$n_k=7*10^{k+1}+1$ a)$7*10^{k+1}+1=7(10^{k+1}-11) \pmod {13}$ So we need find such $k$ that $10^{k+1}=11 \pmod {13}$. But we can easy check, that $10^t \equiv 1,3,4,9,10,12 \pmod {13}$ b)$7*10^{k+1}+1=7*10^{k+1}+1-17*10^{k+1}=1-10^{k+2} \pmod {17}$ If $16|k+2(k=16t-2)$ then $17|10^{k+2}-1$

How is $ 7*10^{k+1}+1 = 7(10^{k+1}-2) (mod 13) $ ? When I do the calculations I get that $ 7(10^{k+1}-2) = 7*10^{k+1}-14 = 7*10^{k+1}-1 (mod 13) $, which is different from $ 7*10^{k+1}+1 $ I believe that $ 7*10^{k+1}+1 $ should actually be $ 7(10^{k+1}-11) (mod 13) $, which still actually applies

Yes, my fault.