Bunuel wrote:
If x and y are integers, is xy divisible by x^2?
(1) x divides into y^2 with no remainder.
(2) x is a prime.
x and y are integers.
Is xy divisible by x^2?
x^2 has two x's. xy already has one x. For xy to be divisible by x^2, now all we need is another x to be a factor of y.
If x is a factor of y, xy will be divisible by x^2.
(1) x divides into y^2 with no remainder.
x is a factor y^2 does not mean that x is a factor of y. For example, if x = 4 and y = 2, x is a factor of y^2 but not a factor of y.
Not sufficient.
(2) x is a prime.
x may or may not a factor of y. Not sufficient.
Using both, we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y.
Sufficient.
Answer (C)
When it says x divides into y^2 does it mean y^2 is divisible by x or vice versa. I solved it correctly assuming the former.
So if y = 4 then x can be 2,4,8 and 16 .
If you can please share your insight , we would be grateful.