Let $a_n=23...3$ with $n$ $3$s.
Prove that there are infinitely many $m$ such that $a_{2009}|a_m$
My attempt:
If $a_{2009}|a_{2009+k}$, then $a_{2009}|a_{2009+2k}$, it remains to prove that there exists a $m$ that divides $a_{2009}$. I am stuck on this part. Any suggestions?
Any help appreciated.
Let us try to find a $k$, where $k$ is as stated in your post. Slightly translating your statements, $a_{2009+k} = a_{2009}*10^k + 33\cdots3$, with $k$ 3's. To have it divisible by $a_{2009}$, it suffices to find $k$ such that
$$ a_{2009} \ | \ 33\cdots 3 = \frac13(10^k - 1). $$
As $3$ does not divide $a_{2009}$, you essentially need $a_{2009} \ | \ (10^k - 1)$, or equivalently, $10^k \equiv 1 \pmod{a_{2009}}$. There are infinitely many of them. Why?