@Ashim
with a check that N =n
N can also be less than n
On Wed, Dec 8, 2010 at 6:57 PM, Ashim Kapoor ashimkap...@gmail.com wrote:
Let me try. Any thing involving n would leave no remainder.
so (1 + 2 ! + ... + n ! + + N !) mod n = (1 + 2 ! + ... + (n-1)! )
mod n
This should be computed from a loop. I don't know how to reduce it further.
Ashim.
On Wed, Dec 8, 2010 at 6:49 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=100
n=1000;
please help me write a program for this problem
thanx in advance
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