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

Question

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.

Bunuel · Accepted Answer

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

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.

Answer: A.

Hope it's clear.

AbhayPrasanna · Answer

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.

ANSWER TO PROMPT: Maybe. 
INSUFFICIENT.

Pick A.

sunaimshadmani · Answer

Bunuel plz help. I m stuck here, how does st (1) ensure that just +ve multiples of 3 are in set P? For instance if it has -6, than 3 + -6 =-3, is also in that set, so the statement holds true but it has -ve multiples within the set. So I answered E due to the condition of "+ve multiples"

Bunuel · Answer

Please pay attention to the part in red:
 If P is a set of integers and 3 is in P, is every positive multiple of 3 in P?

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.

Answer: A.

The question does NOT ask whether P consists ONLY of positive multiples of 3. It asks whether every positive multiple of 3 in P.

sunaimshadmani · Answer

Thanks. The last line made it crystal clear

hoangphuc · Answer

I'm not very clear with this answer. If your statement 2 can state like that, how didn't you question statement 1 in the same way? It means that we're not sure about whether set P contains min number like 0, 3, 6. Set P can start from 500 for example. In that case not every multiple of 3 is in the set P. Insufficient

Bunuel · Answer

We know that 3 is in the set. From (1) we also know that if any integer is in the set, then (that integer) + 3 is also in the set. Since 3 is in the set, then so must be 3 + 3 = 6. If 6 is in the set, then so must be 6 + 3 = 9, and so on.

gvrk_77 · Answer

Hi Bunnel,

I understand how A is right, but I think B should also works fine. Because the second statement clearly mentions for every 
Integer minus three results in that set. So now for. the given question 3 is already present, and that we need 6 in the set to get the result 3. So we only need multiples of 3 in positives too.

Bunuel · Answer

Not so. You cannot say that if 3 is there than 6 must also be there. What if the set is {3, 0, -3, -6, -9, -12, ... } So, basically what if 3 is the source integer? Similar questions to practice: https://gmatclub.com/forum/for-a-certai ... 36580.html https://gmatclub.com/forum/a-set-of-num ... 98829.html https://gmatclub.com/forum/k-is-a-set-o ... 03005.html https://gmatclub.com/forum/k-is-a-set-o ... 96907.html https://gmatclub.com/forum/for-a-certain ... 36580.html https://gmatclub.com/forum/for-a-certai ... 61920.html Hope this helps.

gvrk_77 · Answer

Hi bunnel, 
So now as you’ve mentioned that what if the set consists of {3, 0, -3, -6, -9, -12, ... } is possible solution, then there is only one positive number which is multiple of 3 and that itself will satisfy the condition. We can’t write the set as {3, 2, 1, 0, -3, -6, -9, -12, ... }. Please help what I’m missing

Thanks, 
Rajesh

Bunuel · Answer

The question asks: is every positive multiple of 3 in P? So, basically the question asks whether 3, 6, 9, 12, 15, ... ALL positive multiple of 3 are is set P. It's certainly possible ALL of them to be in the set, if the set is {..., -12, -9, -6, -3, 0, 3, 6, 9, 12, ...} nothing prevents this set to be an actual one (answer YES) but it's also possible the set to be {3, 0, -3, -6, -9, -12, ... } (answer NO).

patriciadfer · Answer

I would go for E as well...

Imagine that the set mentioned is the set of all integers:
 
Or it is a set of integers n, where n = 0+3k or n = 1+3k, k being any intenger:

In this case, the set satisfy statement a and b, but not every number in the set is a multiple of 3
so statement a and b are insufficient

Bunuel · Answer

The question does not ask whether all the numbers in the set are multiple of 3. The question asks whether all positive multiples of 3 are in the set. The correct answer to the question is A, not E. Please read the discussion above carefully and follow the links provided.

Hope it helps.

Timebomb · Answer

Bunuel

I am not clear how 2nd statement is insufficient. According to me second statement should be sufficient as it has all the elements that are Element-3,

If there is an infinitely large number that is a multiple of 3 say 3333333333 then it will have 3333333330 in the set but it will not have 3333333336 in the set. If it has 3333333336 in the set it will not have 3333333339 in the set.

It does not guarantee that all positive multiple of 3 are there in the set as all positive multiple can tend to infinity.

Hence the statement 2 should be sufficient.

Posted from my mobile device

amanvermagmat · Answer

Hi

In your example, if 3333333333 is present, then its sure that 3333333330 is also present. 
and now since 3333333330 is present, we are sure that 3333333327 is also present.
Similarly 3333333324, 3333333321, .........., 9, 3, 0, -3, -6, -9, .... are ALL present.

But we cannot be sure about multiples of 3 greater than 3333333333. 3333333336 might or might not be present. Statement 2 doesn't say that 'element + 3' cannot be present, it says that 'element - 3' has to be present.

So we cannot be sure whether ALL positive multiples of 3 are present or not.

Thats why statement 2 is NOT sufficient.

Timebomb · Answer

Hi,
Thanks for the reply.
But what validates statement 1 is that for every integer , t, t+3 exists in the set, and as solution says, the set is infinite. It DOES ENSURE that all positive multiples of 3 are in the set.
The question stem reads " is every positive multiple of 3 in P?" Its doesn't say all the elements of P are multiple of 3. Domain of " all possible multiples of 3" is not same as the domain of set P, instead its much larger.

I hope I am able to convey the message.

Posted from my mobile device

amanvermagmat · Answer

Hi

As you yourself said, "statement 1 DOES ENSURE that all positive multiples of 3 are in the set". Hence it is sufficient to answer the Question with a YES

But statement 2 just ensures that  all negative multiples of 3  are in P"..because its given that 3 is in P, so 3-3=0 also must be there, so 0-3 =-3 also must be there, so -3-3= -6 also must be there and so on... But we simply cannot say whether positive multiples of 3 are present in P or not. There could be some positive multiples of 3 in P, there could be ALL positive multiples of 3 in P who knows.. we just cannot answer the asked question with either a YES or a NO.. thats why statement 2 is not sufficient.

MHIKER · Answer

(1) We can take any integer of set P, 3 is in P. Then 3+3=6; 6+3= 9... and so on. Sufficient.

(2) We can take any integer of set P, 3 is in P. Then 3-3=0; 0-3=-3 and so on. So, P has (3, 0, -3). Thus P has a Positive number, non-negative, and negative. Insufficient.

The answer is A.

gmatuser1239 · Answer

Regarding Statement 1, what if 1 is in P ---> 1+3=4 which is not multiple of 3 (Same goes for 2,4,5.....) =>Insufficient

Regarding statement 2, What if 10 is in P--> 10-3=7 which is not multiple of 3=>Insufficient

Taking 1 and 2 together suppose 10 is in P --> 10-3=7 and 10+3=13 in P and both 7 and 13 are not multiple of 3 =>Insufficient 
E

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

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

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