2011 AMC 10B Problems/Problem 23

Problem

What is the hundreds digit of

Solution 1

Since we know that

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} (Error compiling LaTeX. Unknown error_msg)

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

\begin{aligned}11^{2011} &= 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} (Error compiling LaTeX. Unknown error_msg)

Therefore, the hundreds digit is

Solution 2

We need to compute By the Chinese Remainder Theorem, it suffices to compute and

In modulo we have by Euler's Theorem, and also so we have

In modulo we have by Euler's Theorem, and also 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} (Error compiling LaTeX. Unknown error_msg)

After finding the solution we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is

See Also

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