DS

Yes, the answer should be A. 1) tells us that m and p are even, so the remainder will be even when p is divided by m. Since we're told the remainder is not zero (since p is not a factor of m), the remainder must be 2 or greater. Statement 2 is not sufficient, but:

stern wrote: 2) 3, 10, 15, 6, 5 are all that could have 30 as a common multiple. 10/3 leaves remainder 1 and 5/3 leaves 2 which is greater that 1 --Insufficient

you cannot use 5 and 3 here; the LCM of 5 and 3 is 15, not 30. You can, however, use m = 6 and p = 10.

[edited to fix the typo pointed out by ricaototti]

ricaototti wrote:Dear Ian,

You made a mistake. It can not be true that m=10 and p=6 since in the stem it is stated that 2<m<p.

Thus, I think that answer is D. But you are the GMAT genius here (this was just a careless mistake ) . Please is the answer D or A?

Thanks for catching that, ricaototti- I simply had m and p in the wrong order. It should have read m=6 and p=10. The solution is correct otherwise; using only Statement 2, it's possible that the remainder is greater than 1 (if, say, m=6 and p=10) or equal to 1 (if, say, m=5 and p=6). So the second statement is not sufficient on its own.