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If $(m, 10) = 1$, choose $b$ so that $10 b \equiv 1 \pmod m$. Then $n \equiv 0 \pmod m$ if and only if $n' + ba_0 \equiv 0 \pmod m$, where $a_0$ is the unit's digit of $n$, and $n'=(n-a_0)/10$. First generalize this and tell me how to extend this theorem to general divisibility tests of other numbers by a single formula or method or procedure.

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What are n' and $a_0$? –  user7530 Oct 4 '11 at 7:23
    
I suspect that $a_0$ is the last (one’s) digit of $n$ and that $n'=(n-a_0)/10$, the number that you get when you erase the last digit of $n$; is that correct? –  Brian M. Scott Oct 4 '11 at 7:49
    
@Brain M. Scott! Your suspected one is very right. The last digit is $a_0$. –  gandhi Oct 4 '11 at 8:09
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Edited to include Brian's interpretations. –  Gerry Myerson Oct 4 '11 at 12:11

2 Answers 2

I suppose a generalization would be, if $\gcd(m,r)=1$, choose $b$ so that $rb\equiv1\pmod m$. Then $n\equiv0\pmod m$ if and only if $n'+ba_0\equiv0\pmod m$, where $a_0$ is the unit's digit of $n$ when $n$ is written in base $r$, and $n'=(n-a_0)/r$.

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Can you go little further from your consideration. –  gandhi Oct 4 '11 at 17:49
    
Can you be a little more specific in your request? –  Gerry Myerson Oct 4 '11 at 21:44

HINT $\ $ In radix $\rm\:d:\ \ n'\:d + a_0 \equiv 0\ \iff n' + a_0/d \equiv 0\pmod{m}\ \:$ when $\rm\:\ (d,m) = 1\:. $

This amounts to "simplifying" an equation by cancelling some unit factor $\rm\:d\:.\:$ Because $\rm\:d\:$ is a unit (i.e. invertible), this is an invertible transformation, i.e. $\rm\:d\:x\equiv d\:y\iff\ x\equiv y\:.$ One encounters such simplifications (or normalizations) quite frequently, e.g. normalizing polynomial equations to be monic, i.e. scaling them so that the leading coefficient $= 1\:.$

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Good. Thank you sir. –  gandhi Oct 4 '11 at 17:48

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