From the question stem, we see our set P looks something like this: {...a, b, c, d, 3, e, f...}. That's my way of denoting that there is the number 3, and then some other numbers. Some may be bigger, some may be smaller... but right now we don't know. We also don't know if the set is finite or infinite.

Coming to statement (1), we see that for any number in the set P, the number + 3 is also in the set P. Right now, the only number that we know is in the set is 3. So, 3 + 3 = 6 must also be in the set. Since 6 is in the set, then 6 + 3 = 9 must also be in the set. You can see where this is going. The set {3, 6, 9, 12, ....} must be a subset of P. I'm careful by saying subset, because there is nothing that would prevent, say, 7 from being in P (and thus 10, 13, 16, etc.)

Ok, so from statement (1) we see that P has to contain every positive multiple of 3. Thus, statement (1) is sufficient.

Coming to statement (2), we see that for any number in the set P, the number - 3 is also in the set P. Right now, the only number that we know is in the set is 3. So, 3 - 3 = 0 must also be in the set. Since 0 is in the set, then 0 - 3 = - 3 must also be in the set. You can see where this is going. The set {3, 0, -3, -6, -9, ....} must be a subset of P. This does NOT guarantee that all positive multiples of 3 are in the set. It only makes a guarantee about the NEGATIVE multiples and 3 itself. So, statement (2) is not sufficient.

Since statement (1) is sufficient and statement (2) is not, the answer is A.

Hope that helped,

Tatiana

Is the answer A?

Here is what I think

(1)

P is a set of integers and 3 is in P
&
for any integer in P, the sum of 3 and that integer is also in P.

All that we know is there are a bunch of integers in set P and 3 is the only one we know. Statement 1 says sum of 3 and that integer is also in P

we know that 3 is in P. Sum of 3 +3 =6

That makes 3+6 =9 also belong to P. So does 3+9, 3+12 and so on

we can say that every positive multiple of 3 is in P . BCE out, Ans between A & D


(2)


P is a set of integers and 3 is in P
&
for any integer in P, that integer minus 3 is also in P.

By similiar logic, we can say 3-3 =0 is in P, 0-3,=-3 is in P
-3 -3 =-6 is in P and so on..

Now we know that P contains 3, 0 -3 -6 , -9 and so on along with bunch of other numbers. 3X1 =3, so one of the multiple of 3 comes into set P . Now it gets tricky. I don't know whether 6, 9,12 .. happen to be in the set by chance or not. They can be and they possibly are not INSUFF

Let us know if A is correct or not

Ans is A.

Thank you all. It's clearer to me now.