2014 AMC 12B Problems/Problem 15

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. Unknown error_msg)

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2014 AMC 12B Problems/Problem 15

Problem

Solution

See also