2007 AMC 10A Problems/Problem 17

Problem

Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$. What is the minimum possible value of $m + n$?

$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 5700$

Solution

$3 \cdot 5^2m$ must be a perfect cube, so each power of a prime in the factorization for $3 \cdot 5^2m$ must be divisible by $3$. Thus the minimum value of $m$ is $3^2 \cdot 5 = 45$, which makes $n = \sqrt[3]{3^3 \cdot 5^3} = 15$. The minimum possible value for the sum of $m$ and $n$ is $\boxed {(D)60}.$

Solution 2

First, we need to prime factorize $75$. $75$ = $5^2 \cdot 3$. We need $75m$ to be in the form $x^3y^3$. Therefore, the smallest $m$ is $5 \cdot 3^2$. $m$ = 45, and since $5^3 \cdot 3^3 = 15^3$, our answer is $45 + 15$ = $\boxed {(D)60}$

~Arcticturn

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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