Difference between revisions of "2011 AMC 10B Problems/Problem 23"
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To compute this, we use a clever application of the [[binomial theorem]]. | To compute this, we use a clever application of the [[binomial theorem]]. | ||
| − | < | + | <math>\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}</math> |
In all of the other terms, the power of <math>10</math> is greater than <math>3</math> and so is equivalent to <math>0</math> modulo <math>1000,</math> which means we can ignore it. We have: | In all of the other terms, the power of <math>10</math> is greater than <math>3</math> and so is equivalent to <math>0</math> modulo <math>1000,</math> which means we can ignore it. We have: | ||
| − | < | + | <math>\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}</math> |
Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}.</math> | Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}.</math> | ||
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== Solution 2 == | == Solution 2 == | ||
Revision as of 14:26, 29 March 2015
Contents
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.
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:
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
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
