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 2007-04-02, 17:23 #1 davar55     May 2004 New York City 102138 Posts 2n+1 Integers Let a1,a2,...,a2n+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 a1 = a2 = ... = a2n+1.
 2007-04-03, 03:01 #2 axn     Jun 2003 14F516 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 right-shifted 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.
 2007-04-03, 07:45 #3 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts I don't see that (2) is true unless the two groups each contain n numbers. Last fiddled with by davieddy on 2007-04-03 at 07:54
2007-04-03, 08:39   #4
axn

Jun 2003

5·29·37 Posts

Quote:
 Originally Posted by davieddy I don't see that (2) is true unless the two groups each contain n numbers.
Quote:
 Originally Posted by davar55 the remaining ones can be separated into two collections of n integers with equal sums.

Without it, 1,1,3,3,5 is a trivial counter-example:

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 2007-04-03 at 08:49

 2007-04-03, 08:46 #5 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts THX. At least I have demonstrated that this stipulation was necessary.
2007-04-03, 10:53   #6
S485122

"Jacob"
Sep 2006
Brussels, Belgium

111000110002 Posts

The two groups had to be of n integers :
Quote:
 Originally Posted by davar55 Let a1,a2,...,a2n+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 a1 = a2 = ... = a2n+1.

2007-04-03, 12:18   #7
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

25×72 Posts

Quote:
 Originally Posted by davar55 Let a1,a2,...,a2n+1 be a collection of integers such that...
It is also true for (2*n+1) real numbers.

2007-06-05, 15:58   #8
m_f_h

Feb 2007

1101100002 Posts

Quote:
 Originally Posted by R. Gerbicz It is also true for (2*n+1) real numbers.
Proof: its true for a collection of rationals, which can be converted to integers by multiplying by the common denominator.

The extension to reals follows by.... "continuity" ?
Another (simpler(?)) proof ?

2007-06-06, 13:24   #9
victor

Oct 2005
Fribourg, Switzerlan

22×32×7 Posts

Quote:
 Originally Posted by m_f_h Another (simpler(?)) proof ?
I agree. A different and simpler proof would be welcome

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