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If P is a set of integers and 3 is in P, is every positive [#permalink]
02 Jul 2010, 09:24

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Difficulty:

45% (medium)

Question Stats:

36% (01:00) correct
64% (00:56) wrong based on 150 sessions

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.

I had difficulty with this question because of the wording, I wasn't sure what they were looking for exactly, and I didn't find the explanation in the book to be sufficient. If anyone can break it down into an easier explanation I'd apprecaite it.

Re: DS P is a set of integers... [#permalink]
02 Jul 2010, 10:00

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We can write P as a set of an undetermined number of integers that contains the number 3.

P = {l , m , n, ..... , 3 , x , y , z, ....}

Is every positive multiple of 3 in P ? In effect the question is asking you is every number in this infinite series : 3,6,9,12,15,....... is present in P, or not. A yes or no answer will suffice.

Statement 1:

For any integer "q" in P, "q+3" is also in P.

Since we know that 3 is in P, 3+3 = 6 is also in P. Since we know that 6 is in P, 6+3 = 9 is also in P. Since we know that 9 is in P, 9+3 = 12 is also in P. AND SO ON.... Clearly this will go on forever, ensuring that EVERY positive multiple of 3 is in P. ANSWER to PROMPT - Yes

SUFFICIENT.

Statement 2:

For any integer "q" in P, "q-3" is also in P.

Since we know that 3 is in P, 3-3 = 0 is also in P Since we know that 0 is in P, 0-3 = -3 is also in P. Since we know that -3 is in P, -3-3 = -6 is also in P. AND SO ON.... Clearly this will go on forever, ensuring all NEGATIVE multiples of 3 are in P.

What can we say about the POSITIVE multiples, remember an answer of No will suffice, but CAREFUL:

Two things: 1. 2 statements will never contradict eachother, so either this one is going to answer the question as "yes" just as Statement 1 did, or it is going to be insufficient. Since we don't seem to reach a clear yes, it is probably insufficient.

2. We don't know what other numbers were in the set P other than 3. Consider that P contained the highest positive multiple of 3. This is ofcourse a hypothetical situation since this number would be akin to infinity. But it is theoretically possible that this set contained that maximum positive multiple of 3. Thus, stepping down by 3 from this number as we have above, would result in obtaining all positive multiples of 3. Thus it is possible, but we cannot be sure of this fact from statement 2 since we do not know if this hypothetical number exists in the set or not.

Re: DS P is a set of integers... [#permalink]
02 Jul 2010, 10:01

3

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Expert's post

Caffmeister 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.

I had difficulty with this question because of the wording, I wasn't sure what they were looking for exactly, and I didn't find the explanation in the book to be sufficient. If anyone can break it down into an easier explanation I'd apprecaite it.

Positive multiples of 3 are: 3, 6, 9, 12, 15, ... The question asks whether ALL these numbers are in the set P, taking into account that 3 is in this set.

(1) For any integer in P, the sum of 3 and that integer is also in P --> if x is in the set, so is x+3 --> we know 3 is in P, hence 3+3=6 is also in, and as 6 is in so is 6+3=9, and so on. Which means that ALL positive multiples of 3 are in the set P. Sufficient.

Side note: above does not mean that only positive multiples of 3 are in P, there can be other numbers but we are only interested in them.

(2) For any integer in P, that integer minus 3 is also in P --> if x is in the set, so is x-3 --> we know 3 is in P, hence 3-3=0 is also in and as 0 is in, so is 0-3=-3, and so on. So we are not sure whether all positive multiples of 3 are in P, all we know that there will be following numbers: 3, 0, -3, -6, -9, -12, ... Not sufficient.