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.
 A: We have $$A=3\cdot \frac{10^{165}-1}{9}$$ 
We have $10^{165}\equiv 10^3\equiv 12\mod 19$ , so we have $10^{165}-1\equiv 11 \mod 19$
Modulo $19$ we have $9^{-1}=17$, hence $A\equiv 3\cdot 17\cdot 11\equiv 10\mod 19$
A: 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 $19$?
A: ${\rm mod}\ 19\!:\,\ \dfrac{\color{#c00}{10^{\large 33}}\!-1}3\equiv \dfrac{\color{#c00}8\!-\!1}3\equiv\dfrac{-12}3\equiv -4,\ $ by $\,\ \overbrace{\color{#c00}{10^{\large 33}}\equiv\dfrac{1}{10^{\large 3}}}^{\Large \overbrace{10^{18N}\ \equiv\ 1}^{\rm Fermat}\ }\equiv \dfrac{2^{\large 3}}{20^{\large 3}}\equiv \color{#c00}{\dfrac{8}1}$
A: 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 $3$.
A: 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
 $(10^{165}-1)/3 \equiv (12-1)\cdot 13 \equiv -8\cdot -6 \equiv 48 \equiv 10 \bmod 19$
A: 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 $1\bmod{n}$).
A: We know $333\dots 18$ times will be divisible by $19$.
So we check for the remaining $333\dots 15times$.
We can write it as $3 * 111,111,111,111,111$
Further we can write it as $3 * 111 * 1,001,001,001,001$
Further writing it as $3 * 111 * (1,001,001,001,000 + 1)$
$$3 * 111 * (1001 * 1000 * 1,000,001  +  1)$$
Now dividing it by $19$ we get remainders as
$$3 * -3 * ( -6 * -7 * -7  +  1)$$
$$-9 * (-6 * -8  +  1)$$
$$-9 * 49$$
$$-9 * -8$$
$$72$$
$$72/19$$
Remainder is $15$.
