CIO
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Re: Mat Set 22. Q15 [#permalink]
17 Feb 2008, 08:53
After you get very comfortable with the gmat, you will see that there is a pattern with that algebra:
n, n+1, n1 are three consecutive numbers. Arranged in order:
n1, n, n+1
This is really common.
A few things about this:  Often these questions have to do with even/odd  notice that each triplets alternates between two odd and one even and vice versa  Any time there are two evens, the product of all three will be a multiple of 8  in every triplet, the product of all three will always be divisible by 3 and by 6
This problem deals with 24, which has a lot to do with 6, 8, 24.
Now, you can simply make a list of a bunch of consecutive triplets and knock off the statements. In this problem, we are multiplying the first and last of each triplet, so you can do that quickly, too:
123: 3 234: 8 345: 15 456: 24 567: 35 678: 48 789: 63 8910: 80
Notice already that some of these are multiples of 24.
Statement 1 says that n is not even (i'm paraphrasing). So cut out all the ones where the middle number is even, ie, (n1)(n+1) is not going to be odd. Still not enough, since there are different even ones.
Statement 2 says that n is not a multiple of 3. Cut those out, still not enough info.
Together, we now know that n is not even, and it is not a multiple of 3, so cut out 2,3,4; 8,9,10, etc..
WHat's left? All multiples of 24.
So the answer is C.
You can do this really quickly on paper.
