In this one 450=2x3^2x5^2
if we substitute --> (2x3^2x5^2)y=n^3 --> y=n^3/(2x3^2x5^2)

So in 1- [n^3/(2x3^2x5^2)]/3x2^2x5 =n^3/2x3x5
in 2- [n^3/(2x3^2x5^2)]/3^2x2x5= n^3/5
in 3- [n^3/(2x3^2x5^2)]/3x2x5^2= n^3/3

I dont understand why only 1 is correct. If y and n are positive integers and 450y=n^3 then n^3 has to be a multiple of 2,3^2 and 5^2, so more over of 2,3,5. That is why all of them seems to be correct.

I think it's a prime factor issue.

Since 450y = n^3

All the prime factors of n^3 must be in pairs of 3s.... i.e (3, 6, 9 ,etc)

Prime factors of 450 = 3^2 * 5^2 * 2

Therefore y -- MUST have, at LEAST, one 3, 1 5 and 2 2s to complement the prime factors of 450.

= 3 * 2^2 * 5, y is definitely a factor of this product.

The Guest has got it. The easiest way to approach it is to realize that y MUST contain at least two 2's, one 3, and one 5 (though it could contain other things, the problem only asks us what MUST be true). Therefore, only roman numeral I is guaranteed to have its denominator cancel out. II and III could be integers, but they don't have to be.

Also, Luci, the manipulations you showed aren't done correctly. For example, you said:


But that's not the right simplification.
[n^3/(2*3^2*5^2)] / (3*2^2*5) =
[n^3/(2*3^2*5^2)] * [1/(3*2^2*5)]
So all of those 2's and 3's and 5's end up in the denominator. You cancelled them out against each other. It really simplifies to:
n^3 / (2^3*3^3*5^3)

Since I can also say that n must contain 2^3, 3^3, and 5^3, everything in the denominator will cancel out, so I can say this one MUST be an integer.

The second option is:
[n^3/(2*3^2*5^2)]/(3^2*2*5) =
[n^3/(2*3^2*5^2)] * [1/(3^2*2*5)] =
n^3 / (2^2*3^4*5^3)
Here, I can cancel out all of the 2's and 5's, but I've got an extra 3 in that denominator that I may or may not be able to cancel out. Doesn't pass the MUST criterion. You can use the same logic to eliminate option 3.