Averages in a sequence III
William McWorter, Jr.
Mon, 23 January 2006
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.
First assume that the xi are rational numbers. Multiplying each term by an appropriate fixed integer, we get a sequence of 100 integers still satisfying the average condition. This integer sequence is constant exactly when the rational sequence it came from is constant. So assume the xi are integers.
Let S be the sum of any 8 of the xi. Then there is a subsequence of 9 of the xi, with sum T with the same average. Hence
Now let xi and xj be any two terms of the 100 terms and let M be the sum of any 7 terms other than xi and xj. Then
Suppose, by way of contradiction, that not all of the xi are equal and let
Thus, when all the xi's are integers, the xi's must be equal. Therefore, when all the xi's are rational, the xi's must be equal.
But now we can claim that all the xi's are equal EVEN IF THE xi's ARE REAL! For, the average condition defines a rational homogeneous linear system of equations which the xi must satisfy. Our argument above shows that this linear system has rank 99 as a linear system of equations with rational coefficients. Hence the system has rank 99, even allowing real xi's!
Evolution of a Solution
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