Averages in a sequence II
William McWorter, Jr.
Sat, 16 July 2005
Let x1, ..., x100 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 xi are equal.
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 xi are rational satisfying the average condition, then all xi are equal. This has been established previously.
Next, let x1, ..., x100 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 x1, ..., x100 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
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