2011 AMC 10B Problem 23

The hundreds digit only depends on the last three digits. When you multiply, the last three digits of the answer only depend on the last three digits of the numbers. $2011$ ends in $011$, so $2011^{2011}$ ends in the same three digits as $11^{2011}$.

Step A: find the pattern for powers of $11$.

1. Small powers of $11$ follow the rows of Pascal's triangle:
- $11^2 = 121$ (row $1, 2, 1$)
- $11^3 = 1331$ (row $1, 3, 3, 1$)
- $11^4 = 14641$ (row $1, 4, 6, 4, 1$)
2. Read the row from the right end. The ones digit gets $1$. The tens place gets $n$. The hundreds place gets the third number of the row, which is $\frac{n(n-1)}{2}$. Everything further left is in the thousands or higher, so we drop it. When $n$ is big these numbers carry into higher places, so write it as a sum:
$$\text{last three digits of } 11^n = \text{last three digits of } 1 + 10n + 100 \times \frac{n(n-1)}{2}.$$
3. Check with $n = 4$: $1 + 40 + 100 \times 6 = 641$, and $11^4 = 14641$. It matches.

Step B: use $n = 2011$.

1. $10n = 20110$, which ends in $110$.
2. $\frac{n(n-1)}{2} = \frac{2011 \times 2010}{2} = 2011 \times 1005$. We only need its last digit, because multiplying by $100$ pushes everything else past the hundreds place. The last digit is $1 \times 5 = 5$. So $100 \times (\ldots 5)$ ends in $500$.
3. Add: $1 + 110 + 500 = 611$.

So $2011^{2011}$ ends in $611$, and the hundreds digit is $6$.

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

A number that is $1$ more than a multiple of $10$, like $11$, raised to a power follows Pascal's triangle, and only the first few numbers of the row reach the last few digits. Anytime you need the last digits of a huge power, first trim the base to its last digits, then look for a shortcut like this one.

In the language of the binomial theorem, $(1 + 10)^n = 1 + n \cdot 10 + \binom{n}{2} \cdot 10^2 + \binom{n}{3} \cdot 10^3 + \cdots$, and every term from $10^3$ on ends in $000$. So only the first three terms matter, exactly as in Step A.