Problem:
Given that 3!((3!)!)!=k⋅n!, where k and n are positive integers and n is as large as possible, find k+n.
Solution:
Note that
3!((3!)!)!=6(6!)!=6720!=6720⋅719!=120⋅719!
Because 120⋅719!<720!, conclude that n must be less than 720, so the maximum value of n is 719. The requested value of k+n is therefore 120+719=839.
The problems on this page are the property of the MAA's American Mathematics Competitions