# Difference between revisions of "2011 AMC 10B Problems/Problem 23"

## Problem

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

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

## Solution 1

Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$

To compute this, we use a clever application of the binomial theorem.

\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}

In all of the other terms, the power of $10$ is greater than $3$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:

\begin{aligned}11^{2011} &\equiv 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}

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

Sidenote: By Euler's Totient Theorem, $a^{\phi (1000)} \equiv a \pmod{1000}$ for any $a$, so $a^{400} \equiv a \pmod{1000}$ and $11^{2011} \equiv 11^{11} \pmod{1000}$. We can then proceed using the clever application of the Binomial Theorem.

## Solution 2

We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem, it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$

In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have $$2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.$$

In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have \begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}

After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

## Solution 3

Notice that the hundreds digit of 2011^2011 won't be affected by 2000. Essentially we could solve the problem by finding the hundreds digit of 11^2011. Powers of 11 are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of 11 can be found by reading rows of the triangle and adding extra numbers up. (add source) For example, the sixth row of the triangle is 1, 5, 10, 10, 5, and 1. Adding all numbers from right to left, we get 161051, which is also 11^5. In other words, each number is 10^n steps from the right side of the row. The hundreds digit is 0. We can do the same for 11^2011, but we only need to find the 3 digits from the right. Observing, every 3 number from the right is 1 + 2 + 3... + n. So to find the third number from the right on the row of 11^2011, f(11^n) = 1 + 2 + 3... + (n-1), or (2010 * 2011)/2, or 2021055. The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously 2011. We must carry the 1 in 2011's tens digit to the 5 in 2021055's unit digit to get $\boxed{\textbf{(D) } 6}$. The one at the very end of the row doesn't affect anything, so we can leave it alone.

## See Also

 2011 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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