When you combine 14 and 12, you should get an extra 2 to conclude that p is divisible by 168. A '2' factor in 12 could have come from 14. That is the reason why we need to look for an extra '2'

This is a pretty important concept - the GMAT loves to test factorization, probably because, as I've mentioned in another thread on here recently, it replicates a pretty important business skill of breaking complicated problems down to their essential elements.

The rule here is that, to find the Least Common Multiple of two numbers, you take the set of prime factors for each:

14 = 2, 7
12 = 2, 2, 3

And use each unique prime factor the maximum number of times it appears in any one number:

2 - appears once for 14 and twice for 12, so we need two 2s: 2, 2
3 - appears once for 12, so we need one 3: 3
7 - appears once for 14, so we need one 7: 7

The Least Common Multiple of 12 and 14 is thus: 2 * 2 * 3 * 7, or 84.

If you're trying to plug in numbers as you suggested, it's a tough path to follow. Essentially, the numbers that will work are those that you create by taking the product of the two numbers (12*14 = 168) and dividing by the redundant factor(s) (168/2 is 84). But if you know to do that, you'll probably find that it's easier to build from the ground up as above, by noting the prime factors.

IMO E

plz specify OA..

ta

MrE2All wrote:Is p divisible by 168?

(1) p is divisible by 14
(2) p is divisible by 12

When we factor 168 we get [2, 2, 2, 3, 7]
When we factor 14 we get [2, 7]
When we factor 12 we get [2, 2, 3]

Combining the the factorization of 14 and 12 seems to give us sufficient info to say p is divisible is by 168. Yet, when we look at a 84 we see that it is divisible by both 14 and 12 BUT NOT by 168. The problem obviously lies in a redundancy of factors.

Other than testing numbers, how do we pick up on this redundancy?

So whats the OA here? [spoiler]E?[/spoiler]