If n is a positive integer, is n(n^2 - 1) is divisible by 24?

we can write n(n^2-1) as n(n-1)(n+1) as n^2-1 = (n-1)(n+1)
So now we have product of 3 consecutive integers.
As we know, a product of 3 consecutive integers will always be divisible by 3 and 2.
But in order to be divisible by 24, the product should be divisible by 8 and 3.

Stament 1: Not Sufficient
As we know n is divisible by 3 so let us test this condition.
If n = 3 than 2*3*4 = 24 and we get a yes.
if n = 6 than 5*6*7 = 210 and we get a No
Hence this is not sufficient.

Statment 2: Sufficient
n is odd. Let us test this.
If n =1 then 0*1*2 = 0 and 0 is divisible by any integer so yes.
If n = 3 then 2*3*4 is divisible by 24.
If n = 5 then 4*5*6 is divisible by 24.
If n = 7 then 6*7*8 is divisible by 24.
So far we are getting only yes. So now let us look at the logic behind this. If n is odd then both (n-1) and (n+1) will be even. Also in two consecutive even integers, one is always a multiple of 4. For eg: 2&4 or 4&6 or 6&8. So we will always have a 2 and a 4 in the product of both these numbers. This makes (n-1)(n+1) divisible by 8 provided n is odd.
And as we know (n-1)n(n+1) is already divisible by 3 hence it will also be divisible by 3*8 or 24

Answer:B