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

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 234
Joined: Fri Oct 01, 2010 7:28 pm
Location: chennai
Thanked: 5 times
Followed by:4 members
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 IS A

why is statement 2 not sufficient ?
Source: — Data Sufficiency |

User avatar
Senior | Next Rank: 100 Posts
Posts: 30
Joined: Fri Jan 30, 2009 12:51 am
Location: Pune

by ameya85 » Mon Mar 19, 2012 12:31 am
From A we can safely assume that set will always have Positive numbers. Be it {3,6,9...} or any other combination. However, from B we get Negative numbers as well as 0 which is not a multiple of 3.
For e.g. let 6 be there in the set, then we alos have 6-3 in the set. So we will also have 3-3 in the set which is 0 so B cannot be sufficient.

At lease this is my understanding. I am open to counter view on this.

Ameya

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 273
Joined: Tue Sep 21, 2010 5:37 am
Location: Durham, NC
Thanked: 154 times
Followed by:74 members
GMAT Score:770

by Whitney Garner » Fri Mar 23, 2012 4:15 pm
pappueshwar 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 IS A

why is statement 2 not sufficient ?
Hi pappueshwar!

Let's take a look at what we know from the stem. We have a set, and we only know that the number 3 is in that set.

Now, the question wants to know if all other POSITIVE multiples of 3 are in that set: 6, 9, 12, 15, .... Clear enough - let's jump in!

Statement 1: For any integer in P, the sum of 3 and that integer is also in P.

Well, we can only start with what we know: 3 is a number in the set. And it says, for any integers in the set, 3+that number is also in the set.

So if 3 is there, then 3+3=6 is there. Oh, so if 6 is there 3+6=9 is there. AH, and if 9 is there, 9+3=12 is there and so on and so on and so on! This means that ALL positive multiples of 3 are going to me in my set! SUFFICIENT

Statement 2: For any integer in P, that integer minus 3 is also in P.

Well once again, we can only start with what we know: 3 is a number in the set. And it says, for any integers in the set, 3-that number is also in the set.

So if 3 is there, then 3-3=0 is there. Oh, so if 0 is there 0-3=-3 is there. AH, and if -3 is there, -3-3=-6 is there and so on and so on and so on! This means that ALL negative multiples of 3 are going to me in my set! But wait, what about the positive ones??

3 is the biggest (and only positive) number in the set that I am aware of, and I can't prove that there is anything bigger (I can only take what I know and apply it)

Take a look at this set:
{.....,-12,-9,-6,-3,0,3}

Is it true that for EVERY integer in the set, the integer #-3 is there as well?? It sure is, and there is only 1 positive multiple of 3 in there...
ameya85 wrote: as well as 0 which is not a multiple of 3.
Please note that 0 IS A multiple of 3...its actually a multiple of EVERY number (a multiple of 3 just must be able to be written in the form 3k where k is an integer - which can be k=0), it just isn't positive or negative!!

So, I don't know if EVERY positive multiple of 3 is going to be in there...[spoiler]NOT Sufficient![/spoiler]

Hope this clears up the confusion!
:)
Whit
Whitney Garner
GMAT/GRE/EA Instructor & Anxiety/Accommodations Coach
www.whitneygarner.com

Contributor to Beat The GMAT!

Math is a lot like love - a simple idea that can easily get complicated :heart-eyes: