Bunuel
If a positive even number n is not divisible by 3 or 4, then the product (n + 6)(n + 8)(n + 10) must be divisible by which of the following?
I. 24
II. 32
III. 96
A. None
B. I only
C. II only
D. I and II only
E. I, II, and III
VERITAS PREP OFFICIAL SOLUTION:This question offers several constraints for the possible values of n, and missing any one of those key traits will likely cost you the question. N must be:
* Positive
* Even
* Not divisible by 3
* Not divisible by 4
As this question was discussed by aspiring MBA students in a forum thread earlier this week, the most common reason for someone to miss it was that they missed that small word “even”; they’d plug in a number like 1 or 5 for n – numbers that ARE NOT possible given the constraints – and incorrectly believe that the answer is E. In actuality, however, these constraints taken together mean that the only possible values for n are numbers like:
2, 10, 14, 22, 34…
And for these numbers, the product of (n+6)(n+8)(n+10) is divisible by all of the above. We are assured that, if n is even but not divisible by 4, then n + 6 and n + 10 WILL BE divisible by 4 (try it: 2, 6, 10, 14…all of these numbers are even but not divisible by 4; add 6 and you get 8, 12, 16, 20 —> all divisible by 4). So the product will give us:
(n + 6)(n + 8)(n + 10)
which is equivalent to:
Divisible by 4 * Even * Divisible by 4
Providing prime factors of (at minimum):
(2*2)(2)(2*2) –> the product must be divisible by 2^5.
And because we have three consecutive even integers, exactly one will be divisible by 3 (try it: 2, 4, 6, 8, 10, 12, 14, 16, 18… – every third value is divisible by 3).
So we know that the product is divisible by 3*2^5, and to be divisible by I, II, and III we need:
24: 3*2^3 (check!)
32: 2^5 (check!)
96: 3*2^5 (check!)