20070402, 17:23  #1 
May 2004
New York City
108B_{16} Posts 
2n+1 Integers
Let a_{1},a_{2},...,a_{2n+1} be a collection of integers such that if any one element in the collection is removed, the remaining ones can be separated into two collections of n integers with equal sums.
Prove a_{1} = a_{2} = ... = a_{2n+1}. 
20070403, 03:01  #2 
Jun 2003
2^{4}·7·47 Posts 
1. If one of the numbers is odd, then all the numbers are odd. Similarly, if one of them is even, then all of them are even. Proof: Suppose that there is atleast one odd and one even number in the collection. If you can make equal sums while leaving out the odd number, then you can't do the same while leaving out the even (change of parity). 2. If the given set of numbers, a[1]..a[2n+1] is a solution, then so is a[1]+c..a[2n+1]+c. (ie all numbers added/subtracted by a constant integer). Similarly a[1]*c..a[2n+1]*c (where c is a constant _real_ number) is also a solution, assuming that the results of multiplication/division are all integers. 3. Since LSB of all the numbers must be the same (by 1), we can say that (a[1]LSB)/2 .. (a[2n+1]LSB)/2 (ie all numbers rightshifted by 1) will also be a solution to the problem (by 2). 4. Repeat 1 & 3 until all the bits are shown to be equal, starting from LSB to MSB. 
20070403, 07:45  #3 
"Lucan"
Dec 2006
England
1100101001010_{2} Posts 
I don't see that (2) is true unless the two groups
each contain n numbers. Last fiddled with by davieddy on 20070403 at 07:54 
20070403, 08:39  #4  
Jun 2003
12220_{8} Posts 
Quote:
Quote:
Without it, 1,1,3,3,5 is a trivial counterexample: Leaving out 5: 3+1 = 3+1 Leaving out 3: 5 = 3+1+1 Leaving out 1: 5+1 = 3+3 Last fiddled with by axn on 20070403 at 08:49 

20070403, 08:46  #5 
"Lucan"
Dec 2006
England
6474_{10} Posts 
THX. At least I have demonstrated that this stipulation was necessary.

20070403, 10:53  #6 
"Jacob"
Sep 2006
Brussels, Belgium
2×887 Posts 
The two groups had to be of n integers :

20070403, 12:18  #7 
"Robert Gerbicz"
Oct 2005
Hungary
1530_{10} Posts 

20070605, 15:58  #8 
Feb 2007
2^{4}×3^{3} Posts 

20070606, 13:24  #9 
Oct 2005
Fribourg, Switzerlan
374_{8} Posts 

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