Don't know how to continue modulo reduction Ok , so the question is to find the last three digits of $2013^{2012}$. After some reduction using Euler's Theorem I got $13^{12}$(mod 1000). I tried dividing it in 8 and 125 and later use the CRT , but this didn't help (or maybe I did it wrong) .
What is the shortest way to get the answer ? (ofc. without using very big numbers)

HINT:
$$13^{12}=(10+3)^{12}=3^{12}+12\cdot3^{11}10+\binom{12}2\cdot3^{10}10^2\pmod{1000}$$
Now, $\displaystyle3^{10}=9^5\equiv(-1)^5\pmod{10}\equiv-1$
$\displaystyle\implies\binom{12}2\cdot3^{10}10^2\equiv\binom{12}2\cdot(-1)10^2\pmod{100}$
and  $\displaystyle3^{11}=3\cdot9^5=3(10-1)^5\equiv3(-1+\binom51\cdot10^1)\pmod{100}$
Finally, $3^{12}=(3^2)^6=9^6=(10-1)^6=1-\binom6110+\binom6210^2\pmod{1000}$
