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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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