# When an integer is disible by 17,19,23, or 41?

Let $n=a_m10^m++a_{m-1}10^{m-1}+\dots + a_{2}10^2+a_{1}10+a_0$ where $a_k$ are integers and $0\leq a_k \leq 9,k=0,1,\dots,m$ be the decimal representation of a positive integer $n$.
Let $S=a_0+a_1+\dots + a_m, T=a_0-a_1+ \dots + (-1)^m a_m$. Then
1.$n$ is divisible by $2$ iff a_0 is divisible by $2$.
2. $n$ is divisible by 11 iff $T$ is divisible by 11.
3. $n$ is divisible by 3 iff $S$ is divisible by 3. I know the proof of these results. I shall prove first one and the proof of rest two are similar.
Now my question : Is there any similar rule which will tell us whether an integer
$n$ is divisible by the primes 17,19,23,37,41 etc or not?

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Hope the following proof help to solve my question –  Md Kutubuddin Sardar Mar 30 '13 at 12:04
@labbhattachajee thanks for giving this link. But there is no solution for the test of 23,37,41 and higher primes. –  Md Kutubuddin Sardar Mar 30 '13 at 12:12
please let me know if you could not follow $(3)$ of the answer : " any number $y$ co-prime with $10$" –  lab bhattacharjee Mar 30 '13 at 15:49

$f(x)=a_m x^m++a_{m-1}x^{m-1}+\dots + a_{2}x^2+a_{1}x+a_0$.
We have $10 \equiv 0(mod~ 2)$.
Therefore $f(10) \cong f(0)(mod ~ 2)$.
But $f(10)=n$ and $f(0)=a_0$.
Therefore $2 | (n-a_))$.
Hence $2 |n$ iff $2 | a_0$