20th Putnam 1959

Given any finite ordered tuple of real numbers X, define a real number [X], so that for all x_i, α:

(1)  [X] is unchanged if we permute the order of the numbers in the tuple X;

(2)  [(x1 + α, x2 + α, ... , xn + α)] = [(x1, x2, ... , xn)] + α; 

(3)  [(-x1, -x2, ... , -xn)] = - [(x1, x2, ... , xn)]; 

(4)  for y1 = y2 = ... = yn = [(x1, x2, ... , xn)], we have [(y1, y2, ... , yn, xn+1)] = [(x1, x2, ... , xn+1)]. 

Solution

It is convenient to write [(x₁, ... , x_n)] simply as [x₁, ... , x_n]. We use induction on n. For n = 1, we have [0] = - [0], so [0] = 0. Then (3) gives that [x] = x, which establishes the result for n = 1.

[x, 0] = - [-x, 0] = - [0, -x] = - [x, 0] + x, so [x, 0] = x/2. Hence [x, y] = y + [x - y, 0] = y + (x - y)/2 = (x + y)/2, which establishes the result for n = 2. Now suppose it is true for n.

We have that [x, -x, 0, 0, ... ,0] = - [-x, x, 0, 0, ... , 0] by (3), = - [x, -x, 0, 0, ... ,0] by (1), so [x, -x, 0, 0, ... ,0] = 0. Now suppose x₁ + x₂ + ... + x_n+1 = 0, then the n numbers x₁, x₂, ... , x_n and the n numbers 0, 0, ... , 0, -x_n+1 have the same sum, so by (4) we have for any z that [x₁, ... , x_n, z] = [0, 0, ... , 0, -x_n+1, z]. Take z = x_n+1, then [0, ... 0, -x_n+1, x_n+1] = 0, so [x₁, ... , x_n, x_n+1] = 0.

Finally, given arbitrary x₁, ... , x_n+1, let k be their mean, then x₁ - k, x₂ - k, ... , x_n+1 - k have zero sum, so [x₁ - k, x₂ - k, ... , x_n+1 - k] = 0. Hence [x₁, ... , x_n+1] = k, as required.

20th Putnam 1959

© John Scholes
jscholes@kalva.demon.co.uk
15 Feb 2002