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

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Answer 1 · 2016-04-30T08:03:49

See explanation...

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

$n = 3 q + r$ $n = 3q+r$

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

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

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

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

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

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

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Similarly for $2$ $2$ ...

$n = 2 q_{1} + r_{1}$ $n = 2q_1+r_1$

for some integer quotient $q_{1}$ $q_1$ and remainder $r_{1} \in {0, 1}$ $r_1 in {0, 1}$

If $r_{1} = 0$ $r_1=0$ then $n$ $n$ is divisible by $2$ $2$

If $r_{1} = 1$ $r_1=1$ then $n + 1$ $n+1$ is divisible by $2$ $2$

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

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

How do you show that the product of 33 consecutive integers is always divisible by 66 ?