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    MHF Contributor Bruno J.'s Avatar
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    Show that all of these numbers are zero

    Suppose you have 2n numbers (n > 1) which have the following property : whenever you remove one of them (any one), you can split the remaining ones into two sets having equal sum.

    Show that all of these numbers are zero.
    Last edited by Bruno J.; December 17th 2009 at 09:32 PM.
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    Quote Originally Posted by Bruno J. View Post
    Suppose you have 2n-1 numbers (n > 1) which have the following property : whenever you remove one of them (any one), you can split the remaining ones into two sets having equal sum.

    Show that all of these numbers are zero.
    by "zero" did you mean "equal"? also, each of those two sets with equal sum must have exactly n-1 elements. otherwise, the claim would be false.
    Last edited by NonCommAlg; December 17th 2009 at 09:04 PM.
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    MHF Contributor Bruno J.'s Avatar
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    Oh, yeah. I'm sorry! I made a mistake in the statement of the problem! I have fixed my post.
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    Quote Originally Posted by Bruno J. View Post
    Suppose you have 2n numbers (n > 1) which have the following property : whenever you remove one of them (any one), you can split the remaining ones into two sets having equal sum.

    Show that all of these numbers are zero.
    a weaker property: whenever you remove one of them (any one), either the sum of the remaining ones is zero or you can split the remaining ones into two sets having equal sum.

    this problem is equivalen to this claim that the 2n \times 2n matrix A=[a_{ij}] with a_{ii}=0, \ a_{ij}=\pm 1, \ \forall i \neq j, is invertible. to prove this claim we'll show that \det A \neq 0:

    \det A = \sum_{\sigma \in S_{2n}} \text{sign}(\sigma) \prod_{i=1}^{2n} a_{i \sigma(i)}=\sum_{\sigma \in D} \text{sign}(\sigma) \prod_{i=1}^{2n} a_{i \sigma(i)}, where D = \{\sigma \in S_{2n}: \ \ \sigma(i) \neq i, \ \forall i \}, because we're given that a_{ii}=0. so D is the set of derangements of \{1,2, \cdots , 2n \}.

    but we know that the number of derangements of a set with even number of elements is odd. so \det A is a sum of odd number of terms where each term is \pm 1. clearly this sum can

    never be zero and hence A is invertible. Q.E.D.
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