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

Also consider the following case:

n-1 = 0, n=1, n+1 = 2; (coz, "0" is an "even integer")

i.e (n-1)(n)(n+1) = 0, which is divisible by any number, hence a multiple of any number.

The answer is "A" .

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

Those are 3 consecutive positive integers.

1) (n-1) is an even integer

As per statement we have the situation: Even, ODD, Even

Any three consecutive integers, mean we have guarantee that we at least 3.

Two even integers, we can guarantee that we have 8.

So answer to the question ins always yes.

Another way is plug numbers.

n-1 =2 & n=3 & n+1=4............2*3*4=24

n-1 =4 & n=5 & n+1=46............4*5*6=120

n-1 =6 & n=7 & n+1=8............6*7*8=336

You can notice above rule mentioned in first solution.

Sufficient

2) n is a multiple of 3

Plug in numbers.

n-1 =2 & n=3 & n+1=4............2*3*4=24............Answer is Yes

n-1 =5 & n=6 & n+1=7............5*6*7=210..........Answer is No

insufficient

Answer: A

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