Difference between revisions of "2011 AMC 10B Problems/Problem 23"
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9</math> | ||
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Since <math>2011 \equiv 11 \pmod{1000},</math> we know that <math>2011^{2011} \equiv 11^{2011} \pmod{1000}.</math> | Since <math>2011 \equiv 11 \pmod{1000},</math> we know that <math>2011^{2011} \equiv 11^{2011} \pmod{1000}.</math> | ||
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− | + | == Solution 2 == | |
We need to compute <math>2011^{2011} \pmod{1000}.</math> By the Chinese Remainder Theorem, it suffices to compute <math>2011^{2011} \pmod{8}</math> and <math>2011^{2011} \pmod{125}.</math> | We need to compute <math>2011^{2011} \pmod{1000}.</math> By the Chinese Remainder Theorem, it suffices to compute <math>2011^{2011} \pmod{8}</math> and <math>2011^{2011} \pmod{125}.</math> |
Revision as of 14:37, 4 January 2014
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.
\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
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.