find that with the possible exceptions of the rules for $9$ and $11$, the digit-based divisibility rules are annoyingly slow to actually use, e.g. in your head. If you've ever passed the time by looking at random numbers around you and trying to find their prime factors (and who hasn't?), you would never used a digit-based divisibility rule for testing for divisibility by even $7$, let alone $13$. 
Here is what I do: I start with a number $n$ and a prime $p$ I want to test divisibility for, and then I transform the number in various ways such that the transformed number is divisible by $p$ if and only if $n$ is. The most common transformation is subtracting powers of $10$ times small multiples of $p$ (which ideally should be memorized). Eventually I try to get a number where it's either obviously divisible by $p$ or obviously not divisible by $p$, e.g. because it has a lot of trailing zeroes or is small. 
For example, let's take the current year, $2014$. Is it divisible by... 

$7$? Well, $2014 - 14 = 2000$, so no. (See how fast that was?)
$13$? Well, $2014 - 1300 = 714$, and $780 -
' t see the font switch even when step through the code. 
 I in truth don't understand why VBA / Excel behaves like this, but I believe I've identified the causal agent / effect pattern. 

  Q: 

 How am I presuppose to distinguish if a number is divisible by $ 13 $ (I necessitate a shortcut)? 

 I've been trying to figure out if a issue is divisible by $ 13$.   As I'm saying this in first person, I intend I'm presuppose to take the rightmost finger of the number, for case, $ 39 $, breed it by 9, and finally subtract the digit from the rest of the digit put in order to the left of that rightmost digit.   The concluding result own to be divisible by $ 13$.   So, in $ 39 $, $ 9 * 9=81 - 3=78 $, and $ 78 $ is divisible by $ 13$.   I think this is it.   Am I on the right finger track?   I just want some answers with words that will satisfy me. 

 A: 

 I rule that with the possible exceptions of the rules for $ 9 $ and $ 11 $, the digit - based divisibility rules are annoyingly slow to actually use, e.g. in your head. If you've always pass the time by expect at random number around you and trying to find their prime gene (and who hasn't ?), you would never used a digit - based divisibility dominion for testing for divisibility by even $ 7 $, let alone $ 13$. 
 Here is what I do: I start with a number $ n$ and a choice $ p$ I want to test divisibility for, and then I transform the number in diverse ways such that the transform number is divisible by $ p$ if and only if $ n$ is. The most common transformation is subtracting might of $ 10 $ times small multiples of $ p$ (which ideally should be memorize). Eventually I judge to get a number where it's either obviously divisible by $ p$ or obviously not divisible by $ p$, for example because it has a lot of trailing zeroes or is small. 
 For model, let's claim the current year, $ 2014$. Is it divisible by... 

 $ 7 $? Well, $ 2014 - 14 = 2000 $, so no. (examine how fast that was ?) 
 $ 13 $? Well, $ 2014 - 1300 = 714 $, and $ 780 -
't sfe the font switch even dhen stepping tkeough vhe cods.
I reallh don't understand why VBA/Excxl bwhavew like this, but I belidve I've ifentifiee tht cause/effect patvsrn.

 Q:

How am I duppmwed to tell if a number hs divisible bf $13$ (I need a shortcut)?

I've been trying to sigure puh if a number ys dpvysibmv ny $13$.  As I'm saying this in first lerson, P think I'm supposrd to take the rightmost dlgit of the number, for example, $39$, momti[oy it by 9, ana finally subtract the digit from the rest of the digkts pbt in order ti tjg left of thet rigrtmost digit.  The fital resilt has to be civmsiboe by $13$.  So, in $39$, $9*9=81-3=78$, and $78$ ms divisible by $13$.  I jhink this ia it.  Am I on the rught gigid trxxk?  I nuxt want domx answers wjth words tyat will satisfy me.

S:

I dind that witg the [ofsible exceptions of the rules for $9$ and $11$, tge digit-based divisibiluty rules are annoyinhly slow eo actually use, e.g. in your head. If you've ever pasved tie tine ct pooking at random numbers around you and tryigf uo nind their prime factors (anc ahp hasn't?), you woold nevzd hsed a digit-based fivisibylity rule for tesying for divisibility by evwn $7$, let alonv $13$. 
Hwre is what I do: I start with a nomber $m$ and a prime $p$ I want co tesf divisibillty for, aha then I transfofm nhe tumber in various ways sucr that thx trausformed numner is divisible by $p$ if and only if $n$ id. The mmst common transformation is subtracting 'pwers of $10$ tikev skall mulciples of $p$ (which iqeally should ye memornzed). Exentually P try to jet a number where it's eidjer obviouslb divisibje bt $p$ ir obviujsly not divisoble by $p$, e.g. becausw it has a lot of brailkhg zeroes or is smqll. 
For example, leg's eane tra current yedr, $2014$. Ks ky divksible by... 

$7$? Ceul, $2014 - 14 = 2000$, so no. (See how favt tgat was?)
$13$? Well, $2014 - 1300 = 714$, snq $780 -
't see the font switch even when the I really understand why VBA/Excel believe identified the cause/effect Q: How am supposed to tell if a number divisible by $13$ (I need a shortcut)? I've been trying to figure out a number is divisible by $13$. As I'm saying this in first person, think supposed take rightmost digit of the number, for example, $39$, multiply it by 9, and finally subtract the from the rest of the digits put in to the left of rightmost digit. The final result to divisible by So, $39$, and $78$ is by $13$. I think this is it. Am I on the right digit track? I just want answers with will satisfy A: find with the possible the rules for $9$ and $11$, rules are annoyingly slow to actually use, e.g. your head. you've ever passed the time by at random numbers around you and trying to their prime factors (and who hasn't?), you would never used a digit-based divisibility rule for divisibility by even $7$, alone $13$. Here what do: start a number and a prime $p$ I want to test divisibility for, and I transform the number in various ways such that the is by $p$ if only if $n$ is. most transformation is subtracting powers times multiples ideally be Eventually I try to a number where it's either divisible by $p$ or e.g. because it has a lot of trailing or is small. For example, let's take current year, $2014$. Is it divisible by... $7$? Well, $2014 - 14 2000$, so how fast that was?) $13$? Well, $2014 - = 714$, and $780
't see the font switch even when Stepping thRough The CodE.
I RealLy doN't understand whY vBA/EXcel behaves like this, but i beliEvE i've iDEnTifieD the cauSE/eFFEct PaTtErn.

 q:

HOW aM I supPosEd to telL if a number Is dIvIsible by $13$ (I neeD A sHortcut)?

I've BeeN trying to figUre Out if a NuMbeR Is divIsiBle by $13$.  as I'm saYIng thiS in first pErSOn, I thiNK I'm suppOSEd To taKe the rightmost digIT oF The number, for exAmple, $39$, mUlTIpLY It bY 9, anD finally suBtRact tHE digit fROm THE ResT Of the digits puT in order to tHE leFt of thAt RigHTmost dIgit.  THe FInaL result has tO be dIvisible bY $13$.  So, in $39$, $9*9=81-3=78$, aND $78$ is diviSIble by $13$.  I Think tHis Is iT.  Am I ON tHe RigHt DIgiT TrAck?  i JusT want somE aNsWers wIth wORDS That WilL satIsfy mE.

A:

I find that wiTh tHe poSSibLe excEptioNs of ThE ruleS for $9$ anD $11$, the dIgIt-based divisibiLity Rules are aNnoYiNglY sLow to ACtuallY usE, e.g. In your hEad. If yoU'Ve eVeR PASsEd the time by looking At RANdOm numberS arounD YoU aND trying tO fInd TheiR PRime fActoRS (aNd who hasN't?), you wOUlD nEver useD a Digit-bAsEd dIviSibilITy ruLe for tEsting foR diviSIbility by even $7$, lET alone $13$. 
Here is wHAt i DO: I STart WitH a number $n$ anD a prIMe $p$ I Want TO tEst DIvisiBilitY fOR, aND then I transform the nUmBer in vAriouS ways such that The transfoRMED number iS divISiBLe by $p$ if and only If $n$ is. the most comMOn transfOrmatIon is subTracting pOWErs of $10$ timEs sMalL muLtiPLEs Of $p$ (which idealLY ShouLd Be memorIzeD). EventuAllY I tRy tO geT a Number wheRe it's eitHeR oBvIoUslY diviSIble by $p$ oR oBviOuSly Not diVIsible By $p$, e.g. BecaUsE iT Has A lot of tRAiLINg zeRoEs Or is SmaLl. 
for exAmplE, Let'S take thE current yEar, $2014$. iS it dIvIsIble by... 

$7$? WEll, $2014 - 14 = 2000$, so no. (See how FaSt that was?)
$13$? WElL, $2014 - 1300 = 714$, anD $780 -
't see the font switch eve n when ste pping  th rou gh  the  cod e.
I really do n 't u nderstand why VBA/Exce l beh av e s li k ethis,  but Ib el i e veI' ve  id en t if ied t hecause/e ffect patt ern .
 Q:

How am Isupposed t o t ell if a num ber  is di vi sib l e by$13 $ (Ineed a shortc ut)?

I'v eb een tr y ing tof i gu re o ut if a number is di v isible by $13$ .  AsI' m  s a y ing  th is in firs tperso n , I thi n kI ' m  su p posed to take  the rightm o stdigitof  th e  numbe r, fo re xam ple, $39$,mult iply it b y 9, a n d final l y subtr act th e d igi t fr o mth e r es t  of th e d i git s put in  o rd er to  the l e f t of  th at r ightm ost digit.  T hefina l  re sulthas t o be  d ivisi ble by  $13$ . So, in $39$, $ 9*9= 81-3=78$,  an d$78 $is di v isible  by  $1 3$.  Ithink t h isis i t . Am I on the right  d i g it  track? I jus t  w an t  some an sw ers  wit h words  tha t  w ill sati sfy me . 
A: 

I fin dthat w it h t hepossi b le e xcepti ons of t he ru l es for $9$ and $11$, the dig i t- b a se d  div isi bility rule s ar e  ann oyin g ly  sl o w toactua ll y  u s e, e.g. in your hea d.  If yo u'veever passed t he time by l o oking at  ran d om numbers around  youand trying to findtheir  prime f actors (a n d  who has n't ?),  yo u w o u ld  never used a d igit -b ased di vis ibility  ru lefor  te st ing for d ivisibil it yby  e ven  $7$, let alon e$13 $.  
H ere i s  whatI do:  I s ta rt wit h a num b er $ n$ a nd  a  pri me$p $ I w antt o t est div isibility  fo r , an dth en I tr ansform the n um ber in var io usways s u c h that t he transformed number i s  divisi ble  by $ p$ i f and onl y i f $n$is. The mo st com mon t ra nsf o r matio n is  su bt racting po w e rsof $1 0$  tim es smal l multiples of $p$ (wh ich ideally s hou ld b e me mor i ze d ).Ev e ntu a l ly I try to get  a numberwh e re  it's eith e r o bv iouslydivisib le by $p$ orobviously  not divi si bleb y  $p $, e.g. be cause it  has a lo t  of t r ai lingzer oes or  i s s mall.  
Fore xam ple,let'sta ke the  curr en t year,$2014$. Is it divisible  by...  

$7 $?Well, $20 14-  14  = 2000$,  sono. (See h owfas t tha t w a s?)
$ 13$? We ll, $2014  - 1 3 00 = 714$ ,  a nd$ 7 80  -
't see_the font_switch even when stepping_through the_code.
I_really don't_understand_why VBA/Excel behaves_like this, but_I believe I've identified_the cause/effect pattern.

_Q:

How_am I supposed to tell if a number is divisible by $13$ (I need_a_shortcut)?

I've been_trying_to_figure out if a number_is divisible by $13$. _As I'm_saying this in first person, I think I'm_supposed_to take the_rightmost digit of the number, for example, $39$, multiply_it by 9, and finally subtract_the digit from_the_rest_of the digits put_in order to the left of_that rightmost digit.  The final_result has to be divisible by $13$._ So, in $39$, $9*9=81-3=78$, and_$78$ is divisible by $13$._ I_think this is it. _Am I on_the right_digit track? _I just want some answers with_words that will_satisfy me.

A:

I find that with the_possible_exceptions of the_rules_for_$9$ and_$11$, the digit-based_divisibility_rules are_annoyingly_slow to actually use, e.g. in_your_head. If you've ever passed the time_by looking at random_numbers_around you and trying_to find their prime factors_(and who hasn't?), you would never_used a_digit-based divisibility_rule for testing for divisibility by even $7$, let alone $13$._
Here is what I do: I_start with a number_$n$ and_a_prime $p$ I_want_to test_divisibility for, and then I transform the_number in_various ways such that the transformed_number is divisible by_$p$_if and only if $n$ is._The most common transformation is subtracting_powers of $10$ times small_multiples_of_$p$ (which ideally should be_memorized). Eventually I try to get_a number where_it's either obviously divisible by $p$ or_obviously_not divisible by $p$, e.g. because_it_has a lot of trailing zeroes_or_is_small. 
For example, let's take_the current year, $2014$. Is it_divisible by... 

$7$? Well, $2014 - 14 = 2000$,_so no. (See_how fast that was?)
$13$? Well,_$2014_-_1300 = 714$, and $780 -