0 is divisible by every integer (except 0 itself).
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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

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

Right. Now see the mistake.
Thanks!

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?

JUST realised that the two first instances that I listed up are NOT positive integers hence they cannot be N. Ooops. Forget my question

Oh! Thanks, that´s good to know :D

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.

i dont understand why b is not sufficient,
if k-1 is a multiple of 3, then k or k+1 should be a multiple of 2...
and so, it should be sufficient as well....
am i missing something?
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You are correct to say that if k-1 =3p, then k or k+1 will be =2r. But what is the connection between k and n? You are using information from statement 1 to check for sufficiency of statement 2. This absolutely wrong. You need to isolate both statements completely to determine their respective sufficiency. Only once they are not sufficient ALONE, you go to combining the 2 statements.
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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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Hi Engr2012,
irrespective of what K-1 is, one of k or k+1 will always be divisible by 2 as k is a positive integer..

Yes that is what I have written above. Additionally, the question asks about n and NOT k. Without any relation between n and k , this information is of no use.
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Hi,
I am not questioning the solution. it is very clear and fairly simple...
i am just pointing to the highlighted portion that it is not because k-1=3p that k or k+1 will be div by 2 but one of k or k+1 will always be div by 2...

Hi Bunuel

I just realized something while solving this question, please let me know if im correct on this one:

Product of n consecutive integers is always divisible by
-all integers between 1 and n inclusive.
-n!
-all the factors of n!

Can we make this generalization??

hi ,
if you mean first n consecutive integers..
then product of n consecutive integers is nothing but n! itself...

Nope i dont mean that.. I mean 'ANY n consecutive Integers'. What do you say about that?

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

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

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

(2) k – 1 is a multiple of 3

This is a recently common type of question.
There are 2 variables (n,k) and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer. Looking at the conditions individually,
for condition 1, n becomes a product of 3 consecutive integers, so this always makes it a multiple of 6, so this is sufficient.
for condition 2, there is not information about n, so this is insufficient, and the answer becomes (A).
We cannot look at the conditions together as this is a commonly made mistake type 4(A).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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