Question

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

Answer 1

See explanation...

Explanation:

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

`n = 3q+r`

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

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

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Similarly for `2`...

`n = 2q_1+r_1`

for some integer quotient `q_1` and remainder `r_1 in {0, 1}`

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