Find the remainder when $1111^{2019}$ is divided by $11111$ I honestly have no clue what to do. I thought I'd start by defining $11111$ as $x$ and then re-writing the expression as $\frac{(\frac{x-1}{10})^{2019}}{x}$ but I do not what to do from there.
 A: $$
1111^{2019} = (11111-10000)^{2019}\equiv -(10000)^{2019} \\= - 10^{4*2019} =
-10^{5*1615}*10 = -100000^{1615}*10 \\= -(1 + 9*11111)^{1615}*10 \equiv -10 \equiv 11101 \pmod{11111}
$$
A: The answer of @Exodd is already the correct one, here a fun fact (and more involved answer). It is possible to show that 
$$
\{ 1111^{n + 10} \, \% \, 11111 \}_{n=0...9} \, = \, \{ 1, 1111, 1000, 11011, 10, 11110, 10000, 10111, 100, 11101  \}
$$ 
So for $2019$ you have to consider $1111^{9}$, which is the last value in the list above.   
A: Note that $11111=41\cdot271$.

Mod $41$
$$
\begin{align}
1111^{2019}
&\equiv4^{19}&\pmod{41}\tag1\\
&\equiv4^9&\pmod{41}\tag2\\
&\equiv31&\pmod{41}\tag3
\end{align}
$$
Explanation:
$(1)$: $1111\equiv4\pmod{41}$ and $2019\equiv19\pmod{40}$
$(2)$: $4^{10}\equiv1\pmod{41}$
$(3)$: square and multiply

Mod $271$
$$
\begin{align}
1111^{2019}
&\equiv27^{129}&\pmod{271}\tag4\\
&\equiv27^9&\pmod{271}\tag5\\
&\equiv261&\pmod{271}\tag6
\end{align}
$$
Explanation:
$(4)$: $1111\equiv27\pmod{271}$ and $2019\equiv129\pmod{270}$
$(5)$: $27^{10}\equiv1\pmod{271}$
$(6)$: square and multiply

Since $31\equiv-10\pmod{41}$ and $261\equiv-10\pmod{271}$, we have $1111^{2019}\equiv-10\pmod{11111}$. So the remainder is $11101$.
