Difference between revisions of "2013 AMC 10B Problems/Problem 9"

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===Problem===
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==Problem==
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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==
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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>
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== See also ==
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{{AMC10 box|year=2013|ab=B|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 14:27, 9 August 2015

Problem

Three positive integers are each greater than $1$, have a product of $27000$, and are pairwise relatively prime. What is their sum?

$\textbf{(A)}\ 100\qquad\textbf{(B)}\ 137\qquad\textbf{(C)}\ 156\qquad\textbf{(D)}\ 160\qquad\textbf{(E)}\ 165$

Solution

The prime factorization of $27000$ is $2^3*3^3*5^3$. These three factors are pairwise relatively prime, so the sum is $2^3+3^3+5^3=8+27+125=$ $\boxed{\textbf{(D) }160}$

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 10 Problems and Solutions

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