it says 450y=n^3 correct? Also n,y>0

So prime factorization of 450=2^1*3^2*5^2 and we know that 450y is n^3 where n is +ve integer. The only way 450y can be equal to n^3 is when the missing parts in 450 itself can be represented as (...)^3 OR put this 450*(2^2*3^1*5^1). So we must have at least y/(2^2*3^1*5^1) condition, where this ratio is both +ve and integer (at least). Otherwise 450y can't be equal to n^3 or a^3, actually can't be derived the root power of 3

2^2*3^1*5^1 is the part of y here, which completes 410 and help our statement 410y=n^3, because Root^3(410) isn't good, and we look for possible Root^3(410y)...

ok?

tito1545 wrote:Could anyone please give me a step by step approach to this question:

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

tito1545 wrote:Could anyone please give me a step by step approach to this question:

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

450 = 2 x 3^2 x 5^2

450y=n^3

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

n = (2*3^2*5^2*y)^1/3

since n is integer, so should the RHS of above equation be.

For (2*3^2*5^2*y)^1/3 to be integer, y must be at least 2^2*3*5 [ so that (2*3^2*5^2*y)^1/3 becomes (2^3*3^3*5^3)^1/3 = 2*3*5]

Thus I is the only one that must be an integer. => Answer B