Alternate way to Prove or disprove $6\mid n(n+1)(n+2)$

Question

This is my proof, I'm wondering if I'm correct, and how to do without induction.

My Work

Basis Step

$$\frac{(1)(2)(3)}{6} = 1$$

Inductive Hypothesis

Assume that $\dfrac{k(k+1)(k+2)}{6} = d$ where $d \in \mathbb Z$

Induction Step

We must show $\frac{(k+1)(k+2)(k+3)}{6}$ is an integer

$$\frac{(k+1)(k+2)(k+3)}{6} = \frac{(k+1)(k+2)}{6} \cdot (k+3)$$

We know that:

$\dfrac{(k+1)(k+2)}{6} = \dfrac{1}{k}\cdot\dfrac{k(k+1)(k+2)}{6}$ by our inductive hypothesis we know this to be an integer

Therefore, by the closure property of integers under multiplication and addition, $\dfrac{(k+1)(k+2)}{6} \cdot (k+3)$ must be integer.

Answer 1

Without induction:

At least one of $n, n+1, n+2$ is even and exactly one is divisible by three. Hence $2 \ | \ n(n+1)(n+2)$ and $3 \ | \ n(n+1)(n+2)$. As $2$ and $3$ are co-prime, this means $6 \ | \ n(n+1)(n+2)$.

Answer 2

If you insist on using induction:

Let $6|n(n+1)(n+2)$.

It is clear that one of the factors $n+1$, $n+2$ is even, so that $2|(n+1)(n+2)$.

If also $3|(n+1)(n+2)$ then $6|(n+1)(n+2)$ and consequently $6|(n+1)(n+2)(n+3)$ and we are ready.

If $3\nmid\left(n+1\right)\left(n+2\right)$ then we can conclude that $3\mid n$ . Then also $3\mid n+3$ resulting again in $6\mid\left(n+1\right)\left(n+2\right)\left(n+3\right)$.