KCC's Quiz AQQ300 about Divisibility by 6

n^5-n = n(n-1)(n+1)(n^2+1) 
n-1, n and n+1 are 3 consecutive numbers

for n = 0,1; product will always be zero (which is 6*0)
for all other n, one number amongst n, (n-1) and (n-2) has to be even number (multiple of 2) and one number has to be multiple of 3
Thus their product has to be divisible by 6

n^5-n = n(n-1)(n+1)(n^2+1) 
n-1, n and n+1 are 3 consecutive numbers

for n = 0,1; product will always be zero (which is 6*0)
for all other n, one number amongst n, (n-1) and (n-2) has to be even number (multiple of 2) and one number has to be multiple of 3
Thus their product has to be divisible by 6

Right Sriramg ! The key point is to have observed the product contains 3 consecutive numbers which must have a muliple of 2 and a multiple of 3 in it!