Three positive integers $a$, $b$, and $c$ satisfy $abc=8!$ where $a<b<c$. What is the smallest possible value of $c-a$?
I know that $40,320=8!=8*7*6*5*4*3*2*1=7*5*3^2*2^7*1$.
I'm trying to see how I can choose $a$, $b$, and $c$ such that $abc=8!$ and $a<b<c$. I don't see it yet though.