How to prove: 6 | x(x+1)(x+2)

Question

How can I prove that $6$ divide $x(x+1)(x+2)$? I try with $(x-2)$ and $(x-3)$ but that is not ok. Thank you for your help.

Answer 1

Out of two consecutive integers, one of them must be divisible by 2.

Out of three consecutive integers, one of them must be divisible by 3.

$x(x+1)(x+2)$ includes both cases, thus is divisible by 2 and 3, and consequently by 6.

Answer 2

Hint: consider the 3 cases $$x=3k\\x=3k+1\\x=3k+2$$

Answer 3

Induction:

1) $n=1$:

$6|(1)(2)(3) √$.

2) Hypothesis:

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

3)Step for $(n+1)$:

$(n+1)(n+2)(n+3) =$

$n(n+1)(n+2) +3(n+1)(n+2).$

Note :

$(n+1)(n+2)=2k$, is even , since

one of the factors $(n+1), (n+2)$ is even.

Hence:

$3(n+1)(n+2)=3\cdot 2k=6k$, i.e.

$6|3(n+1)(n+2).$

Putting together:

$6|n(n+1)(n+2)$, hypothesis,

and $6| 3(n+1)(n+2).$

Hence $6|(n+1)(n+2)(n+3).$

Answer 4

For the choices of $x$, at least one of $x, x+1, x+2$ is even and we can have the following cases:-

$$x \equiv 0 \mod{3}$$ $$x \equiv 1 \mod{3}$$ $$x \equiv 2 \mod{3}$$

If $x \equiv 0 \mod{3}$, then $x = 3k$ for some integer $k$.

If $x \equiv 1 \mod{3}$, then $x + 2 = 3k$ for some integer $k$.

If $x \equiv 2 \mod{3}$, then $x + 1 = 3k$ for some integer $k$.

Thus, at least one of $x, x + 1, x + 2$ is a multiple of $2$ and one is a multiple of $3$. So, their product is a multiple of $6$. Hence, $6 \vert x \left( x + 1 \right) \left( x + 2 \right)$.

Answer 5

As the others have mentioned, a simple explanation is that among any $3$ consecutive integers, there must be either $1$ or $2$ even integers, so the product will be even and divisible by $2$. There will also be a multiple of $3$, so the product will also be divisible by $3$. Thus, the product must be a multiple of $6$, and thus, divisible by $6$.

For a more formal proof, we are looking to prove that

$$n(n+1)(n+2) = 6m; \quad m \in \mathbb{Z}$$

For $n = 1$, this is obvious, as $1(2)(3) = 6$. Now, let $n = k$, getting $k(k+1)(k+2) = 6m$, the induction hypothesis. Next, let $k = n+1$.

$$(n+1)(n+2)(n+3) = \underbrace{n(n+1)(n+2)}_{= 6m}+\underbrace{3(n+1)(n+2)}_{\text{$(n+1)(n+2)$ is even, so let $(n+1)(n+2) = 2q$.}}$$

From here, it should be simple to take on.

Answer 6

Intuitively:

Dividing successive integers by an integer $k$ gives a repeating cycle of $k$ possible remainders $0\dots k-1$.

Wherever you start in the cycle, after $k$ successive integers you've used all $k$ remainders. So a set of $k$ or more successive integers must include at least one with a remainder of $0$, ie divisible by $k$.

Your set of $3$ successive integers is obviously also a set of $2$ or more, so it must contain at least one divisible by $3$ and one divisible by $2$, making the product of all three divisible by $6$.

Answer 7

All answers above give perfect explanation, why $x(x+1)(x+2) \equiv 0\ (\textrm{mod}\ 6)$. But I would like to suggest an alternative approach. $$ x(x+1)(x+2) = x^3 + 3x^2+2x $$ If $x \equiv r\ (mod\ 6)$, where $r \in \{ 0,1,2,3,4,5 \}$, $x^3 + 3x^2+2x \equiv r^3 + 3r^2 + 2r\ (mod\ 6)$. Here, $r^3 + 3r^2 + 2r \equiv 0\ (mod\ 6)$ for all $r \in \{ 0,1,2,3,4,5 \}$. Hence, 6 always divides $x(x+1)(x+2)$.