### 20th Putnam 1959

Problem B7

Given any finite ordered tuple of real numbers X, define a real number [X], so that for all xi, 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)]. ```
Show that [(x1, x2, ... , xn)] = (x1 + x2 + ... + xn)/n.

Solution

It is convenient to write [(x1, ... , xn)] simply as [x1, ... , xn]. 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 + ... + xn+1 = 0, then the n numbers x1, x2, ... , xn and the n numbers 0, 0, ... , 0, -xn+1 have the same sum, so by (4) we have for any z that [x1, ... , xn, z] = [0, 0, ... , 0, -xn+1, z]. Take z = xn+1, then [0, ... 0, -xn+1, xn+1] = 0, so [x1, ... , xn, xn+1] = 0.

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