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If p and q are consecutive positive integers, is p a multiple of 3 ? [#permalink]
13 Dec 2011, 09:51
45% (01:28) correct
55% (00:17) wrong based on 22 sessions
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If p and q are consecutive positive integers, is p a multiple of 3 ?
(1) \(q\) is not a multiple of 3. (2) \(q - 1\) is not a multiple of 3.
The question is easy; however, I have a doubt in relation to the number 0. Should we consider 0 a multiple of 3? I think we should because a multiple of an integer is that integer multiplied by other integer. So , if 0 is that other integer, \(3 x 0 = 0\); thereofore, 0 is a multiple of 3. In other words, 0 would be a multiple of every number.
But I don't know whether it is the reasoning of the GMAT. Or, do they consider only positive multiples? In other words, they don't consider 0 "the other integer" to create a multiple. If they think in that way, how is the answer for this question affected? For example, in statement (2), could \(q\) be 1? In that sense, \(q-1\) would be 0, and if they only consider positive multiples, \(q-1\) would not be a multiple of 3. I know that the answer to my question is not necessary to solve this problem, but I prefer to solve that doubt for future problems.
We know that p and q are consecutive positive integers. So p could be q+1 or it could be q-1
From statement 1 we know that q is not a multiple of 3. So we know that either q+1 or q-1 is a multiple of 3, since 1 of every 3 consecutive integers must be a multiple of 3. Statement 2 tells us that q-1 is not a multiple of 3, so combining this with Statement 1, we now know that q+1 is a multiple of 3. However, we still don't know if p=q+1 or if p=q-1.
So combining both statements is not sufficient. _________________
Bhavin Parikh Magoosh Test Prep
Re: If p and q are consecutive positive integers, is p a multiple of 3 ?
14 Dec 2011, 11:38
Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...