2011 AMC 10B Problems/Problem 23
Problem
What is the hundreds digit of
Solution
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