Really difficult one i cant figure out

[jlaipple]

If n and y are positive integers and 450y=n^3, which of the following must be an integer

I Y/(3 * 2^2 * 5)
II Y/(3^2 * 2 * 5)
III Y/(3 * 2 * 5^2)

A None
B I only
C II only
D III only
E I, II, and III


I can't for the life of me figure this one out... It's from the official practice test.... I'll post the OE when someone shows how to solve it correctl.y

[Stuart@KaplanGMAT]

If n and y are positive integers and 450y=n^3, which of the following must be an integer

I Y/(3 * 2^2 * 5)
II Y/(3^2 * 2 * 5)
III Y/(3 * 2 * 5^2)

A None
B I only
C II only
D III only
E I, II, and III


I can't for the life of me figure this one out... It's from the official practice test.... I'll post the OE when someone shows how to solve it correctl.y

We're told that 450y is a perfect cube.

Just as perfect squares are comprised of pairs of prime factors, perfect cubes are comprised of trios of prime factors. Let's begin by prime factoring 450.

450 = 9 * 50 = 3*3*5*10 = 3*3*5*5*2

So, to create a perfect cube, we need to add one 3, one 5 and two 2s. Therefore, the minimum possible value for y is:

2^2 * 3 * 5

All of the roman numerals occur with equal frequency, so let's start at the top:

I Y/(3 * 2^2 * 5)

An exact match for our prediction! Since y must contain 2^2 * 3 * 5, I will always be an integer.

Eliminate A, C and D. We can test either II or III to determine the final answer.

II Y/(3^2 * 2 * 5)

While Y must have one factor of 3, Y does not need to be a multiple of 3^2. Therefore, II doesn't have to be an integer.

Eliminate E. Only B is left, no need to test III.

[jlaipple]

Thanks for the quick thorough response... I wasn't aware of the little tidbit that 2 pairs of prime numbers are perfect squares and multiple sets are perfect cubes. That is good info to have. Still not sure if I could have gotten this one in under the 2 minutes knowing that. Thanks though! You were exactly correct with your answer.