# $m \mid n$ if and only if $R_m \mid R_n$

Let $$R_n = \underbrace{ 1\dots1}_{\text{n times}}$$.

I read that it's easy to show that $$m \mid n$$ if and only if $$R_m \mid R_n$$.

If $$m \mid n$$ then $$n = k\cdot m$$ and we have that $$R_n \div R_m = 1\underbrace{0\dots0}_{\text{m-1 times}}10\dots01$$, therefore, $$R_m \mid R_n$$

But I don't see why the other direction is true.

• You can write $$R_n = {10^n-1\over 9}$$
• Use the fact that $$a^n-1\mid a^m-1 \iff n\mid m$$ (proof of this you can find here on MSE, proved many times)