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) $

 A: I would say that you're basically there. All that remains is to prove (by counterexample) that (E) is not the correct answer.
A: Looks good. The multiplication table for three, is $3,6,9,12,15$ etc any consecutive three integers has to include one of these because there are only two integers between them.
A: One formal proof is by case analysis.

Let $n$ be an integer.

Then $n$ is congruent to one of $0,1,2\;$mod $3$.



*

*If $n\equiv 0\;(\text{mod}\;3)$, then $n$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\\[4pt]$

*If $n\equiv 1\;(\text{mod}\;3)$, then $n+2$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.$\\[4pt]$

*If $n\equiv 2\;(\text{mod}\;3)$, then $n+1$ is a multiple of $3$, hence the product $n(n+1)(n+2)$ is also a multiple of $3$.


Thus, in all three cases, $n(n+1)(n+2)$ is a multiple of $3$.

Showing that the product of two consecutive integers is even can be done in the same way, using mod $2$.

Then since the product of the three factors is a multiple of both $2$ and $3$, it must be a multiple of their least common multiple, which is $6$.

Note that if $n=1$, the product is $6$, hence there can't be any integer larger than $6$ which must be a divisor of $n(n+1)(n+2)$, for every integer $n$. In particular, choice E can be rejected.
A: Assume $$a\equiv k \mod 3$$ where $0\le k\le 2$.
$$k=0\implies \text{$a$ is divisible by $3$}$$
$$k=1\implies \text{$a+2$ is divisible by $3$}$$
$$k=2\implies \text{$a+1$ is divisible by $3$}$$
A: 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 $2\cdot3\cdot4=24$.
A: Since at least one of the integers must be even, the product cannot be (A) odd. You and all of the other answerers have shown that since at least one of the integers must be even, and at least one of the integers must be divisible by three, the product must be (D) divisible by $6$. The product often will be, but need not be (B) divisible by $4$, (C) divisible by $5$, and/or (E) divisible by $12$. Note that if the product is divisible by $4$, it will perforce be divisible by $12$ (since one of the integers must be divisible by $3$), so B and E are jointly both true or both false in particular examples. The integers $4,5,6$ multiply to $120$ which satisfies B, C, and E as well as D. To the extent the question asks what must be true, the answer is D. But one or more of B, C, and E will be true unless the three integers begin and end with an odd integer, the even integer has only one factor of $2$, and none of the integers is divisible by $5$.
