The product of three consecutive integers is ...? Odd? Divisible by $4$? by $5$? by $6$? by $12$? If i have the product of three consecutive integers:
$n(n+1)(n+2)$, so the result is:
$A)$ Odd
$B)$ Divisible by $4$
$C)$ Divisible by $5$
$D)$ Divisible by $6$
$E)$ Divisible by $12$
My thought was:
$i)$ If we have three consecutive numbers, $a, (a + 1), (a + 2)$, one of these three numbers must be divisible by $3$.
$ii)$ If we have two consecutive numbers, $a, (a + 1)$, one of these two numbers must be divisible by $2$, also one of these numbers will be even and the other will be odd.
$iii)$ Any number is divisible by $6$, when is divisible by $3$ and $2$ at the same time.
So, the correct answer must be $D)$
Well, I would like to know if:


*

*My answer is correct

*What is the formal proof of what I said in $ i) $


To prove your statement $i)$ you can use the technique of induction.
First consider $a=1$ therefore the product becomes 
$$1\cdot(1+1)\cdot(1+2)=1\cdot2\cdot3$$
which is clearly divisible by $3$ since one of the factors is $3$ itself. Next, by setting $a=n$ and $a=n+1$, we get
$$\begin{align}
n(n+1)(n+2)&\equiv0\mod(3)\\
(n+1)(n+2)(n+3)&\equiv0\mod(3)
\end{align}$$
To derive the $n+1$-statement from the truth of the $n$-statement just split up the last factor which yields to
$$(n+1)(n+2)(n+3)=n(n+1)(n+2)+3(n+1)(n+2)$$
where the first term is the given hypotesis and the second one has a $3$ in it and so your $i)$ is right.

For the overall answer for the divisibilty by $6$ it gets a little bit tricky and so I am not quite sure about my own approach. Howsoever:
Again lets assume the cases $a=n$ and $a=n+1$ which yields to the equations
$$\begin{align}
n(n+1)(n+2)&\equiv0\mod(6)\\
(n+1)(n+2)(n+3)&\equiv0\mod(6)
\end{align}$$
This time the rewrite part is a little bit different
$$\begin{align}
(n+1)(n+2)(n+3)&=(n+1)[n^2+5n+6]\\
&=(6)(n+1)+n(n+1)(n+5)\\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)\\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(3)+n(n+1)(3)-n(n+1)(3)\\
&=(6)(n+1)+n(n+1)(n+2)+n(n+1)(6)-n(n+1)(3)\\
&=(6)(n+1)+(n-1)n(n+1)+n(n+1)(6)\\
\end{align}$$
Two of these terms contain a $6$ and are so by defintion divisible by $6$. The remaining term is again the sum of three consecutive integers but in a different form than the given statement. I guess this should be valid nevertheless but I am not sure at all.

Besides the trivial $1\cdot2\cdot3=6$ the divisibilty by $12$ and $4$ fails with $5\cdot6\cdot7=210$. The divisibility by $5$ fails for example with $2\cdot3\cdot4=24$.
