2011 AMC 10B Problems/Problem 23
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 ![\[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.\]](http://latex.artofproblemsolving.com/0/0/0/0005cb6d3a60d8bfe27bc0ad83d3686a04cb0ff2.png)
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
