given, 5^2 * 3^2 *2 *y = n^3

To make it a perfect cube y should be a multiple of 5 * 3* 2^2 = 60 * x, then both sides will be a perfect cube.

The only condition given for y is that it is positive, so the equation can be satisfied for any value of y such that it is 60x.
So all the values given can be an integer.

Is the answer E?

pkw209 wrote:Hey all,

Couldn't figure this one out. Would appreciate a brief explanation. Thanks!

67) n and y are positive integers and 450y = n^3, which is an integer?

y / 3 * 2^2 * 5

y / 3^2 * 2 * 5

y / 3 * 2 * 5^2

a. None
b. I
c. II
d. III
e. I, II, III

I've cleaned up the question - in the future, please make sure you write the information correctly.

Here the question is "which of the following is an integer"... which means not just that it could be an integer, but that it's always an integer.

If we isolate y in the original, we have:

y = n^3/450

y = n^3/2*3*3*5*5

Now, since y and n are both integers, we know that n^3 is a perfect cube. Therefore, all the prime factors of n appear in groups of 3.

So, in order for n^3/450 to be an integer, the minimum value for n is 2^3 * 3^3 * 5^5.

Therefore, the minimum possible value for y is:

2*2*2*3*3*3*5*5*5/2*3*3*5*5 = 2*2*3*5

So, of the 3 choices, only the first one MUST be an integer: choose (B) I only.

Thanks so much Stuart! I'm sorry about the messy set up. I copied and pasted from a gmat prep doc and didn't double check to see if everything tranferred over. I'll check in the future.