@Ankit: So how does that work with, e.g., N = n = 997? I.e., what is the calculation?
Dave On Dec 8, 11:33 am, ankit sablok <ankit4...@gmail.com> wrote: > @ all the authors thanx for the suggestions actually wt i know about > the problem is i think we can solve the problem mathematically if we > know about congruences > > for instance > if N=100 > 1! + 2! + ......... + 100! > and n=12 > > we find that > 4!mod24=0 > > hence the above equation reduces to the > (1!+2!+3!)mod 12 =9 > hence the answer is 9 > > so can anyone write a program for this logic > > On Dec 8, 6:19 pm, ankit sablok <ankit4...@gmail.com> wrote: > > > > > Q) can anyboy find me the solution to this problem > > > Given an integer N and an another integer n we have to write a program > > to find the remainder of the following problems > > (1! + 2! + 3! + 4! + ..................... + N!)mod(n) > > > N<=1000000 > > n<=1000; > > > please help me write a program for this problem > > thanx in advance- Hide quoted text - > > - Show quoted text - -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
