Results 1 to 4 of 4

Thread: An easy problem!

  1. #1
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1

    An easy problem!

    Let $\displaystyle M(n)=\{-1,\dots, -n\}$. Define the empty product to be $\displaystyle 1$. For every subset of $\displaystyle M(n)$, multiply its elements together and add up the resulting $\displaystyle 2^n$ numbers; what is the result?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Oct 2009
    Posts
    68
    Spoiler:
    Given a subset $\displaystyle S$ of $\displaystyle M(n),$ define its twin subset $\displaystyle S^*$ to be $\displaystyle S^*=S\cup\{-1\}$ if $\displaystyle -1\notin S$ and $\displaystyle S^*=S\setminus\{-1\}$ if $\displaystyle -1\in S.$ Then it is clear that the twin subset of $\displaystyle S^*$ is $\displaystyle S$ itself, that all the $\displaystyle 2^n$ subsets of $\displaystyle M(n)$ (provided $\displaystyle n\ne0)$ can be partitioned into $\displaystyle 2^{n-1}$ pairs of mutually twin subsets, and that if the product of all the elements of $\displaystyle S$ is $\displaystyle k$ then the product of all the elements of $\displaystyle S^*$ is $\displaystyle -k.$ Hence the answer to the problem is $\displaystyle 0.$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Good!

    Here's my solution :

    Spoiler:

    Expand $\displaystyle 0=(1-1)(1-2)(1-3)\dots(1-n)$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    From
    MARYLAND
    Posts
    12
    Thanks Bruno.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Feb 9th 2013, 08:59 PM
  2. Easy Log Problem. Help
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 2nd 2010, 11:45 AM
  3. An easy Problem(or is it?)
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Oct 15th 2009, 04:38 AM
  4. Fun easy problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 29th 2008, 10:59 AM
  5. Should be a easy problem...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 16th 2008, 05:34 PM

Search Tags


/mathhelpforum @mathhelpforum