The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
 $32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a gmat exam question.

From the definition $A=(10^{33}-1)/3$
(if that is what you meant - a $33$-digit number, all threes)
$\bmod 19$, we have
$10^2\equiv 5$,
$10^3\equiv 12$,
$10^4\equiv 6$,
$10^5\equiv 3$,
$10^6 \equiv 11$,
$10^9\equiv 12\cdot 11 \equiv 18\equiv -1$,
$10^{18}\equiv (10^9)^2\equiv 1$,
$10^{33}\equiv 10^{15}\equiv -1\cdot 11 \equiv 8$. 
Then since $10^{5}\equiv 3$,
$3^{-1}\equiv 10^{18-5}\equiv 10^{13}\equiv 10^910^4 \equiv -6\equiv 13$ and
  $(10^{33}-1)/3 \equiv (8-1)\cdot 13 \equiv 91 \equiv 15 \bmod 19$

If you actually meant a $165$-digit number, all threes, then the above preparatory work can support that easily:
$10^{165} \equiv 10^{18\cdot 9}10^{3} \equiv 10^{3} \equiv 12\bmod 19$ and
 $(10^{165}-1)/3 \equiv (12-1)\cdot 13 \equiv -8\cdot -6 \equiv 48 \equiv 10 \bmod 19$
