broy wrote:1. is zero a positive integer? sometimes, i think gmat considers zero positive.
2. when n =1, it's not divisible by 6. Also, when n = 0, it is not divisible then as 0/6 is there.

so, combining 1 and 2, it has to be E. If we consider the case of zero outside of the solution set, the answer would be A. I would have selected A if this was on the real test, under time pressure.

Anybody else got their 2 cents?

k is stated to be just an integer. so 0 should rightfully be among the choices. If k were 0, n=1.
when (1) and (2) are combined, every number that is plugged in makes n^3 - n divisble by 4.
i still feel the answer choice is C.

0 is divisible by 4. 4 is not divisble by 0. A is the right answer.

If n is a positive integer, is n^3 - n divisible by 4?

1) n=2k + 1, where k is an integer.
2) n^2 + n is divisible by 6.

n^3 - n = n(n^2 -1) = n(n+1)(n-1) = (n-1)(n)(n+1) i.e product of three consecutive integers.

Now, 1) says n is odd, so (n-1) and (n+1) got to be even, so clearly the product must be a multiple of 4 (because having two even numbers would mean having the "factor 2" multiplied twice and thats 4)
Considering k=0: n=1, (n-1) =0 and (n+1)= 2
So the product is 0, which is no doubt divisible by 4.
0 is neighther +ve nor -ve.
Note n-1 is 0 here and not n (n=1), which means the original condition of "n" being a positive integer still holds good.

So the answer is "A" without any ambiguity. Hope I've made myself clear

I'm having the same doubt as the OP. If k can be zero, then clearly (n^3-n) would be zero and thus statement 1 would be sufficinet, wouldn't it?

mbaapp07 wrote: Considering k=0: n=1, (n-1) =0 and (n+1)= 2
So the product is 0, which is no doubt divisible by 4.

Is this really true? Is 0 divisible by any number?

BlueMentor

Thanks Ian.

BM

I can manage to sort out the first part without getting to the numerator 0.

for (n³-n)/4 = n(n²-1)/4

For n=2k+1

(2k+1)((2k+1)² - 1)/4 =

(2k+1)(4k²+4k+1-1)/4 =

(2k+1)(4)(k²+k)/4

The 4 on the numerator makes it divisible by 4...

Guys the stem comes down to is k>3n??
A is suff.
I have doubt about stmt 2

n+k>3n = k>2n.
If k>2n does'nt it automatically mean it is greater thatn 3n??????

ratindasgupta wrote:If n is a positive integer, is n^3 - n divisible by 4?

1) n=2k + 1, where k is an integer.
2) n^2 + n is divisible by 6.

(1) okay, first off, we are able to know that K>0. This is because of "if n is a positive integer" means that n>0. If k<0, then n would equal a negative number. we also know that n is odd. this is because of the "+1" and the facts that k is multiple by 2, and any number multiplied by an even number will result in an even number. adding one insures that n will be odd, because an odd and an even added together create an odd number. However, this alone is insufficient

(2) gives us a couple of ideas about what n maybe be. a good way of thinking about this is breaking 6 into it's factors or 3 and 2. we know that 2 would work for 6, but so would 3, and so would 4 and so would 6, and so forth. so insufficient as well.

together we know that n is odd, so n cannot be a multiple of 2, however, it could be an odd multiple of of 3.

is n^3 - n divisible by 4?

look at it as an (oddxoddxodd)-odd

three odds multiplied by each other = an odd-odd=an even. I would plug in 3,9,15

3^3-3=24/4=6
9^3-9=261-9=252/4-=63
i would probably stop here and feel comfortable with saying C

I see what I did wrong here, looking at A, i should have quickly plugged in 1,2,3 for K just to check. and seeing that n would equal to 3,5,7, each of those would be sufficient.