(n^2-2n)(n^2-1) = (n-1)n(n+1)(n^2-2)
ii. Product of 3 consecutive integers is always divisible by 3 and since one of n-1,n,n+1 is even => The product is divisible by 6
i. n= odd => n-1 and n+1 are even, so the product is divisible by 4
n= even => n and n^2-2 are even, so the product is divisible by 4
iii. for the expression to be divisible by 18, the product should have 3,3,2
lets consider n = 100 and n= 101
n=100, 99*100*101*9998 => 99 has two threes and overall expression
...
ii. Product of 3 consecutive integers is always divisible by 3 and since one of n-1,n,n+1 is even => The product is divisible by 6
i. n= odd => n-1 and n+1 are even, so the product is divisible by 4
n= even => n and n^2-2 are even, so the product is divisible by 4
iii. for the expression to be divisible by 18, the product should have 3,3,2
lets consider n = 100 and n= 101
n=100, 99*100*101*9998 => 99 has two threes and overall expression
...
