Does the difference of two integer powers of 10 divide any number? I was trying to think through how I can prove any rational number has a circulating (or terminating) decimal representation. Of course, the division process is itself a proof, but I'm looking for something else.
So I ended here: the statement would be true if for any positive integer $n$, there exists $a,b\in\mathbb Z_+$ such that $n\mid(10^a-10^b)$ (i.e., can be represented with a fraction with denominator $99\cdots900\cdots0$). I couldn't factor the difference enough to proceed with the proof. Any cues? Thanks.

There is indeed such powers of $10$.

Claim: Let $M$ be any positive integer. Then by there exist nonnegative integers $n_1,n_2$; $M \ge n_2 > n_1$; such that
$M|(10^{n_1}-10^{n_2})$.

Indeed, $10^n \pmod M; n=1,2,3,\ldots$ can take on at most $M$ values. So by the pigeonhole principle, there exist nonnegative integers $n_1,n_2 \le M$ such that $10^{n_1} \pmod M = 10^{n_2} \pmod M$. But then note that this implies $M|(10^{n_2}-10^{n_1})$, which is what we want to prove.
In fact with a bit more work you can show that $M-1 \le n_1,n_2$ for all integers $M$. [If there is an integer $n$ such that $10^n \pmod M$ is $1$ then $(10,M)=1$, and if there is an integer $n$ such that $10^n \pmod M$ is $0$ then $(10,M)=0$. The point is that for any integer $M$, there is an integer $n_1$ such that $10^{n_1} \pmod M =1$, then there cannnot be an integer $n_2$ such that $10^{n_2} \pmod M = 0$. So $10^n \pmod M; n=0,1,2,\ldots$ can range over at most $M-1$ integers in $\{0,1,\ldots, M-1\}$.] And this bound of $M-1 \le n_1,n_2$ is tight for some integers $M$; for example for $m=7$, in the fraction $\frac{1}{7}$ the same sequence of $6$ digits repeat.
