2011 AMC 10B Problems/Problem 23

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Problem

What is the hundreds digit of $2011^{2011}$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution

Since $2011 \equiv 11  (\text{mod }1000),$ we know $2011^{2011} \equiv 11^{2011}   (\text{mod }1000).$

To compute this, write it as $(1+10)^{2011}$ and use the binomial theorem.

\[1^{2011} + 2011 \cdot 1^{2010}10^1 + \frac{2011 \cdot 2010}{2} 1^{2009}10^{2} + \cdots\] From then on the power of $10$ is greater than $3$ and cancel out with $\text{mod }1000.$ \begin{align*} 11^{2011} &\equiv 1 + 20110 + 100\frac{11 \cdot 10}{2}\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\ &=611 \end{align*}

Therefore, the hundreds digit is $\boxed{\textbf{(D) } 6}$

See Also

2011 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions
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