Problem Solving

psm12se wrote: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

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