How to quickly compute $2014 ^{2015} \pmod{11}$ Without using Fermat's Little Theorem, how can I quickly solve $2014 ^{2015} \pmod {11}$?
 A: This can be computed in a few seconds of mental arithmetic as follows. First we cast out $11$'s to compute $2014\pmod{11}$ as its alternating digit sum:
${\rm mod}\ 11\!:\ \color{#0a0}{10\equiv -1}\,\Rightarrow\, \color{#c00}{2014}\equiv 2(\color{#0a0}{10})^3\!+\!\color{#0a0}{10}\!+\!4\equiv 2(\color{#0a0}{-1})^3\!\color{#0a0}{-\!1}\!+\!4\equiv\color{#c00} 1\ $ [cast $11$'s - see below]    
Therefore $\ \color{#c00}{2014}^n\equiv \color{#c00}1^n\equiv 1\ $ by the Congruence Power Rule.
Remark $\ $ The first line is a special case of casting out elevens, the analog of casting out nines, i.e. $\, \color{#0a0}{10\equiv -1}\,\Rightarrow\,f(\color{#0a0}{10})\equiv f(\color{#0a0}{-1}) = $ alternating digit sum, where $\,f(x)\,$ is the decimal radix polynomial, e.g. above $\, f(x) = 2x^3+x+4,\ f(10) = 2014.\,$ See here for further discussion.
A: The fundamental fact about modular arithmetic is that
$$ ab \bmod n = (a\bmod n)(b\bmod n)\bmod n $$
(and the analogous fact for addition). Apply this rule 2014 times to
$$ \underbrace{2014\cdot 2014\cdot 2014 \cdots 2014}_{2015\text{ factors}} $$
(and then 2014 more times in reverse to get rid of the intermediate "$\bmod 11$"s) and you get
$$ 2014^{2015} \bmod 11 = (2014 \bmod 11)^{2015} \bmod 11 = 1^{2015} \bmod 11 = 1 $$
A: 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 :-)
