Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 May 2017, 15:42

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# I'm flipping through the princeton review 2007 study guide

Author Message
Intern
Joined: 10 Feb 2007
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

I'm flipping through the princeton review 2007 study guide [#permalink]

### Show Tags

10 Feb 2007, 14:32
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I'm flipping through the princeton review 2007 study guide and come to this question:

Data sufficiency problem
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

The answer given is A, but I think it's D, here's why.

Given Statement 1, since 3 is in P, then 3+3=6 is in P, so P = {3,6,9,12......3n}
The answer to the stated question is "Yes" so Statement 1 is sufficient information to answer the question.

Given Statement 2, since 3 is in P, then 3-3 = 0 etc, so P = {3, 0, -3, -6.....-3n}
In which case we can still answer the question "No, all positive multiples of 3 are not in P"
We can still answer the question, and Statement 2 is sufficient information to do that.

Tell me if you agree or disagree, and why?

Thanx

Last edited by crankharder on 10 Feb 2007, 14:47, edited 1 time in total.
SVP
Joined: 05 Jul 2006
Posts: 1747
Followers: 6

Kudos [?]: 358 [0], given: 49

### Show Tags

10 Feb 2007, 14:44
OK , WHY DO U ASSUME IN STATMENT TWO THAT THE SET STARTS WITH 3.......................... Got it
Intern
Joined: 10 Feb 2007
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

10 Feb 2007, 14:46
yezz wrote:
OK , WHY DO U ASSUME IN STATMENT TWO THAT THE SET STARTS WITH 3.......................... Got it

Because it's given in the question "3 is in P"
SVP
Joined: 01 May 2006
Posts: 1796
Followers: 9

Kudos [?]: 154 [0], given: 0

### Show Tags

10 Feb 2007, 14:52
crankharder wrote:
I'm flipping through the princeton review 2007 study guide and come to this question:

Data sufficiency problem
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

The answer given is A, but I think it's D, here's why.

Given Statement 1, since 3 is in P, then 3+3=6 is in P, so P = {3,6,9,12......3n}
The answer to the stated question is "Yes" so Statement 1 is sufficient information to answer the question.

Given Statement 2, since 3 is in P, then 3-3 = 0 etc, so P = {3, 0, -3, -6.....-3n}
In which case we can still answer the question "No, all positive multiples of 3 are not in P"
We can still answer the question, and Statement 2 is sufficient information to do that.

Tell me if you agree or disagree, and why?

Thanx

The bold is wrong .... U have taken the extrem case in one side... only 3 is a positive multiple of 3 in P.... but what about the extremum case in the other side.... all can be in P ...

Intern
Joined: 02 Jan 2007
Posts: 41
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

10 Feb 2007, 16:51
What the other posters are suggesting is that along with what you wrote, this also satisfies statement 2:

3n, 3(n-1), 3(n2),..., 12, 9, 6, 3

Which *does* include all multiple of 3 and, thus, along with your reasoning for statement 2, makes this not sufficient.
_________________

Beginning with the end in mind. Aiming to join the 700+ club.

SVP
Joined: 05 Jul 2006
Posts: 1747
Followers: 6

Kudos [?]: 358 [0], given: 49

### Show Tags

10 Feb 2007, 17:00
man it says 3 is in P this doent mean it has to start with 3
Senior Manager
Joined: 04 Jan 2006
Posts: 279
Followers: 1

Kudos [?]: 38 [0], given: 0

### Show Tags

10 Feb 2007, 19:37
Can set P in (1) be

P = {-3n, -3(n-1), ..., -9, -6, -3, 0, 3, 6, 9,...3n} ?

P = {..., 3(-3), 3(-2), 3(-1), 3(0), 3(1), 3(2), ...}

If true, wouldn't (1) INSUFF?
10 Feb 2007, 19:37
Display posts from previous: Sort by