Multiple of 3

[svishal1123]

Hello, Here's a question from Princeton Review:

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 ant integer in P, that integer minus 3 is also in P.

The answer is A.
My answer was E based on the following argument.

3 is in P. Taking first statement into consideration, lets take a number, say x and assume that it is in P. So we will have x+3 also in P. We dont have any information whether x or x+3 is a multiple of 3. If x = 6 then both ARE multiples, however if x = 5, both will not be. We know that a number and its sum with 3 are in the set but we have no clue, what are the other numbers in the set.

Taking second statement into consideration, it will work absolutely the same way as first.

Can somebody help me understand the reason behind the correct answer or is my answer correct?

As far as SC is concerned is there any flaw in my question above - Can somebody help me understand the reason... I like to practice SC this way.

[cramya]

Stmt I

For 3 to be in P 0 must have been in P since sum of 3 and intger 0 makes 3 to be in P.

If 3 is in p 3+3 is in P etc....

SUFF

Stmt II

For ant integer in P, that integer minus 3 is also in P.


For 3 to be in P 6 must have been in P. The other positive multiples may or may not be in P

INSUFF

Hence I think the oa is what it is.

[svishal1123]

I am not really sure I understand that. For 0 to be in P, the statement has to be reflexive. So if n is in P, there is integer in P which is n-3.

Lets take an example to understand this.
# Lets take a number which is in P, say 11 so 14 is also in P.
# If 11-3=8, is not in P, it does not contradict the question statement or the first statement. Hence 11 can be in P without 8.

This above was only for statement 1.

[svishal1123]

On thinking further I understood how the answer is.

3 is in P (from the question) so from statement 1, 6 should also be in P, hence 9 should also be in P. There could be other numbers as well, but 3,6,9... are there for sure. hence all the POSITIVE multiples are there

From statement 2 since 3 is already there, 0 is there, hence -3 is there, -6 is there. and so on... but these are not positive multiples. There could be other numbers, but 0,-3,-6 are there for sure, but these are not positive. We dont whether the positive multiples are there from this statement alone.
Hence the answer.