Math Question - Factors - Data Sufficiency
I am having trouble with the following question - It seems easy, but these are one of the few types of questions I seem to frequently get wrong.
“If y is an integer, is y^3 (y cubed) divisible by 9?
(1) y is divisible by 4
(2) y is divisible by 6
The answer is B (only 2 is sufficient) because y^3 is divisible by 9 only when y is divisible by 3. I know that is the right answer, but I don’t understand the reasoning behind it. If y is divisible by 6, then it is divisible by both 3 AND 2. The only logical explanation that I can come up with is that if y is divisible by 6, then y is a multiple of 6 (i.e., 6n, where n = 1,2,3…) . This means that y^3 = 6^3 = (3*2)*(3*2)(3*2)(n^3), which is always divisible by 9 since the 3’s will cancel out no matter what. But is there a more intuitive reasoning behind it? I feel that I am missing the key concept that is supposed to make this kind of question easy. Just saying that “y^3 will be divisible by 9 if the integer y is divisible by 3″ isn’t enough of an explanation for me.
Thanks for your help.
Written by gmatnewbie on June 26th, 2008 with
1 comment.
Read more articles on GMAT Question of the Day.
- [+] Digg: Feature this article
- [+] Del.icio.us: Bookmark this article
- [+] Furl: Bookmark this article
#1. June 26th, 2008, at 9:23 PM.
I think the best way to tackle this question is by evaluating the prime boxes for #1 and #2.
#1: y is divisible by 4. That means that y’s prime box will have 2 2s in it. As a result y^3 will have 6 2s in its prime box. In order for y^3 to be divisible by 9, however, it must be divisible by 2 3s (3×3=9). That means #1 is not sufficient, and A and D are out.
#2: y is divisible by 6. That means that y’s prime box will have a 2 and a 3 in it. As a result y^3 will have 3 2s and 3 3s in its prime box. Since we only need 2 3s to ensure that y^3 is divisible by 9, #2 is sufficient, and the answer is B.
Rich