How to find the prime divisors of repunit $R_{13}$  and $R_{79}$? 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.
$10^{13}-1$ 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.
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 )