<s>[INST] 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.
 [/INST] Here's something crude but it takes less than the whole $33$ steps:
$$
\begin{array}{rccccccccccccccccc}
& & & 1 & 7 & 5 & 4 &  \\[12pt]
& 19) & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & \ldots  \\
& & 1 & 9 \\[12pt]
\text{remainder} = 14 \rightarrow & & 1 & 4 & 3 \\
& & 1 & 3 & 3 \\[12pt]
\text{remainder} = 10 \rightarrow & & & 1 & 0 & 3 \\
& & & & 9 & 5 \\[12pt]
\text{remainder} = 8 \rightarrow & & & & & 8 & 3 \\
& & & & & 7 & 6 \\[12pt]
\text{remainder} = 7 \rightarrow & & & & & & 7 & 3
\end{array}
$$
If we ever get $0$ as a remainder -- say after $n$ steps -- then we can discard the first $n$ digits, and the the next set of $n$ digits if that many remain, then the next set of $n$ digits after that if that many still remain, and so on. In that way we reduce the problem to one with fewer than $n$ iterations of the digit $3$.
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