$E_{33}=\frac{10^{33}-1}{9}=$ 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 $E_{33}$ ?
 A: $67$ 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 $10^{33}\pmod{67}$.
A: 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 $67 \left|\; \dfrac {(10^{33}-1)}{9}\right.$
A: Look to modular arithmetic to help you here. If 67 divides $E_{33}$ then $E_{33} \equiv 0 \mod(67)$. Start with a few small powers of 10, that is, calculate $E_{3}\mod(67)$ or $E_{4}\mod(67)$ and see if you can find the pattern.
A: $10^3 = 15 \cdot 67 - 5
\implies
10^{33} = (15 \cdot 67 - 5)^{11} = 67 a -5^{11} = 67 b + 1$, because $5^{11}=67 \cdot 728778-1$.
A: Actually for all $a$ with $(a,67)=1$ one has $$\frac{10^{33}-1}{a}\text{ is divisible by } 67\iff\frac{10^{33}-1}{a}=0\text{ in }\mathbb F_{67} $$
It is easy to calculate directly $10^{33}=1$ in $\mathbb F_{67}$:
$$10^{33}= (10^{10})^3\cdot 10^3=23^3\cdot 10^3=(67\cdot81+40)\cdot 10^3=40000=67\cdot597+1$$ 
Thus, from Fermat's Little Theorem, $10^{66}-1=(10^{33}-1)(10^{33}+1)=0$ we get $$\frac{10^{33}-1}{a}=0$$ because $10^{33}-1=2$ and $a$ is invertible in $\mathbb F_{67}$ 
