@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<=1000000
>> n<=1000;
>>
>> please help me write a program for this problem
>> thanx in advance
>>
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-- 
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee

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