2013 AMC 10B Problems/Problem 9

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
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All AMC 10 Problems and Solutions

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