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
-
0What are n' and $a_0$? – 2011-10-04
-
0I 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? – 2011-10-04
-
0@Brain M. Scott! Your suspected one is very right. The last digit is $a_0$. – 2011-10-04
-
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 $n'=(n-a_0)/r$.
-
0Can you go little further from your consideration. – 2011-10-04
-
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 $= 1\:.$