Technically if it is n*(n+1)*(n-1). It would be divisible by 3.
as n-4 and n-1 have difference of 3, the remainder would still be the same. Ig the best way is to look it in term of cases
case 1: n is greater than six and is of the format 3k
Here as n=3k, the product is divisible by 3
Case 2: n is greater than six is of format 3k+1
Then n*(n+1)(n-4) becomes (3k+1)(3k+2)*(3k-3). Again last term is divisible by 3
Case 3: n=3k+2
Product becomes: (3k+2)*(3k+3)(3k-2)
Here middle term is divisible by 3
So technically the number is divisible by 3