How do you find an integer n such that $((n+7)!)/(n!(n+6)(n+4))$ ?

Question

6289 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-16T00:20:53

See below.

$((n+7)!)/(n!(n+6)(n+4))=((n+1)(n+2)cdots(n+7))/((n+6)(n+4))=$
$=(n+1)(n+2)(n+3)(n+5)(n+7)$

As we can see

$((n+7)!)/(n!(n+6)(n+4))=(n+1)(n+2)(n+3)(n+5)(n+7)$ and the division always gives an integer as result. In other words

$((n+7)!)/(n!(n+6)(n+4))$ is integer for any $n in NN, n ge 0$

How do you find an integer n such that ((n+7)!)/(n!(n+6)(n+4))((n+7)!)/(n!(n+6)(n+4))?