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

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I'm flipping through the princeton review 2007 study guide [#permalink]  10 Feb 2007, 13:32
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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, 13:47, edited 1 time in total.
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OK , WHY DO U ASSUME IN STATMENT TWO THAT THE SET STARTS WITH 3.......................... Got it
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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"
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Re: DS: sets [#permalink]  10 Feb 2007, 13:52
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 ...

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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.
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man it says 3 is in P this doent mean it has to start with 3
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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?
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