@ 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 -- 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.
