Divisibily coincidence

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Goistein

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Oct 8, 2006
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I know that 999999999999 (12 9's) is divisible by 13, and with the exception of 2 and 5, given a prime number n, if any number is reapeated n-1 times, it's divisible by n. Coincidence, or is there a proof?
 
This line of thought is from my son Rishi.

We need to prove 1111...11 - (p-1) terms - is divisible by 'p(rime) other than 2,3 and 5.

\(\displaystyle 9999...99 = 10^{p-1} -1\)

Now Little Fermat's Theorem can be used to prove that

\(\displaystyle (10^{p-1} -1)|p = 0\)
 
Cool! But doubling 999999 gives me 1999998, instead of 181818181818, which still divides 7 (6 repeated 18's). Is this provable using fermat's little theorom?
 
Since 181818181818 is not divisible by 111111 - the proof above will not work directly.
 
Is there something related to Fermat's Little Therom that can help me or do I just have to divide the long way for numbers like these?
 
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