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]
Last edited by
Ian Stewart on Mon Sep 01, 2008 9:44 am, edited 1 time in total.
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