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Source: — Data Sufficiency |

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by Rahul@gurome » Sun Jan 23, 2011 9:44 pm
(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]
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by Night reader » Sun Jan 23, 2011 10:06 pm
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
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