# How do you show that the product of 3 consecutive integers is always divisible by 6 ?

Apr 30, 2016

See explanation...

#### Explanation:

Consider three consecutive integers: $n$, $n + 1$ and $n + 2$.

$n = 3 q + r$

for some integer quotient $q$ and remainder $r \in \left\{0 , 1 , 2\right\}$

If $r = 0$ then $n$ is divisible by $3$

If $r = 1$ then $n + 2$ is divisible by $3$

If $r = 2$ then $n + 1$ is divisible by $3$

So at least (in fact exactly) one of $n$, $n + 1$ and $n + 2$ will be divisible by $3$.

Hence $n \left(n + 1\right) \left(n + 2\right)$ will be divisible by $3$ too.

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Similarly for $2$...

$n = 2 {q}_{1} + {r}_{1}$

for some integer quotient ${q}_{1}$ and remainder ${r}_{1} \in \left\{0 , 1\right\}$

If ${r}_{1} = 0$ then $n$ is divisible by $2$

If ${r}_{1} = 1$ then $n + 1$ is divisible by $2$

So at least one of $n$ and $n + 1$ will be divisible by $2$.

Hence $n \left(n + 1\right)$ (and $n \left(n + 1\right) \left(n + 2\right)$) will be divisible by $2$ too.