If n and k are positive integers, is n divisible by 6?

1.) Sufficient

Always remember, when 3 consecutive integers are multiplied, then one should be a multiple of 3 and atleast one (or max 2) number is even. Also, if the multiple of 3 is also the even number like 6, 12, 18 then it is divisible by 6 always..

For ex:- 1*2*3 ; -4*-5*-6 ; 4*5*6 ; 11*12*13..


In any case; the product would be divisible by 6.

2.) insufficient.. does not say anything about n..

Ans. A

0 is divisible by every integer (except 0 itself).
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My answer would be A.
Since n = product of 3 consecutive numbers, it will be divisible by 3. We should also remember that 0 is also a multiple of 6. For eg : if k = 1,
n = 1*2*0 => n = 0 which IS divisible by 6.

I hope I am right. "0 is divisible by any number" right?

I slightly differ on your explination......specially if k = 1...

If k = 1 then n = 0*1*2 then n = 0 ... zero is not a positive integer hence all values of k will start from 2 onwards
hence values of n will start from 6 onwards.....but yes your ans is right A it is

Unsure how A is sufficient (I know that's the answer but I'm hoping someone could help show me why)

For me A is insufficient because:

if n= k(k+1)(k-1)

we can get several cases, such as

k= 0
0*1*(-1) = -1 <-- not divisible by 6

k=1
1*2*0 = 0 <-- not divisible by 6

k=2
2*3*1 = 6 <--- divisible by 6

And so on. Statement 2 could supplement to my dilemma here, but since the answer is A, does anyone have a dumbed down explanation?

Here's my way to solve this:

Statement (1):
Rewrite n = k(k + 1)(k - 1) to: \(k^3-k\) and plug in various positive integers to see that the result will alsways be divisible by 3. Therefore AC 1 = Sufficient

Statement (2)
Gives you just an idea about the term k-1, but nothing about k or k+1 itself. Therefore clearly insufficient.

reto, a couple of points about your solution (although you were able to get to the correct answer):

For statement 1, k(k-1)(k+1) will ALWAYS be divisible by 6. Think of this way. For any number to be divisible by 6, it needs to be divisible by BOTH 2 and 3 at the same time.

Consider 2 cases, if k =2p, then k+1 or k-1 MUST be disivible by 3. Thus the product k(k-1)(k+1) will always be divisible by 6.

Case 2: if k = 3p, then k-1 or k+1 MUST be divisble by 2. Thus the product k(k-1)(k+1) will always be divisible by 6.

As for your interpretation of statement 2, we do not need to even look at the statement and start computing numbers as the question stem asks us whether "n" is divisible by 6 and NOT whether "k" is divisible by 6. Statement 2 does not provide any relation between k and n and hence straightaway you can reject this statement.

Also you are incorrect to say that as k-1=3p, you have no information about k+1 or k. This is not correct. If k-1=3p., then 1 of k or k+1 MUST be divisible by 2 (you can check with some numbers). Thus you do get some actionable information (albeit not pertinent to the original question asked!).
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we need to check whether or not n/6 is and integer
Lets look at statements
Statement 1
n=k(k+1)(k-1)
n=(k-1)k(k+1)
There is an interesting property which can be applied here incase you don't want to use the test cases approach

Property => PRODUCT OF N CONSECUTIVES IS ALWAYS DIVISIBLE BY N!

Hence n which is a product of 3 consecutive must be divisible by 3! i.e. it must be divisible by 6.
Bingo.That is what we needed.
hence sufficient

Statement 2
This statement provides no clue of n
hence not sufficient.

Hence A
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Yes, and also K cannot be 0 according to the question stem.

Source: Gmat-Prep

Bunuel

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Updated the tag. Thank you!
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