If P is a set of integers and 3 is in P, is every positive m

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If P is a set of integers and 3 is in P, is every positive multiple of 3 in P?

(1) For any integer in P, the sum of 3 and that integer is also in P.
(2) For any integer in P, that integer minus 3 is also in P.

What's the best way to determine which statement is sufficient? Can any experts show?

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by DavidG@VeritasPrep » Sun Mar 04, 2018 10:58 am
ardz24 wrote:If P is a set of integers and 3 is in P, is every positive multiple of 3 in P?

(1) For any integer in P, the sum of 3 and that integer is also in P.
(2) For any integer in P, that integer minus 3 is also in P.

What's the best way to determine which statement is sufficient? Can any experts show?
Statement 1: If we know that 3 is in the set, and that for every integer in the set, the sum of 3 plus that integer is also included, we know that 3 + 3 = 6 is in the set too. Similarly, we'd know that 6 + 3 = 9 is in the set and so forth. (More abstractly, if x is in the set, then x + 3 is in the set. If x + 3 is in the set, then x + 3 + 3 is in the set, and so forth. If x = 3, then we'd include every positive multiple of 3 you can conjure.) Thus we know that all positive multiples of 3 would be included. (Of course, this raises interesting philosophical questions about infinite sets, but you get the idea.) Statement 1 alone is sufficient

Statement 2: Well, we know that if 3 is in the set, then 3 - 3 = 0 is in the set. And we'd know that if 0 is in the set, then 0 - 3 = -3 is in the set. and so forth. But what if 3 is the largest element in the set? There's no way to know. All we know for sure is that 3, 0, -3, -6.... are included. This statement is not sufficient.

The answer is A
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