TITLE: How to quickly compute $2014 ^{2015} \pmod{11}$
QUESTION [5 upvotes]: Without using Fermat's Little Theorem, how can I quickly solve $2014 ^{2015} \pmod {11}$?

REPLY [2 votes]: I like Bill Dubuque's answer, but I definitely didn't use his approach to find $2014 \bmod 11$, just used a discarding version of long division:
$20 \bmod 11 = \color{red}{9} \\
\to \color{red}{9}1 \bmod 11 = \color{green}{3} \\
\to \color{green}{3}4 \bmod 11 = 1$
And then of course $2014^{2015} \equiv 1^{2015} \equiv 1 \bmod 11$.
If $2014$ had been a bigger number, I would probably have used the alternating digits rule for divisibility by $11$, $(0+4)-(2+1) = 1$, because I would have stopped to think before starting the division process :-)