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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.

2 Answers 2

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$10^{13}-1$ 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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    @Quixotic, it doesn't. All I claimed was that it was worth checking.2012-07-11
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Let p be a non-divisor prime number of ( (a\pm b) )and let
gcd(a,b) = 1; then the prime factors of ( (a^p \pm b^p) ) that are not contained in ( (a\pm b) ) , can only be of the form ( 2mp + 1 )