OG - DS #63

[jenwindy]

I am not clear with the answer explanation to the following problem:


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

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



Answer explanation:
In order for y^3 to be divisible by 9, the integer y must also be divisible by 3. WHY IS THIS? Can anybody help clarify?

(1) Not all multiples of 4 are divisible by 3 (e.g. y=12 is divisible by 3 but y=16 is not divisible by 3; NOT sufficient)
(2) Any number divisible by 6 is also divisible by 3; SUFFICIENT.

The correct answer is B; statement 2 along is sufficient.

[StaceyKoprince]

y^3 = y*y*y

In order for y^3 to be divisible by any number, the components of that number must be contained in those three y's. To check if something is divisible by 9, I first break that down into its prime factors, 3*3. So I need two factors of 3 to be contained in those three y's (if it is divisible by 9).

I can't break 3 down any further because it's already prime. So, if one y contains a 3, that will guarantee me the two factors of 3 I need to say that y*y*y is divisible by 9 (and, actually, it will end up giving me three factors of 3, since each y contains a 3).

[jenwindy]

I understand it now! Thanks Stacey!!

[RonPurewal]