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The question is very simple:

Find a prime divisor of $\frac{(10^{13}-1)}{9}$ , i.e. $11\cdots11$($13$ ones), also known as $R^{(10)}_{13}$ or $R_{13}$. Same question for $R_{79}$.

Of course, calculating the answer using a calculator is simple, but I have no idea how to tackle it. Furthermore, we know by Fermatss little theorem that $10^{12}=1 \pmod{13}$, but I can't seem to apply this to this problem. Thanks in advance.

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1 Answer 1

$10^{13}-1$ is a divisor of $10^{52}-1$, so any prime divisor of $10^{13}-1$ is also a prime divisor of $10^{52}-1$. By Fermat, 53 is a prime divisor of $10^{52}-1$, so it's worth checking to see whether 53 might be a prime divisor of $10^{13}-1$.

Similarly for $10^{79}-1$ and 317.

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Interesting approach (+1). But how does this conclude that $53$ and $317$ will be the prime divisors of $R^{(10)}_{13}$ and $R^{(10)}_{79}$ respectively? –  Quixotic Jul 11 '12 at 7:47
    
@Quixotic, it doesn't. All I claimed was that it was worth checking. –  Gerry Myerson Jul 11 '12 at 9:05

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