If n is a positive integer, is (n - 1)(n)(n + 1) a multiple of 24?

Answer is A.
Question stem says (n-1), n and (n+1) are consecutive positive integers. Hence, one of them is a multiple of 3

Stat1: alone is sufficient
(n-1) is even implies that (n-1) is a multiple of 2 or 4
if (n-1) is multiple of 2, then (n+1) is a multiple of 4
if (n-1) is multiple of 4, then (n+1) is a multiple of 2

There is a multiple of 2 and a multiple of 4 in the set {(n-1) ; (n+1)}

as per question stem, There is a multiple of 3 in the set {(n-1) ; n ; (n+1)}
Therefore, (n-1)*n *(n+1) is multiple of 2*4*3 = 24

Stat2 is not sufficient
n=3 gives YES answer
n=6 gives NO answer

since n-1 is even, n+1 has to even too. the product of n-1 and n+1 should be multiple of 8 [2m*(2m+2))=4m(m+1),m is positive int]
as we know the product of three consecutive positive int should be multiple of 3, so in this case it should be multiple of 24

If (n-1)(n)(n+1) a multiple of 24, there are 2 cases possible.

Case 1. n is odd
Case 2. n is multiple of 8.

Statement 1. It tells us that n is odd. Sufficient
Statement 2. n can be odd or even. Insufficient

Asked: If n is a positive integer, is (n-1)(n)(n+1) a multiple of 24?

1) (n-1) is an even integer
n is an odd integer = 2k +1 form where k is an integer
(n-1)(n)(n+1) = 2k(2k+1)(2k+2) = 4k(k+1)(2k+1)
2k(2k+1)(2k+2) is a multiple of 3 since they are 3 consecutive integers
2k(2k+1)(2k+2) = 4k(k+1)(2k+1) is a multiple of 8 since k(k+1) is a multiple of 2
(n-1)(n)(n+1) = 2k(2k+1)(2k+2) = 4k(k+1)(2k+1) is a multiple of 8*3 = 24
SUFFICIENT

2) n is a multiple of 3
n = 3k where k is an integer
(n-1)(n)(n+1) = (3k-1)3k(3k+1) = 3(3k-1)(3k+1)
If k = even ; 3k = even: 3k-1 & 3k+1 are odd; (n-1)(n)(n+1) is NOT a multiple of 24
If k = odd ; 3k = odd: 3k-1 & 3k+1 are even; one of them is a multiple of 2 and another one is a multiple of 4; (n-1)(n)(n+1) is a multiple of 3*8 = 24
NOT SUFFICIENT

IMO A

The product of x consecutive integers is always divisible by x!.

Thus (n-1)(n)(n+1) is divisible by 3! = 6.

Now we must find out whether any of the integers is divisible by 4, or in other words, if n is odd or even?

If n is odd, (n-1) or (n+1) will be divisible by 4. If n is even, it depends on whether n itself is divisible by 4.

Posted from my mobile device

WHAT IF N=1
then the answer would not be A
( because even 0 is an even integer)
Am I missing something?

If n=1, then (n-1)=0 and if 0 is one of the factors, then (n-1)n(n+1) would not be a multiple of 24, and we can answer the question with a definite No.

If (n-1)=2, the product would be 2*3*4 = 24, and the answer is Yes.

So A is not the answer.


Edit: Sorry for confusion. And thank you for your interesting doubts.

I think the answer is that 0 is defined as a multiple of everything. It is the "zero-multiple" of 24. And also the "zero-multiple" of 6373,927.

Here is a discussion about this: https://gmatclub.com/forum/is-0-zero-to ... 04179.html

Posted from my mobile device

thanks alot

I cannot agree with the above. "0" is also an even number, so if "n-1" = 0, so the the product will not be a multiple of 24"

0 IS a multiple of 24.

Integer \(a\) is a multiple of integer \(b\) means that \(a=kb\), for some integer \(k\) (so, \(a\) is a multiple of \(b\) if we can express \(a\) as \(a = kb\)). As we can express 0 as \(0 = 0*b\), for ANY integer b, then 0 is a multiple of every integer.

We can see
N-1 = 2k
N = 2K+1
N+1 = 2K+2

(N-1)N(N+1) = 2K(2K+1)(2k+2) = 4K(k+1)(2k+1)
this shows that it is divisible by 4

and on the fact we know that n consecutive numbers always divisible by n!, means 3 consecutive integers always divisible by 3! = 6

hence A is sufficient.


for B take the example of 3 and 6
insufficient