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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.

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Hint:

  • 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)
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