Results 1 to 3 of 3

Thread: function

  1. #1
    Day
    Day is offline
    Newbie
    Joined
    Mar 2009
    Posts
    2

    function

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

    $\displaystyle 2n! \over n!n!2^k$

    For :

    $\displaystyle 2n! \over n!n!2^{k+1}$
    Result isn't integer value.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,714
    Thanks
    3002
    Quote Originally Posted by Day View Post
    $\displaystyle n\in<1,1004>$
    Find how many n exist when k = 7

    $\displaystyle 2n! \over n!n!2^k$

    For :

    $\displaystyle 2n! \over n!n!2^{k+1}$
    Result isn't integer value.
    I not clear what your question is. Do you mean that n must be an integer such that $\displaystyle 1\le n\le 1004$ such that $\displaystyle \frac{2n!}{n!n!2^7}$ is an integer but $\displaystyle \frac{2n!}{n!n!2^{8}$ 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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Day
    Day is offline
    Newbie
    Joined
    Mar 2009
    Posts
    2
    n is integer.
    how many results have $\displaystyle 2n! \over n!n!2^k$. How many is n in this function? This is question.

    All is up.

    $\displaystyle

    2n! \over n!n!2^k$ = integer


    $\displaystyle 2n! \over n!n!2^{k+1}$ isn't integer
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 20
    Last Post: Nov 27th 2012, 05:28 AM
  2. Replies: 0
    Last Post: Oct 19th 2011, 04:49 AM
  3. Replies: 4
    Last Post: Oct 27th 2010, 05:41 AM
  4. Replies: 3
    Last Post: Sep 14th 2010, 02:46 PM
  5. Replies: 2
    Last Post: Sep 2nd 2010, 10:28 AM

Search Tags


/mathhelpforum @mathhelpforum