2011 AMC 10B Problem 23

The hundreds digit depends only on the last three digits, so compute $2011^{2011} \bmod 1000$. Since $2011 \equiv 11 \pmod{1000}$, we need $11^{2011} \bmod 1000$.

Write $11 = 1 + 10$ and expand with the binomial theorem:
$$(1 + 10)^{2011} = 1 + \binom{2011}{1} 10 + \binom{2011}{2} 10^2 + \binom{2011}{3} 10^3 + \cdots$$
Every term from the fourth onward contains $10^3$ and vanishes mod $1000$. The three survivors:

• $1$.
• $2011 \cdot 10 = 20110 \equiv 110 \pmod{1000}$.
• $\binom{2011}{2} \cdot 100$. Here $\binom{2011}{2} = \frac{2011 \cdot 2010}{2} = 2011 \cdot 1005$, whose last digit is $1 \cdot 5 = 5$; multiplied by $100$ this contributes $500 \pmod{1000}$.

Adding: $1 + 110 + 500 = 611$. So $2011^{2011}$ ends in $\ldots 611$, and its hundreds digit is $6$.

The answer is $\boxed{\textbf{(D)}\ 6}$.

Powers of a number that is $1$ more than a multiple of $10$ are tailor-made for the binomial theorem: $(1 + 10)^n$ mod $10^k$ needs only the first $k$ terms, and each term's contribution is easy to reduce. Whenever a base reduces to something like $11$, $101$, or $1001$ modulo the relevant power of $10$, expand rather than hunting for a cycle.

Look for a cycle: $11^{10} \equiv 1 + 100 + \binom{10}{2}100 \equiv 1 + 100 + 500 = 601$, and $11^{50} \equiv 601^5 \equiv 1 + 5 \cdot 600 \equiv 1 \pmod{1000}$ (since $600^2 \equiv 0$). So $11^{2011} = 11^{2000} \cdot 11^{11} \equiv 11^{11} \equiv 11 \cdot 601 = 6611 \equiv 611 \pmod{1000}$.

• Dropping the binomial term $100\binom{2011}{2}$ by assuming only the first two terms survive mod $1000$, which gives $111$ and hundreds digit $1$.
• Computing $\binom{2011}{2}$ mod $10$ as $\frac{1 \cdot 0}{2} = 0$ by reducing before dividing by $2$; divide first, then reduce.
• Reading off the tens digit ($1$) or the units digit instead of the hundreds digit.