Multiple intergers- how to attack this type of problems?

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Taniuca: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Mon Aug 02, 2010 3:25 pm; Thanked: 3 times

by Taniuca » Sat Oct 08, 2011 7:29 am

if P is a ser of intergers and 3 is in p, is every positive multiple of 3 in P?
1) for any interger in P, the sum of 3 and that interger is also in P
2) for any interger in P, that interger minus 3 is also in P.
Correct answer: A
How to attack this kind of word problems? I always tend to fail...

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Source: Beat The GMAT — Data Sufficiency | Sat Oct 08, 2011 7:29 am

sl750: Master | Next Rank: 500 Posts; Posts: 496; Joined: Tue Jun 07, 2011 5:34 amFollowed by: 1 member; Thanked: 38 times

by sl750 » Sat Oct 08, 2011 8:11 am

We have to establish whether the set P has positive multiples of 3 in it or not. Positive multiples
of 3 include {3,6,9,12...}

Statement 1
For any integer in P, (we can consider 3, as it is given in the question stem), 3+(that integer, here it is 3) is in P, so 6 is in P. Similarly 6+=9 is in P and so on. Sufficient

Statement 2
For any integer in P, 3-3=0 is in P 0-3 -3 is in P, you can see here that we have 0,-3 and so on in the set. So our set consists of {3,0,-3..} Insufficient

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Brian@VeritasPrep: GMAT Instructor; Posts: 1031; Joined: Thu Jul 03, 2008 1:23 pmFollowed by: 255 members; Location: Malibu, CA; Thanked: 716 times; GMAT Score:750

by Brian@VeritasPrep » Tue Oct 11, 2011 3:26 pm

Great explanation, sl750!

One other thing I'd add here about problems with this phrasing - I've seen at least a few that define a set with the phrasing: if x is in the set, then ______________ is in the set.

The key to interpreting these is to realize that "x" describes "any number in the set". So in a case like this the set runs in perpetuity. For any value in the set, you can apply that definition to create a new number. So, as sl750 mentioned about Statement 1:

We know that 3 is in the set. And we know that, for any value x, x + 3 is also in the set. So 3 + 3 = 6 is in the set. And then 6 qualifies as "any number in the set" (after all, we've proven that it's a number in the set), so 6 + 3 = 9 is in the set. And now 9 qualifies as a value of x (it's "a number in the set"), so 9 + 3 = 12 is in the set, and so on.

What's often tricky is that people don't see that x means "any number in the set". But that's the definition given in the statements, so any number that we create by that definition also qualifies as a new number to which we can apply the same definition.

Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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