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.

Since $19$ is prime, the first $18$ powers of $10$ (starting with $10^0$) will produce all of the remainders $1, 2, 3, \ldots, 18$ (although not in that order) when each power is divided by $19.$ Hence the sum of these $18$ powers of $10,$ namely $111111111111111111,$ satisfies
\begin{align}
111111111111111111 &\equiv 1 + 2 + 3 + \cdots + 18 \\
&\equiv (1 + 18) + (2 + 17) + \cdots + (9 + 10) \\
&\equiv 0 + 0 + \cdots + 0 \\
&\equiv 0  \pmod{19}.
\end{align}
So $111111111111111111$ is divisible by $19,$ and also any multiple of 
$111111111111111111$ 
(such as $333333333333333333 \times 10^n$ for any non-negative integer $n$)
is also divisible by $19.$
It follows that if a number's decimal representation consists of $165$ digits each of which is $3,$ the first $9 \times 18 = 162$ digits are a number divisible by $19.$ So we just have to deal with the last three digits,
that is, what is the remainder when $333$ is divided by $19$?
