$E_{33}=\frac{10^{33}-1}{9}=$ 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 $E_{33}$ ?

$67$ is a prime and $10$ is a quadratic residue $\!\!\pmod{67}$, since $67\equiv 3\pmod{8}$ and $67\equiv 2\pmod{5}$ give:
$$ \left(\frac{10}{67}\right) = \left(\frac{2}{67}\right)\cdot\left(\frac{5}{67}\right) = (-1)\cdot(-1) = 1. $$
On the other hand, the LHS is $10^{33}\pmod{67}$.
