Yes, you can find the remainder of each, add them up, and reduce again mod 13 to get the remainder of the sum.
Note that 13!, 14!, ..., 100! all leave a remainder of 0 when divided by 13. That should simplify your calculations a lot.
My problem is
What is the remainder when 1!+2!+...+100! is divided by 13?
How do you simplify a factorial to find its remainder?
Also, is the remainder of the summation of factorials the same as the remainder of the sum of the remainders of the factorials?
Thanks.
So any factorial greater than or equal to the divisor is 0, is this right?
This leads into another question that I have
find the remainder when 40!/((2^20) * 20!) is divided by 8.
Can I find the remainder of each separately, or do I have to simply the factorials first?
I'm guessing that I have to simplify the factorial first, because 40!/20! does not contain 8. So, 40!/20! = 40*39*38*...*22*21. Now since 40!/20! is a multiple of 32, and 8 divides 32, then 8 must divide 40!/20! with remainder 0.
But here's where I'm unsure. I think that 2^4 == 0 (mod 8) implies 2^20 == 0 (mod 8). But now I'm left with 0/0. Does this mean the remainder is 0, indeterminate, or did I make a mistake?