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Math Help - 1!+2!+3!...+99!+100!

  1. #1
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    1!+2!+3!...+99!+100!

    What is the remainder when 1!+2!+3!+...+99!+100! is divided by 18?

    Again, I have no idea what to do.
    At first I didnt know what ! meant so I simply did 101*50.
    Now I think its 101!*50, but thats just a guess. The only thing I could conclude was that 1 would be mentioned 100 times, 2 would be mentioed 99 times, etc.
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  2. #2
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    Quote Originally Posted by ceasar_19134 View Post
    What is the remainder when 1!+2!+3!+...+99!+100! is divided by 18?

    Again, I have no idea what to do.
    At first I didnt know what ! meant so I simply did 101*50.
    Now I think its 101!*50, but thats just a guess. The only thing I could conclude was that 1 would be mentioned 100 times, 2 would be mentioed 99 times, etc.
    Any number,
    18!,19!,20!,... leaves remainder zero for they contain 18.

    You need to look at the numbers,
    1!,2!,3!,....,17!

    18 has prime factorization 2*3*3

    From that list find all are even, next see which are divisble by 9.
    All the numbers from 6! and on.
    That leaves you with,
    1!,2!,3!,4!,5!

    Find the remainder of each one add them and find remainder of that.
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  3. #3
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    They're all a remainder beacuse theyre less than 18 so...
    When added together it equals 35. 35/18 leaves R17.
    Is that right?
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  4. #4
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    Hello, ceasar_19134!

    What is the remainder when 1! + 2! + 3!+ ... + 99! + 100! is divided by 18?

    We note that 6! = 720 is divisible by 18.
    Hence, any factorial greater than 6! is also divisible by 18.

    We need to test only: . 1! + 2! + 3! + 4! + 5! .= .153

    And: .153 18 .= .8, remainder 9

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