How can I prove that $(n^2+11n)/6$ has no remainder?

Question

I must prove, that when dividing the following by 6 it yields no remainder: $$n^2+11n $$ Is this done with mathematical induction method? and what other technique can I use?

Update: some may find this question as a potential duplicate of the correct case when we have $n^3$ instead of $n^2$. I want to clarify that this is not true.

Answer 1

If you really mean $n^2+11n$ then the remainder can be different to zero as @OpenBall point out, for $n=2$ we have $26$ is not congruent to zero modulo $6$.

If you mean $n^3+11n$ then it is true because

$$n^3+11n=n(n-1)(n+1)+12n$$

so the remainder is always zero.

Answer 2

The number $(n-1)(n-2)(n-3)$ is always divisible by $6$ because this is a product of three consecutive integers. Now $$(n-1)(n-2)(n-3)=n^3-6n^2+11n-6,$$ whence $$n^3+11n=(n-1)(n-2)(n-3)+6(n^2-1)$$ is divisible by $6$.