## Averages in a sequence II

### William McWorter, Jr.

Sat, 16 July 2005

Let x_{1}, ..., x_{100} be a sequence of real numbers. Suppose that for every subsequence of 8 terms, there exists a subsequence of 9 terms with the same average as that of the 8. Show that all x_{i} are equal.

### Solution

The clever trick of regarding a finite set of real numbers as vectors over the rationals is not needed to solve this problem.

First show that, if the x_{i} are rational satisfying the average condition, then all x_{i} are equal. This has been established previously.

Next, let x_{1}, ..., x_{100} be a sequence of real numbers satisfying the average condition. Then each subsequence of 8 terms has the same average as some subsequence of 9 terms. Thus each subsequence determines a linear homogeneous equation in 100 unknowns with rational coefficients (namely, 83 0's, 8 (1/8)'s, and 9 (-1/9)'s). Let A be the matrix of this system of homogeneous linear equations. Then A has (100 choose 8) rows, 100 columns, and all entries are rational.

The first step in the proof shows that, over the rationals, A has null space of dimension 1 (spanned by the column vector of 1's). Since A has rational entries, the dimension of the null space of A over the reals is also 1. Hence the real numbers x_{1}, ..., x_{100} are all equal!

I never thought of this until I looked at the following problem: There are 11 real numbers with the property that any 10 of them can be split into two 5-subsets with the same sum. Show that all eleven numbers must be equal.

### Evolution of a Solution

- Tue, 25 Mar 2003
- Sat, 16 July 2005
- Mon, 23 January 2006

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