Perfect Squares

Welcome on board!
1. Looking 375y=x^2 tells us that y has to also be 375 in order for x to be a perfect square.
2. Breakdown 375 to its prime factor---- (5*5*5*3)=y
3. Now it's much easier to compare each statement.
I. y/15----> Yes (quotient=25)
II.y/30----> No (quotient=25/2)
III.y^2/25----> Yes (quotient=5625)

Just as 3121gmat, I'm having some difficulty in understanding this problem.

Here's my 2 cents:

y doesn't have to be 375. For example, if we had a statment like 9y = x^2, which will be valid if y=4 & x=6.

In this case, if y=15, x^2 = 375*15,
or x=75.

I will go with (I)

Let me give my shot

\(375y=x^2\)
splitting the 375 = \(5^2*5*3\)
now as 375y is a square of positive integer and y is also a positive integer
implies that y has to be = 5*3*a^2, where a is an integer too

Why? let me show
375y=\(5^2*5*3*[5*3*a^2]\) = \(25^2*3^2*a^2\)

\(a^2\) is added so as to generale the possible values of y.
If you don't understand till here try different values of 'a' such as {2,3,4} to get y and x. It will give you different set of values to understand the reason of using a.

Now start with problem

1) \(y/15 = 15*a^2/15 = a^2\)
definitely a integer that too positive

2) \(y/30 = 15*a^2/30 = a^2/2\)
not necessary an integer if a^2 is not a multiple of 2.

3) \(y^2/25 = 15*15*a^4/25 = 9a^4\)
definitely an integer.

(i) and (iii)

375y= 3.5.5.5.y = x^2 ==> the number of 3 is even number, the number of 5 is even number ==> y contains both 3 and 5.

Clearly, (I) and (III) meet requirement.

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