Difference between revisions of "2013 AMC 10B Problems/Problem 9"
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==Problem== | ==Problem== | ||
| + | Three positive integers are each greater than <math>1</math>, have a product of <math> 27000 </math>, and are pairwise relatively prime. What is their sum? | ||
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| + | <math> \textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}}\ 160\qquad\textbf{(E)}\ 165 </math> | ||
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| + | ==Solution== | ||
| + | The prime factorization of <math>27000</math> is <math>2^3*3^3*5^3</math>. These three factors are pairwise relatively prime, so the sum is <math>2^3+3^3+5^3=8+27+125=</math> <math>\boxed{\textbf{(D) }160}</math> | ||
Revision as of 19:27, 21 February 2013
Problem
Three positive integers are each greater than 
, have a product of 
, and are pairwise relatively prime. What is their sum?
$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}}\ 160\qquad\textbf{(E)}\ 165$ (Error compiling LaTeX. Unknown error_msg)
Solution
The prime factorization of is 
. These three factors are pairwise relatively prime, so the sum is 
