n(n+1)(n+5)=n(n+1)[(n+2)+3]
=n(n+1)(n+2)+3n(n+1)
Now, n(n+1)(n+2) being the product of three consecutive integers is divisible by 3!=6.
n(n+1) being the product of two consecutive integers is divisible by 2!=2
∴3n(n+1) is divisible by 6
∴n(n+1)(n+5)
=n(n+1)(n+2)+3n(n+1) is divisible by 6
i.e., n(n+1)(n+5)=M(6)