Question 1021025
2n! divided by (n-3)! = 84n
<pre>
{{{(2n!)/(n-3)!}}}{{{""=""}}}{{{84n}}}

{{{2n!}}}{{{""=""}}}{{{84n(n-3)!}}}

{{{n!}}}{{{""=""}}}{{{42n(n-3)!}}}

{{{n(n-1)!}}}{{{""=""}}}{{{42n(n-3)!}}}

{{{(n-1)!}}}{{{""=""}}}{{{42(n-3)!}}}

{{{(n-1)!}}}{{{""=""}}}{{{7*6(n-3)!}}}

The left side is a factorial.
For the right side to be a factorial beginning
with 7*6, it must be 7! or 7*6*5*4*3*2*1.
Then (n-3)! must be 5*4*3*2*1 or 5!

Therefore

{{{(n-3)!}}}{{{""=""}}}{{{5!}}}

{{{n-3}}}{{{""=""}}}{{{5}}}

{{{n}}}{{{""=""}}}{{{8}}}

Checking:

{{{(2n!)/(n-3)!}}}{{{""=""}}}{{{84n}}}

{{{(2(8!))/(8-3)!}}}{{{""=""}}}{{{84(8)}}}

{{{2(40320)/(5)!}}}{{{""=""}}}{{{672}}}

{{{2(40320)/120}}}{{{""=""}}}{{{672}}}

{{{672}}}{{{""=""}}}{{{672}}}

So it checks:

Solution: {{{n}}}{{{""=""}}}{{{8}}}

Edwin</pre>