$E_{33}=\frac{10^{33}-1}{9}=$ divisible by $67$ 


Given $E_n =\frac{10^n-1}{9}=1+10+10^2....+10^{n-1}.$ Prove that $\;E_{33}$ is divisible by $67$


$E_{33}$ is such a large number thus one can not "simply" calculate whether $67$ divides $E_{33}$. Can someone give me a tip on how to prove that $67$ divides $E_{33}$ ?

If there is some $a$ such that $a^2 \equiv 10 \bmod 67$ then since $a^{66} \equiv 1 \bmod 67$ (due to Fermat), we would have $10^{33} \equiv 1 \bmod 67$.
A quick check reveals that $2\cdot 67 +10 = 144 = 12^2 \equiv 10 \bmod 67$ and thus $67 \mid (10^{33}-1)$.  Also of course $67 \left|\; \dfrac {(10^{33}-1)}{9}\right.$
