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What is the shortest proof to show \underbrace{{111\cdots}1}_{{\small{p-1} \ 1's}} is divisible by $p$

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    When I looked back my notes, I missed the point $p \geq 7$, and it is all my fault.2012-03-25

1 Answers 1

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\underbrace{{111\cdots}1}_{{\small{p-1} \ 1's}} = \frac{10^{p-\small{1}}-1}{9} \equiv 0 \pmod{p}

because by Fermats little theorem, since gcd$(10,p)=1, 10^{p-1} \equiv 1 \pmod{p}$

NOTE: Assuming $p > 5$. The result is not true for $p=2, 3$ or $5$. In all the excitement to show, I missed important point.

For $p=2$ obviously $1$ is not divisible by $2$ - does not hold, and $p=3$, $11$ is not divisible by $3$ and finally for $p=5$ gcd$(5,10)=2$ and therefore not true as well ($1111$ is not divisible by $5$).

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    Thanks everyone for catching the ignorance so quickly.2012-03-24