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

1. 5
2. \(5^2\)
3. 30
4. \(30^2\)
5. \(30^4\)

Answer:
Solution with Explanation:
Approach Solution (1):

First make the prime factorization of 6,480: 6480 = \(2^4*3^4*5\)

Next, for \(2^4*3^4*5*\sqrt{n}\) to be a perfect cube \(\sqrt {n}\) must complete the powers of 2, 3, and 5 to a multiple of 3, thus the least value of \(\sqrt {n}\) must equal to \(2^2*3^2*5^2 = 500\)

Thus the least value of n is \(900^2 = 30^4\)

Correct Option: E

Approach Solution (2):

6480 = \(2^4*3^4*5*\sqrt{n}\)

Now as we know just like for perfect square where we expect all the prime factors powers to be even [multiple of 2]
Similarly, for perfect cube all the prime factors should be a multiple of 3
Here to make the above one to a perfect cube we need to multiply, \(2^2*3^2*5^2 = 500\) [because number 2’s are only 4, etc…]
So we need to have root n = 900, or root n = \(30^2\)

Therefore n = \(30^4\)

Correct Option: E

Approach Solution (3):

For something to be a perfect cube, all prime factors must have a power that is a multiple of 3
Start by factoring 6,480
= 80 * 81
= \(2^3*10*3^4\)
=\(2^4*3^4*5\)

Okay, so the smallest solution will move powers of 4 to 6 and the power of 1 to 3
i.e. we need \(2^2*3^2*5^2\)

= 900
Thus,
\(\sqrt{n} = 900\)
\(\sqrt{n} = 30^2\)
\(n = 30^4\)

Correct Option: E

“What is the smallest positive integer n such that 6,480 * \(\sqrt{n}\) is a perfect cube?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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