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) $


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.
