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

It's useful to work with the answer choices here. So, I've added them to the question.

It almost always helps to find the prime factorization in these question types where we ask whether a certain rational expression is an integer.

450y = n^3
2*3*3*5*5*y = n^3
For 2*3*3*5*5*y to be a cube, we need the number of 2's, 3's and 5's in the prime factorization to each be divisible by 3.
So, for example, 2*2*2*2*2*2*3*3*3*5*5*5 = (2*2*3*5)^3

For 2*3*3*5*5*y to be a cube, it must be the case that the prime factorization of y includes at least two additional 2's, one additional 3 and one additional 5.
So, y = 2*2*3*5*(other possible numbers)

Now check the option.

I. Must y/(3 * 2^2 * 5) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3 * 2^2 * 5)
= some integer
Since this must be an integer, we can eliminate A, C and D, which leaves us with B or E.

II. Must y/(3^2 * 2 * 5) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3^2 * 2 * 5)
= 2*(other possible numbers)/3
Not necessarily an integer
Since this need not be an integer, we can eliminate E, which leaves us with B.

NOTE: At this point we have the correct answer. But let's check III for "fun"

III. Must y/(3 * 2 * 5^2) be an integer?
Plug in y to get: 2*2*3*5*(other possible numbers)/(3 * 2 * 5^2)
= 2*(other possible numbers)/5
Not necessarily an integer

Answer: B

Cheers,
Brent

ajax001 wrote:Brent,

Couldnt Y also be 3^4*5^4*2^5, so that n^3 is (3^2*5^2*2^2)^3?

In this case, all I,II,III can be integers right?

You're right in that y could equal 3^4 * 5^4 * 2^5, in which case expressions I,II,III would be integers.

However, the question is asking us which of the following must be integers.

So, we need to determine what must be true about y rather than what could be true about y

Cheers,
Brent