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Difference between revisions of "2007 AMC 10A Problems/Problem 17"

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

Revision as of 13:35, 30 June 2022

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 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 10 Problems and Solutions

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