Bunuel wrote:
What is the smallest positive integer \(n\) such that \(6,480*\sqrt{n}\) is a perfect cube?
A. \(5\)
B. \(5^2\)
C. \(30\)
D. \(30^2\)
E. \(30^4\)
For what its worth, I had a very similar problem on the GMAT when I took it.
Here's my solution:
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,
\(\sqrt{n} = 900\)
\(\sqrt{n} = 30^2\)
\(n = 30^4\)
ANSWER E