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

Revision as of 15:18, 4 June 2011

Problem

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

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution

Since $2011 \equiv 11 (\text{mod }1000),$ we know $2011^{2011} \equiv 11^{2011} (\text{mod }1000).$

To compute this, write it as $(1+10)^{2011}$ and use the binomial theorem.

$1^{2011} + 2011 \cdot 1^{2010}10^1 + \frac{2011 \cdot 2010}{2} 1^{2009}10^{2} + \cdots$ From then on the power of $10$ is greater than $3$ and cancel out with $\text{mod }1000.$ $\begin{align*} 11^{2011} &\equiv 1 + 20110 + 100\frac{11 \cdot 10}{2}\\ &= 1 + 20110 + 5500\\ &\equiv 1 + 110 + 500\\ &=611 \end{align*}$

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

@@ Line 3: / Line 3: @@
 What is the hundreds digit of <math>2011^{2011}</math>?
-<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math>
+<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9</math>
 ==Solution==
+Since <math>2011 \equiv 11  (\text{mod }1000),</math> we know <math>2011^{2011} \equiv 11^{2011}   (\text{mod }1000).</math>
-modulus method
+To compute this, write it as <math>(1+10)^{2011}</math> and use the [[binomial theorem]].
+<cmath>1^{2011} + 2011 \cdot 1^{2010}10^1 + \frac{2011 \cdot 2010}{2} 1^{2009}10^{2} + \cdots</cmath>
+From then on the power of <math>10</math> is greater than <math>3</math> and cancel out with <math>\text{mod }1000.</math>
+<cmath>\begin{align*}
+^{2011} &\equiv 1 + 20110 + 100\frac{11 \cdot 10}{2}\\
+&= 1 + 20110 + 5500\\
+&\equiv 1 + 110 + 500\\
+&=611
+\end{align*}</cmath>
-(2000  + 11) ^ 2011 mod 1000 \n
+Therefore, the hundreds digit is <math>\boxed{\textbf{(D) } 6}</math>
-^2011 mod 1000
-(10 + 1)^2011 mod 1000
-C2 * 10^2 + 2011C1 * 10 + 1    mod 1000
-+ 110 + 1   mod 1000
-mod 1000
-So we know the last three digits of 2011 ^ 2011 is 611, and so the hundreds digit is 6 (D).
-pascal's triangle method
-First we try some multiplications. But we will only look at some of the last digits
-First, we see that 2011 ^ 2 gives the 3 last digits 121
-Then 2011 ^ 3 gives 1331
-^ 4 gives 2641
-If we continue, we eventually see that the thousand's term and the hundred's term are part of the triangle numbers, and is part of the pascal triangle.
-The hundred's digit is the last digit of the nth triangle number, which in our case is 2011.
-Therefore, we just do 2011(2012) / 2 => last digit is 6 (D).

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

Revision as of 15:18, 4 June 2011

Problem

Solution