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Difference between revisions of "2014 AMC 12B Problems/Problem 15"
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==Problem==
 
==Problem==
  
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When <math>p = \sum\limits_{k=1}^{6} k \ln{k}</math>, the number <math>e^p</math> is an integer.  What is the largest power of 2 that is a factor of <math>e^p</math>?
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When <math>p = \sum\limits_{k=1}^{6} k \text{ ln }{k}</math>, the number <math>e^p</math> is an integer.  What is the largest power of <math>2</math> that is a factor of <math>e^p</math>?
  
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<math> \textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}}\ 2^{18}\qquad\textbf{(E)}\ 2^{20} </math>
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<math> \textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}\ 2^{18}\qquad\textbf{(E)}\ 2^{20} </math>
  
 
==Solution==
 
==Solution==
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<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>
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<cmath>\ln{(1^1*2^2*3^3*4^4*5^5*6^6)}</cmath>
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<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  
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<cmath>e^p = 1^1*2^2*3^3*4^4*5^5*6^6</cmath>
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<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>
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==Video Solution==
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 +
For those wanting a video: https://www.youtube.com/watch?v=iq2X86GFVBo
 +
== See also ==
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{{AMC12 box|year=2014|ab=B|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 13:09, 4 July 2024

Problem

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

$\textbf{(A)}\ 2^{12}\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}\ 2^{18}\qquad\textbf{(E)}\ 2^{20}$

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 $\boxed{\textbf{(C)}\ 2^{16}}$

Video Solution

For those wanting a video: https://www.youtube.com/watch?v=iq2X86GFVBo

See also

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 Problems and Solutions

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