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.
Divisibility tests for all numbers
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1Edited to include Brian's interpretations. – 2011-10-04
2 Answers
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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0Can you be a little more specific in your request? – 2011-10-04
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 $= 1\:.$