# Proof that the first reappearing remainder when dividing one by a prime number is one

I am trying to proof that the first reappearing remainder when dividing one by a prime number is one.

What I found is that if the expansion of $1/p$ recurs with period $k$ then $10^k-1$ is divisible by $p$.

What I don't see is how this relates to my question although it somehow should, I guess.

Thank you!

EDIT
Is it perhaps possible to proof that the first reappearing remainder when dividing one by a prime number is one directly follows from the condition that if the expansion of $1/p$ recurs with period $k$ then $10^k-1$ is divisible by $p$?

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Adapting Gerry Myerson's answer to your request, if $10^k-1$ is divisible by $p$ then the remainder from dividing $10^{k+c} = (10^k-1)\times 10^c +10^c$ by $p$ is the same as from dividing $10^c$ by $p$, so the remainders reappear.
The first remainder to reappear is $1$, which is the remainder from dividing $1$ by $p$ and from dividing $10^k$ by $p$. If any other remainder reappeared earlier, e.g. from both $10^m$ and $10^n$ with $0<m<n<k$ then $1$ would reappear even earlier than $10^n$ at $10^{n-m}$ and $p$ would divide $10^{n-m}-1$.
The remainders are the successive remainders on dividing $1,10,100,1000,10000,\dots$ by $p$. If $m\lt n$ and you get the same remainder from $10^m$ as from $10^n$ then (try to prove that) you get the same remainder from $10^0$ as from $10^{n-m}$.
Thank you! Is it perhaps possible to proof that the first reappearing remainder when dividing one by a prime number is one directly follows from the condition that if the expansion of $1/p$ recurs with period $k$ then $10^k−1$ is divisible by $p$? I don't see that this is a direct consequence of your answer. – vonjd Jul 21 '11 at 7:59