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Math Help - function

  1. #1
    Day
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    function

    n\in<1,1004>
    Find how many n exist when k = 7

    2n! \over n!n!2^k

    For :

    2n! \over n!n!2^{k+1}
    Result isn't integer value.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Day View Post
    n\in<1,1004>
    Find how many n exist when k = 7

    2n! \over n!n!2^k

    For :

    2n! \over n!n!2^{k+1}
    Result isn't integer value.
    I not clear what your question is. Do you mean that n must be an integer such that 1\le n\le 1004 such that \frac{2n!}{n!n!2^7} is an integer but \frac{2n!}{n!n!2^{8} is not? And is that really "2n!" and not "(2n)!"?
    If so it is asking how many "n"s are such that [tex]\frac{2n!}{n!n!} has 7 factors of 2 but not 8.
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  3. #3
    Day
    Day is offline
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    n is integer.
    how many results have 2n! \over n!n!2^k. How many is n in this function? This is question.

    All is up.

    <br /> <br />
2n! \over n!n!2^k = integer


    2n! \over n!n!2^{k+1} isn't integer
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