If $p$ is a prime other than $2$ or $5$, show that $p$ divides infinitely many numbers of the form:
$11, 111, 1111, 11111, 111111, 1111111,\dots $
Hint: Consider the multiplicative order of $10\pmod p$.
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Another hint: compute by hand the division of $1$ by $p$, and look carefully at the repeating period in the decimal expansion. For $7$, the period is $142857$, what is the value of $142857/999999$? Does $7$ divide $999999$? Why? (look at the last rest, it must be a $1$) What happens if you take $2$ periods? Or more? – Jean-Claude Arbaut Oct 06 '15 at 14:02
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1What did you try? Did you think to Fermat's little theorem? Since $\gcd(p,10)=1$, $10^{p-1}-1$ is a multiple of $p$. – Jack D'Aurizio Oct 06 '15 at 14:02
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@JackD'Aurizio yes I am trying using Little Fermat Theorem – Kevin Oct 06 '15 at 14:03
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So by Little Fermat Theorem we get $p\mid \left(10^{p-1}\right)^k-1=\underbrace{999\cdots 9}_{k(p-1) \text{ 9's}}$ for all $k\in\Bbb Z^+$. – user236182 Oct 06 '15 at 14:06