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

Video Solution

https://youtu.be/MVuQ8G1rCbQ

~savannahsolver

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2007 AMC 10A Problems/Problem 17

Contents

Problem

Solution

Solution 2

Video Solution

See also