IMOA


Since any 3 consecutive integers are divisible by3.

n^2+n = n*(n+1)

Thus (n-1)*n*(n+1) is divisible by 3. Given that n*(n+1) is not divisble by 3, it implies (n-1) has to be divisible b3.

Stmnt 2 doesnt tell us anything

Ok...so to answer this question, let's first state some rules....

3 divides any one number in a set of three consecutive integers so;

3 either divides 1,2 or 3 or 2,3,4 or 3,4,5 (you get the idea)

so 3 divides either n-1,n or n+1

Condition 1: 3 does not divide n^2 + n or n(n+1) so basically it implies that 3 does not divide n or n+1, so 3 MUST divide n-1. The only exceptions are -1,0,1 since 3 does divide 0*1 but does not divide -1 and the other exception is -2,-1,0 since 3 divides -1*0 but 3 does not divide -2. These two exceptions contradict the case for all other cases like 1,2,3 or -1,-2,-3 etc....so this alone does not help us. We need to satisfy some other conditions that eliminates zero as the third integer.

Condition 2: 3n + 5 > k + 8, k is a multiple of 3 so let k = 3c, where c = set of all positive real numbers

so 3n + 5 > 3c + 8
3n > 3c + 3
n > c + 1.
so we know that if k = 3 is the smallest possible multiple of 3 so c = 1. so n>2 which means that we can now positively say that both statements together state that 3 divides n-1.

Hence the answer is C.

great explanation, tx much

crackgmat007 wrote:Given that n is an integer, is n — 1 divisible by 3?
(1) n^2+n is not divisible by 3
(2) 3n + 5 >= k + 8, where k is a positive multiple of 3

OA - C Pls explain.

are u sure that c is the OA?

You're right. A should be the answer not C.

Ian Stewart wrote:

prindaroy wrote:The only exceptions are -1,0,1 since 3 does divide 0*1 but does not divide -1 and the other exception is -2,-1,0 since 3 divides -1*0 but 3 does not divide -2.

There are absolutely no exceptions to the rule that the product of three consecutive integers is divisible by 3. Zero is most certainly divisible by 3 (you get an integer when you divide zero by 3), so if you look, say, at the three consecutive integers -1, 0, 1, that list contains exactly one number which is divisible by 3, just like any other list of three consecutive integers.

The answer to the question in the original post is A, not C -- as adityanarula pointed out above, Statement 2 tells us nothing useful.

thanks

crackgmat007 wrote:great explanation, tx much

for C or A?