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I need help with the following proof-

For $p$ coprime to $10$, let $T$ be the minimal period of the decimal expansion of $1/p$ (For example, $p=11$, so $1/11=0.090909...$ and $T=2$).

Prove that the minimal period of the decimal expansion of $1/p$ is bigger than $log_{10}p$.

I don't know how to approach this question... Any ideas?

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The length $l$ of the minimal period of $1/p$ is the smallest number such that $ p$ divides $10^l-1$. Here $l$ is a divider of Euler's totient function $\varphi(p)$.

We are hunting for big numbers with small $l$.

So clearly $p$ should be a prime which also divides some $10^n-1=99\dots99=9\cdot11\dots11$. That is, $p$ must be a repunit $R_n$, which is also prime. For $R_n$ we clearly have $l=n$, but surely $\log_{10}(R_n)<n$ since $R_n<10^{n}$.

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You can do the following $T=ord_p10 = M$ then $10^M=1modp$ $$10^M -1 = p*k $$ while $10^M>k>0$ and $k \in Z$, thus: $log M-1 = log(pk)$ $$M=log(pk+1)>log(p)$$

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  • $\begingroup$ I don't understand,why $T=ord_{p}10=M$? $\endgroup$ – ChikChak Dec 18 '16 at 18:15

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