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What is the smallest positive integer n such that \(6,480\cdot\sqrt{n}\) is a perfect cube?

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Re: What is the smallest positive integer n such that \(6,480\cdot\sqrt{n}\) is a perfect cube?

by Scott@TargetTestPrep » Mon Sep 28, 2020 11:42 am
AAPL wrote: ↑
Wed Sep 23, 2020 4:33 pm
 Official Guide

What is the smallest positive integer n such that \(6,480\cdot \sqrt{n}\) is a perfect cube?

A. 5
B. 5^2
C. 30
D. 30^2
E. 30^4

OA E
Solution:

First, let’s prime factorize 6,480:

6,480 = 81 x 80 = 3^4 x 2^4 x 5

Recall that, in order for a number to be a perfect cube, all exponents of the prime factors must be positive multiples of 3. We see that √n must be (at least) 3^2 x 2^2 x 5^2 so that 6,480√n is (at least) 3^6 x 2^6 x 5^3, a perfect cube. Therefore, we have

√n = 3^2 x 2^2 x 5^2

n = 3^4 x 2^4 x 5^4 = (3 x 2 x 5)^4 = 30^4

Answer: E

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