We know that \(24=2^3*3\)
So we should check if \(n(n+1)(n+2)\) contains \(2^3\) and \(3\)

(we should keep in mind that when \(n\) or \(n+1\) or \(n+2\) equal to \(0\), then product of all this numbers will be equal to \(0\) and divisible by \(24\))

1) if \(n\) is even than \(n+2\) contain at least \(2^2\) and as \(n\) is even and not zero than \(n+1\) will be contain \(3\). So \(n(n+1)(n+2)\) will be always divisible on \(24\)
Sufficient

2) from this statement we can infer that \((n+1)\) is odd and contain \(3\) and that \(n\) is even and \(n * (n+2)\) contains at least \(2^3\). So \(n(n+1)(n+2)\) will be always divisible on \(24\)
Sufficient

Answer is D
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When n is even n(n+1)(n+2) is always divisible by 8 and 3. Hence it will be divisible by 8*3 = 24
For the explanation look at:
solving-complex-problems-using-number-line-197045.html#p1521230

Statement 1: n is even
Sufficient

Statement 2: n+1 is divisible by 3 but not by 6
-> n+1 is odd
-> n is even
Sufficient

Answer: Option (D)

if n is even
n=2k
n(n+2)= 2k(2k+2)=4.k.(k+1)
k(K+1) must be even because 2 consecutive number
so, n(n+2) is divisible by 8
n+1 is divisible by 3
so, the first condition is divisible by 24

be careful

n(n+1)(n+2)
is 3 consecutive numbers, one of them must be divided by 3.
n is even , then n=2k, n+2=2(K+1)
n(n+2)=2k*2(k+1)=4k(k+1),
k and k+1 is 2 consecutive interger, one of them must be divided by 2
so 4k(k+1) must be divided by 8
so, the total expresstion is divided by 24

we can pick specific numbers and find the answer,

condition 2
n+1 is divided by 3 but not 6. this mean n+1 is odd . this mean n and n+2 is two consecutive even number, product of which are multiple of 8 as I prove above.
sufficient.

the takeaway is that
the product of 2 consecutive even numbers is divided by 8. remember how to prove this
if there are 3 consecutive numbers, one of them must be divided by 3

those concept is tested many times on gmat because it is basic

St 1 : n is even

Minimum value of n = 2

(n) (n+1) (n+2) in this case = 2 * 3 * 4 = 24 Yes

For higher even values of n, (n) (n+1) (n+2) is a higher multiple of 24

Definite Yes

Sufficient

St 2 : (n + 1) is a divisible by 3 but not by 6

Which means (n+1) is a odd multiple of 3

So (n + 1) is odd as 3 is odd

This implies n is even

Definite yes as seen from St 1

Sufficient

Choice D
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Minimum value of n can be zero also.
If n = zero , the entire numerator will be zero and be divisible by 24.
The answer will be YES in that case.

Bunuel VeritasKarishma

Well, n is even means that n is ..., -6, -4, -2, 0, 2, 4, 6, 8, .... So, n can be a negative even number too, not only 0 or positive even number. That changes nothing though, (1) is still sufficient.
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