Nice Factorial Equation - Can You Solve It?

[@graf_paper]

That's an NICE math problem!

[@nicemathproblems]

Haha thanks graf_paper

[@shalvagang951]

My approach was looking for n!-1 units digit for n>4 which is always 9 and hence m>4 it must have at least one zero in end but k .9 leaves zero at 10 multiples but at every 10th multiple m ! Gains one zero and hence it must be one greater than I to accommodate it

[@varunbajoria6126]

I tried a different solution before looking at the answer k!+m!=k!n! n!=k!/k! +m!/k! n!= 1+ m!/k! according to m!>k! since n! as well as n must be a positive integer but if (m!/k!) is 1 less than n! , n! can only be 2right? (since no other factorial has a difference of just one from another factorial) no matter what other fraction u make (m!/k!) equal to it doesnt seem like theres a solution other than 1 so n!=2 n=2