Difference between revisions of "2014 AMC 12B Problems/Problem 15"

(Solution)
Line 9: Line 9:
 
<cmath>1 \ln{1} + 2 \ln{2} + 3 \ln{3} + 4 \ln{4} + 5 \ln{5} + 6 \ln{6} = </cmath>
 
<cmath>1 \ln{1} + 2 \ln{2} + 3 \ln{3} + 4 \ln{4} + 5 \ln{5} + 6 \ln{6} = </cmath>
 
<cmath>\ln{1^1} + \ln{2^2} + \ln{3^3} + \ln {4^4} + \ln{5^5} + \ln {6^6} = </cmath>
 
<cmath>\ln{1^1} + \ln{2^2} + \ln{3^3} + \ln {4^4} + \ln{5^5} + \ln {6^6} = </cmath>
<cmath>\ln{(1^1*2^2*3^3*4^4*5^5*6^6)}</cmath>
+
<cmath>\ln{(1^1\times2^2\times3^3\times4^4\times5^5\times6^6)}</cmath>
 
Raising <math>e</math> to the power of this quantity eliminates the natural logarithm, which leaves us with  
 
Raising <math>e</math> to the power of this quantity eliminates the natural logarithm, which leaves us with  
<cmath>e^p = 1^1*2^2*3^3*4^4*5^5*6^6</cmath>
+
<cmath>e^p = 1^1\times2^2\times3^3\times4^4\times5^5\times6^6</cmath>
 
This product has <math>2</math> powers of <math>2</math> in the <math>2^2</math> factor, <math>4*2=8</math> powers of <math>2</math> in the <math>4^4</math> factor, and <math>6</math> powers of <math>2</math> in the <math>6^6</math> factor.
 
This product has <math>2</math> powers of <math>2</math> in the <math>2^2</math> factor, <math>4*2=8</math> powers of <math>2</math> in the <math>4^4</math> factor, and <math>6</math> powers of <math>2</math> in the <math>6^6</math> factor.
 
This adds up to <math>2+8+6=16</math> powers of two which divide into our quantity, so our answer is <math>\boxed{\textbf{(C)}\ 2^{16}}</math>
 
This adds up to <math>2+8+6=16</math> powers of two which divide into our quantity, so our answer is <math>\boxed{\textbf{(C)}\ 2^{16}}</math>

Revision as of 00:14, 21 February 2014

Problem

When $p = \sum\limits_{k=1}^{6} k \ln{k}$, the number $e^p$ is an integer. What is the largest power of 2 that is a factor of $e^p$?

$\textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}}\ 2^{18}\qquad\textbf{(E)}\ 2^{20}$ (Error compiling LaTeX. ! Extra }, or forgotten $.)

Solution

Let's write out the sum. Our sum is equal to \[1 \ln{1} + 2 \ln{2} + 3 \ln{3} + 4 \ln{4} + 5 \ln{5} + 6 \ln{6} =\] \[\ln{1^1} + \ln{2^2} + \ln{3^3} + \ln {4^4} + \ln{5^5} + \ln {6^6} =\] \[\ln{(1^1\times2^2\times3^3\times4^4\times5^5\times6^6)}\] Raising $e$ to the power of this quantity eliminates the natural logarithm, which leaves us with \[e^p = 1^1\times2^2\times3^3\times4^4\times5^5\times6^6\] This product has $2$ powers of $2$ in the $2^2$ factor, $4*2=8$ powers of $2$ in the $4^4$ factor, and $6$ powers of $2$ in the $6^6$ factor. This adds up to $2+8+6=16$ powers of two which divide into our quantity, so our answer is $\boxed{\textbf{(C)}\ 2^{16}}$

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