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multiples of 3

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multiples of 3

by ankita1709 » Tue May 29, 2012 10:50 am
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.

OA A

The solution that I saw was

(1) If n is in P, then so is n + 3
Well, we know that 3 is in the set. Therefore, 3+3=6 is in the set. Therefore, 6+3=9 is in the set... and so on, and so on, and so on ... Are all the positive multiples of 3 in the set? Definitely YES: sufficient.

But I don't get this. What if we start from 1. The integer series says (1,3,4,7,10,13,...)
There are no multiples of 3 in this.
The option says that sum of 3 and THAT INTEGER is also in P, so there is no compulsion that we have to start with 3+3.

Please correct me if I am thinking in the wrong direction
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by eagleeye » Tue May 29, 2012 11:35 am
ankita1709:

If you start from one, then you will have two series running.

First 1,4,7,10,13.
Since 3 is also in P, it will have its own series (remember for EACH element in P, there is P+3)
so for 3, we have 3,6,9,.... all positive multiples of 3 are there.

So if you had 1 in the series, (and we are told in the question stem that 3 is also there), the set will at least have the following elements
1,3,4,6,7,9,10,12,13 etc. So still sufficient.

Let me know if this helps :)
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by Ashujain » Thu May 31, 2012 5:10 am
ankita1709 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.

OA A

The solution that I saw was

(1) If n is in P, then so is n + 3
Well, we know that 3 is in the set. Therefore, 3+3=6 is in the set. Therefore, 6+3=9 is in the set... and so on, and so on, and so on ... Are all the positive multiples of 3 in the set? Definitely YES: sufficient.

But I don't get this. What if we start from 1. The integer series says (1,3,4,7,10,13,...)
There are no multiples of 3 in this.
The option says that sum of 3 and THAT INTEGER is also in P, so there is no compulsion that we have to start with 3+3.

Please correct me if I am thinking in the wrong direction
@Ankita
You are making a small mistake by taking 1 in P. It is clearly mentioned in question that 3 is in P. So we have to start with 3 only. In questions related to Sets it's very important that we do not assume anything from ourselves. We just have to stick to whatever information is given.

So St1 is sufficient.

But my doubt is with St2: For any integer in P, that integer minus 3 is also in P.
P = (3,0,-3,-6,-9....) from this we can see that every positive multiple of 3 is not in P. So even this statement is sufficient.

So why the answer is not D but A?

Can someone please explain it?
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by eagleeye » Thu May 31, 2012 7:46 pm
Ashujain wrote:
ankita1709 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.

OA A

The solution that I saw was

(1) If n is in P, then so is n + 3
Well, we know that 3 is in the set. Therefore, 3+3=6 is in the set. Therefore, 6+3=9 is in the set... and so on, and so on, and so on ... Are all the positive multiples of 3 in the set? Definitely YES: sufficient.

But I don't get this. What if we start from 1. The integer series says (1,3,4,7,10,13,...)
There are no multiples of 3 in this.
The option says that sum of 3 and THAT INTEGER is also in P, so there is no compulsion that we have to start with 3+3.

Please correct me if I am thinking in the wrong direction
@Ankita
You are making a small mistake by taking 1 in P. It is clearly mentioned in question that 3 is in P. So we have to start with 3 only. In questions related to Sets it's very important that we do not assume anything from ourselves. We just have to stick to whatever information is given.

So St1 is sufficient.

But my doubt is with St2: For any integer in P, that integer minus 3 is also in P.
P = (3,0,-3,-6,-9....) from this we can see that every positive multiple of 3 is not in P. So even this statement is sufficient.

So why the answer is not D but A?

Can someone please explain it?
Hi Ashujain:

A couple of things:

1. Ankita is not "wrong" in starting with 1. P is a set of integers. It can have 1,2, -313, 0 etc. We don't know what else is in it. But we definitely know that 3 is there. For option 1, even if she starts with 1, she will arrive at the right answer as long as she includes 3 and its series of 6,9,12 etc.

2. 2 says that for each integer in P, P-3 is also in P. If you could only start from 3, then 3,0,-3 etc. would be set P and you would be right. However, since P can begin with any number, we can start with positive infinity side. In that case, we will indeed have all positive multiples of 3 in P. Since we are not sure about what else is in P, this statement is Insufficient.

By the way, GMAT is always consistent with its answers. If one option says yes, all multiples of 3 are in P, then the other will never imply the opposite (which is what it would have been if your line of thinking was correct.

Let me know if this helps :)
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