2011 AMC 10B Problem 23

Because $2011\equiv11\pmod{1000}$, it is enough to find $11^{2011}=(10+1)^{2011}$ modulo $1000$. All terms with $10^3$ or higher are divisible by $1000$, so only the first three terms matter: $$1+2011\cdot10+\binom{2011}{2}10^2.$$ Modulo $1000$, this is $$\begin{aligned} &1+11\cdot10+\dfrac{2011\cdot2010}{2}\cdot100 \\ &\quad \equiv1+110+500=611 \end{aligned}$$. The hundreds digit is therefore $6$. Thus, D is the correct answer.