Statement 1: Plug in a few numbers, you can see that for any positive integer k, n is divisible by 6. Sufficient.

Or, for a more formal definition, notice that k(k+1)(k-1) are three consecutive integers, and there is always a number in the sequence that has a multiple of 2, and a subsequent number that is a multiple of 3 (i.e 1,2,3 or 2,3,4 or 7,8,9), so having a multiple of 3 and another multiple of 2 ensures that any number k will be divisible by 6.

Statement 2: No relationship to n. Insufficient.

Answer: A

Why not the answer be D.

if n and k are positive integers, is n divisible by 6

a, n=k(k+1)(k-1)
Yes for all values of K n is divisible by 6
This is true even if K=1 because 0/6=0
i.e 0/any number is 0 , 0 is not divisible only by 0.
Sufficient:

b,k-1 is a multiple of 3

If (k-1) is a multiple of 3 then K can either be even or Odd
Case 1:
If K is even and (k-1) is multiple of 3
K(k+1)(k-1)
even (multiple of 2) * odd * Odd and multiple of 3/ --6 Yes!
Case 2:
If K is odd and (k-1) is multiple of 3 , K cannot be Zero
K(k+1)(k-1)
odd* even (a multiple of 2) * Odd and multiple of 3/ --6 Yes!
Sufficient:

So the answer is D.

A, to my mind.

1) From 1 we know the relationship between n and k.
n is product of 3 consecutive integers. And at least one of them MUST be even (i.e. a multiple of 2) So, the product is divisible by 2 and 3. Hence N is divisible by 6. Sufficient.

2) Knowing that (k-1) is a multiple of 3, we cannot decide whether n is divisible by 6. Insufficient then.

for me too A...whenever we have product of three consecutive integers ...the product is divisible by 6...

-1,0,1
0,1,2
1,2,3

take any combo u want...and the product would be div by 6

It is not that hard.

Just note down this concept:

Product of any n consecutive integers is divisible by n!

Gmatter2.0 wrote:Why not the answer be D.

if n and k are positive integers, is n divisible by 6

a, n=k(k+1)(k-1)
Yes for all values of K n is divisible by 6
This is true even if K=1 because 0/6=0
i.e 0/any number is 0 , 0 is not divisible only by 0.
Sufficient:

b,k-1 is a multiple of 3

If (k-1) is a multiple of 3 then K can either be even or Odd
Case 1:
If K is even and (k-1) is multiple of 3
K(k+1)(k-1)

This is the data sufficiency police: STOP RIGHT THERE AND PUT YOUR MARKER UP!

How is K(k+1)(k-1) relevant to statement (2), taken alone?

You've committed one of the most serious (and common) DS crimes - you're pretending to evaluate statement (2) by itself, but you're stealing information from statement (1).

Remember step 2 of the Kaplan Method for Data Sufficiency: Consider each statement by itself (in conjunction with the stem).

The only way you could make (2) sufficient is by using the information provided in (1), which clearly indicates that (2) isn't sufficient alone.

Looking at statement (2) in a vaccuum, we have information about k, but absolutely no information about n; accordingly, there's no way it can be sufficient to answer a question about n.

Thank's for pointing my error, My big mistake...