Question

Prove that $n(n+1)(n+5)$ is a multiple of $6$.

Answer 1

$n(n+1)(n+5)=n(n+1)(n+6-1)=n(n+1)(n-1+6)=n(n+1)(n-1)+6n(n+1)=(n-1)n(n+1)+6n(n+1)$

Now $6n(n+1)$ is clearly divisible by $6$

$(n-1)n(n+1)$ is the product of three consecutive integers which is always divisible by $6$ whcih can be proved as

When ever a number is divided by $3$ it leaves a remainder $0,1$ or $2$

So the number is of form $n=3p,3p+1,3p+2$

If $n=3p$ then $n$ is divisible by $3$

If $n=3p+1$ then $n-1=3p+1-1=3p$ which is divisible by $3$

If $n=3p+2$ then $n+1=3p+2+1=3p+3=3(p+1)$ which is divisible by $3$

So of of the consecutive terms is always divisible by $3$

Atleast one of the three consecutive numbers will always be even so it will be divisible by $2$

hence proved that $(n-1)n(n+1)$ will be divisible by $6$