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

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

Revision as of 15:42, 23 October 2021

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 $60\ \mathrm{(D)}$.

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)
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Problem 16 		Followed by
Problem 18
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All AMC 10 Problems and Solutions

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