<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] Repeatedly squaring mod $19$ and noting that $3\cdot13\equiv1\pmod{19}$:
$$
\begin{align}
10^2&\equiv5&\pmod{19}\\
10^4&\equiv6&\pmod{19}\\
10^8&\equiv17&\pmod{19}\\
10^{16}&\equiv4&\pmod{19}\\
10^{32}&\equiv16&\pmod{19}\\
10^{33}&\equiv8&\pmod{19}\\
10^{33}-1&\equiv7&\pmod{19}\\
\frac{10^{33}-1}3&\equiv7\cdot13&\pmod{19}\\
&\equiv15&\pmod{19}
\end{align}
$$

Edit After a Clarification to the Question
Above, it is shown that $10^{33}\equiv8\mod{19}$. Squaring twice and multiplying by $10^{33}\equiv8\mod{19}$ gives
$$
\begin{align}
10^{66}&\equiv7&\pmod{19}\\
10^{132}&\equiv11&\pmod{19}\\
10^{165}&\equiv12&\pmod{19}\\
10^{165}-1&\equiv11&\pmod{19}\\
\frac{10^{165}-1}3&\equiv11\cdot13&\pmod{19}\\
&\equiv10&\pmod{19}\\
\end{align}
$$

Computing $\boldsymbol{\frac13\pmod{n}}$
If $n\ne0\pmod3$, then either $n$ or $2n$ is $\equiv-1\pmod3$. This means that either $\frac{n+1}3$ or $\frac{2n+1}3$ is an integer. That integer is $\frac13\pmod{n}$ (that is, multiplying it by $3$ gives $1\bmod{n}$).
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