Given that

$\frac{((3!)!)!}{3!} = k \cdot n!,$
where $k$ and $n$ are [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) and $n$ is as large as possible, find $k + n.$
Note that\[{{\left((3!)!\right)!}\over{3!}}= {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.\]Because $120\cdot719!<720!$, we can conclude that $n < 720$. Thus, the maximum value of $n$ is $719$. The requested value of $k+n$ is therefore $120+719=\boxed{839}$.
~yofro