<s>[INST] Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
 = 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
 = 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
 [/INST] Just to be different (and to rip off Mark Bennett's answer):
$ab = k> 0$ with $0< a \le b$ means $a \le \sqrt {k}$ and $b \ge \sqrt {k}$. (and if $0< a < b$ then $a < \sqrt k$ and $b > \sqrt{k}$)
So $(n+3)(n+4) = 90$ means $n+3 \le \sqrt{90} \approx 9.smallchange \le n+4$
As $n+3$ and $n+4$ are consecutive we must have:
$(n+3) \le \sqrt{90} < (n+3) + 1$.
Or in other words $n + 3 = \lfloor \sqrt{90} \rfloor = 9$.
So $n = 6$.
</s>