+680 Given that \(\frac{((3!)!)!}{3!}\)\(=\text k \times \text n!\) where \(k\) and \(n\) are positive integers and \(n\) is as large as possible, find \(k+n\).
+738 ok so
if we simplify the numerator of the fraction, we get \(((3!)!)!=(6!)!=720!\)
so our new fraction is \(\dfrac{720!}{3!}\) which is equal to \(\dfrac{720 \cdot 719!}{6}\)
we can then cancel out the 720 and 6, and we get \(120\cdot719!\)
so, \(719+120=\boxed{839}\)
