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Math Help - Remainder of Summation and Factorials

  1. #1
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    Remainder of Summation and Factorials

    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.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    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.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    Note that 13!, 14!, ..., 100! all leave a remainder of 0 when divided by 13. That should simplify your calculations a lot.
    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?
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