Question

Show that the product of three consecutive positive integers is always divisible by 6.

Answer 1

(proof by contradiction)

Suppose there exists a product of 3 consecutive positive integers that is not divisible by 6 :
$6\nmid n(n+1)(n+2)$
$\implies~6\nmid [n(n+1)(n+2) - 3n(n+1)]$
$\implies~6\nmid [n(n+1)(n+2-3)]$
$\implies~6\nmid [n(n+1)(n-1)]$
$\implies~6\nmid [(n-1)n(n+1)]$

Repeating the whole argument we see that
$6\nmid [(n-2)(n-1)n]$
$\,\;\vdots$
$6\nmid[1\cdot 2\cdot 3]$

This is a contradiction because $1\cdot 2\cdot 3=6$ and $6\mid 6$.The only resolution to this contradiction is that there doesn't exist a product of 3 consecutive positive integers not divisible by 6. That is, the product of any 3 consecutive positive integers is divisible by 6.