imc

Since $2011\equiv 11(\mathrm{mod}1000),$ we know that $2011^{2011}\equiv 11^{2011}(\mathrm{mod}1000).$ To compute this, we use a clever application of the binomial theorem. In all of the other terms, the power of $10$ is greater than $2$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have: Therefore, the hundreds digit is $\boxed{6}.$ Side note: By Euler's Totient Theorem, $a^{\varphi (1000)}\equiv 1(\mathrm{mod}1000)$ for any $a$ relatively prime with 1000, so $a^{400}\equiv 1(\mathrm{mod}1000)$ and $11^{2011}\equiv 11^{11}(\mathrm{mod}1000)$. We can then proceed using the clever application of the Binomial Theorem, or we can just proceed with solution 6 from here.