Lets begin with some notation. Let # mean not equal to and let > mean greater or equal to.
(1) can be rewritten in the following way n(n+1) # 3. It should be clear that n and n+1 are consecutive numbers and that any three consecutive numbers must contain at least one that has a factor of 3 in it. So, according to our statement both n-1 (and n+2) must be divisible by 3. It works for negative numbers and zero as well. Remember, zero is divisible by 3 since 0/3 = 0. And, neither n or n+1 can be zero since that would make their product divisible by 3. A is sufficient.
Lets look at (2). We can rewrite 3n+5 > k +8. Since k is a multiple of 3, lets substitute 3x for k where x is any non-negative integer, and by subtracting 5 from both sides: 3n > 3x + 3, which implies that n > x + 1. This tells us nothing about the divisibility of n-1. It merely tells us that n-1 is non-negative.
Users browsing this forum: No registered users and 2 guests
You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum
GMAT(TM) and GMAT CAT (TM) are registered trademarks of the Graduate Management Admission Council(TM). The Graduate Management Admission Council(TM) does not endorse, nor is affiliated in any way with the owner or any content of this site.