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Difference between revisions of "2011 AMC 10B Problems/Problem 23"

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In modulo <math>8,</math> we have <math>2011^4 \equiv 1 \pmod{8}</math> by Euler's Theorem, and also <math>2011 \equiv 3 \pmod{8},</math> so we have <cmath>2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.</cmath>
 
In modulo <math>8,</math> we have <math>2011^4 \equiv 1 \pmod{8}</math> by Euler's Theorem, and also <math>2011 \equiv 3 \pmod{8},</math> so we have <cmath>2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.</cmath>
  
−
In modulo <math>125,</math> we have <math>2011^{100} \equiv 1 \pmod{125}</math> by Euler's Theorem, and also <math>2011 \equiv 11 \pmod{125}.</math> Therefore, we have <cmath>\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} </cmath>
+
In modulo <math>125,</math> we have <math>2011^{100} \equiv 1 \pmod{125}</math> by Euler's Theorem, and also <math>2011 \equiv 11 \pmod{125}.</math> Therefore, we have <math>\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} </math>
  
 
After finding the solution <math>2011^{2011} \equiv 611 \pmod{1000},</math> we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is <math>\boxed{\textbf{(D) } 6}.</math>
 
After finding the solution <math>2011^{2011} \equiv 611 \pmod{1000},</math> we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is <math>\boxed{\textbf{(D) } 6}.</math>

Revision as of 15:27, 29 March 2015

Problem

What is the hundreds digit of $2011^{2011}?$

$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$

Solution 1

Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$

To compute this, we use a clever application of the binomial theorem.

$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbinom{2011}{2} \cdot 10^2 + \cdots \end{aligned}$

In all of the other terms, the power of $10$ is greater than $3$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:

$\begin{aligned}11^{2011} &= 1 + 2011\cdot 10 + \dfrac{2011 \cdot 2010}{2} \cdot 100 \\ &\equiv 1+20110 + \dfrac{11\cdot 10}{2} \cdot 100\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\&=611 \pmod{1000} \end{aligned}$

Therefore, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

Solution 2

We need to compute $2011^{2011} \pmod{1000}.$ By the Chinese Remainder Theorem, it suffices to compute $2011^{2011} \pmod{8}$ and $2011^{2011} \pmod{125}.$

In modulo $8,$ we have $2011^4 \equiv 1 \pmod{8}$ by Euler's Theorem, and also $2011 \equiv 3 \pmod{8},$ so we have \[2011^{2011} = (2011^4)^{502} \cdot 2011^3 \equiv 1^{502} \cdot 3^3 \equiv 3 \pmod{8}.\]

In modulo $125,$ we have $2011^{100} \equiv 1 \pmod{125}$ by Euler's Theorem, and also $2011 \equiv 11 \pmod{125}.$ Therefore, we have $\begin{aligned} 2011^{2011} &= (2011^{100})^{20} \cdot 2011^{11} \\ &\equiv 1^{20} \cdot 11^{11} \\ &= 121^5 \cdot 11 \\ &= (-4)^5 \cdot 11 = -1024 \cdot 11 \\ &\equiv -24 \cdot 11 = -264 \\ &\equiv 111 \pmod{125}. \end{aligned}$

After finding the solution $2011^{2011} \equiv 611 \pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\boxed{\textbf{(D) } 6}.$

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

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