If n and y are positive integers and 450y = n³, which of the following must be an integer?

I. y/(3 x 2² x 5)

II. y/(3² x 2 x 5)

III. y/(3 x 2 x 5²)

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

450y is the cube of an integer.

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor:
8 is the cube of an integer because 8 = 2³ = 2*2*2.
27 is the cube of an integer because 27 = 3³ = 3*3*3.

Thus, when we prime-factorize 450y, we need to get at least 3 of every prime factor:
450y = 2 * 3² * 5² * y

Since 450 provides only one 2, two 3's, and two 5's, y must provide the missing prime factors. We need y to provide two more 2's, one more 3, and one more 5.
Thus, the smallest possible value is y = 2² * 3 * 5.

Onto the answer choices:

I. y/(3 x 2² x 5)
(2² * 3 * 5)/(3 x 2² x 5) = 1. The smallest possible value of y yields an integer.
Eliminate every answer choice that does not include I.
Eliminate A, C and D.

II. y/(3² x 2 x 5)
(2² * 3 * 5)/(3² x 2² x 5) = 1/3. Not an integer.
Eliminate every remaining answer choice that includes II.
Eliminate E.

The correct answer is B.

parkman wrote: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)

none
I only
II only
III only
I, II, and III

I am not sure the way to start this question, help would be appreciated

450y=n^3 => (5^2)*(3^2)*2*y=n^3
hence for a perfect cube y must be min 60 ie 5*3*2^2
so for 1 =>y/60=5*3*2^2/60=1 is an integer
for II and III to be integer 450y will not be a perfect cube
hence onle y/60ie I is an integer
hence ans is b

IMO - B

Here it goes....

Since y and n are integers and 450y = n^3, 450y must be a perfect cube.

450 = (3^2)*(5^2)*2
So to make it to nearest perfect cube, least value of y = 3*5*(2^2) - when this is substituted in option I, it gives an integer value 1. So it holds good.

This is the least value of y and hence must hold true.