Finding the remainder when a large number is divided by 13 Let a number $x = 135792468135792468$. Find the remainder when $x$ is divided by $13$.
Is it possible to use Fermat's little theorem on this? I notice that the number is also repeating after $8$. 
Would really appreciate any help, thanks!

Note that $\ 13\mid \overbrace{n\,10^{\large 9}\!+n}^{\textstyle n(\color{#c00}{10^{\large 9}\!+1})\!\!}\,\ $  by $\,\ {\rm mod}\ 13\!:\, \overbrace{\color{#c00}{10^{\large 9}}\equiv ((-3)^{\large 3})^{\large 3}}^{\Large (10\ \ \,\equiv\,\ \ -3)^{\LARGE 9}\quad\ \, }\!\!\equiv (-1)^{\large 3}\equiv\,\color{#c00}{{-}1}$

Remark $\ \ 7,11,19\mid 10^{\large 9}\!+1$ all follow similarly
$\qquad \left.\begin{align}
{\rm mod}\ \ 7\!:&\,\ \color{#c00}{10^{\large 3}}\ \equiv\,\ 3^{\large 3}\,\ \equiv\,\ \color{#c00}{{-}1}\\ \\
{\rm mod}\ 11\!:&\,\ \color{#c00}{10^{\large 3}}\equiv (-1)^{\large 3}\equiv\color{#c00}{-1}
\end{align}\right\}\ \Rightarrow\, 10^{\large 9}\equiv (\color{#c00}{10^{\large 3}})^{\large 3}\equiv (\color{#c00}{-1})^3\equiv -1$
$\qquad {\rm mod}\ 19\!:\ 10^{\large 9}\equiv (-3^{\large 2})^{\large 9}\equiv -3^{\large 18}\equiv -1\ $ by little Fermat
The above shows that $\, 7,11,13\mid 10^{\large 3}+1\mid 10^{9}+1\,$ which leads to the well-known divisibility test: casting out $1001$'s, i.e. the remainder $\bmod 1001\! =\! \color{#c00}{10^{\large 3}}$ is the alternating digit sum in radix $10^3,\,$ which is the analog of casting out $11 =$ elevens in radix $10$ (decimal), i.e.
$\begin{align} \bmod 1001\!:\ \ \ \  &\cdots\, d_8 d_7 d_6\,,\, d_5 d_4 d_3\,,\, d_2 d_1 d_0\ \ \text{in radix $\,10^{\large 3}$}\\ \equiv\, &\cdots\,  d_8 d_7 d_6\! \color{#c00}{-\! d_5 d_4 d_3}\!\! +\! d_2 d_1 d_0\end{align}  $ 
