Rahul@gurome wrote:(1) If p = 13, then 13/3 leaves a remainder of 1.
If p = 16, then 16/3 leaves a remainder of 1.
So, p may or may not be even, no unique answer.
Statement 1 is NOT SUFFICIENT.
(2) If p = 13, then 13/6 leaves a remainder of 1.
If p = 19, then 19/6 leaves a remainder of 1.
It can be seen that p will always be an odd integer in this case.
So, the answer to the main question is "No".
Statement 2 is SUFFICIENT.
[spoiler]The correct answer is B.[/spoiler]
thanks Rahul, I followed this way ...
given: p>0, is p even?
st(1) p/3=i+1/3 where i-integer; p=3i+1 --> p can be either even or odd depending on i (odd or even) Not Sufficient
st(2) p/6=i+1/6 where i-integer; p=6i+1 --> p can be only odd because 6i is always even, and even + 1 = odd Sufficient
answer No, pick B
