linfongyu wrote:If y is an integer, is y^3 divisible by 9?

(1) y is divisible by 4.
(2) y is divisible by 6.

correct answer is B, explanation given below, but i don't understand why...

in order for y^3 to be divisible by 9, the integer y must also be divisible by 3.
(1) not all multiples of 4 are divisible by 3 (e.g., y=12 is, y=16 isn't); NOT SUFFICIENT. i don't have a problem with this.
(2) any number divisible by 6 is also divisible by 3; SUFFICIENT. this one, i do. not all number divisible by 6 is divisible by 9! so why are we content with just divisibility rule for 3?

please explain...

thanks,
Hank

think of this problem, like other divisibility problems, in terms of prime factorizations.

when you cube a number, each of the original prime factors has to appear in groups of three (one from each of the 3 numbers multiplied together to produce the cube in the first place). so, if your original number is divisible by 3, then its cube must be divisible by 3 x 3 x 3 = 27 -- and therefore by 9 as well.
the reference to 9 instead of 27 is just a dirty trick, to make it slightly harder to realize that divisibility of the original number by 3 is the real issue. (i.e., if the problem said something about the cube being divisible by 27, then more test takers would make that connection immediately.)

so that's what you're looking at: saying that the cube is divisible by 9 is equivalent to saying that the original number is divisible by 3. note that there isn't 'poor wording' in the answer choices, nor is there any confusion between divisibility by 3 and by 9: you're talking about two different numbers (the original number and its cube).