Is the total number of divisors of x^2 a multiple of the total number of divisors of y^3?
1. x=4
2. y=6


OA: C

OE: [spoiler]To get the divisors of x and y, we need their respective values.

Statement (1) by itself is insufficient. We can only find the divisors of x .

Statement (2) by itself is insufficient. We can only find the divisors of y .

Statements (1) and (2) combined are sufficient. Combining the statements, we have the divisors of x and y .

The correct answer is C.[/spoiler]

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The way I approach this problem is this:

1. x=4
x^3 = 64 = 2^6
Total no. of factors = 6+1= 7
Now y is a perfect square. The total no. of factors of a perfect square is always in the form 2k+1.
So, question is: Is 7 a multiple of any number of the form 2k+1?
Answer is: it may be (e.g. when k=3) or may not be (e.g. when k=1, 5,....)
So, insufficient


2. y=6
y^2 = 36 = 2^2 * 3^2
Total no. of factors = (2+1)(2+1) = 9
Now x is a perfect cube. The total no. of factors of a perfect cube is always in the form 3k+1
So, question is: Is a number of the form 3k+1 a multiple of 9
Answer is: No, it can never be.
So, sufficient

And hence, B is my answer.
What's wrong in this? And doesn't the OE too simplistic? (In a way, it ignores certain factorization properties of perfect squares and cubes)