## Averages in a sequence III

### William McWorter, Jr.

Mon, 23 January 2006

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

First assume that the x_{i} 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 x_{i} are integers.

Let S be the sum of any 8 of the x_{i}. Then there is a subsequence of 9 of the x_{i}, with sum T with the same average. Hence _{i} has a sum divisible by 8.

Now let x_{i} and x_{j} be any two terms of the 100 terms and let M be the sum of any 7 terms other than x_{i} and x_{j}. Then _{i}_{j}_{i} and x_{j} have the same remainder upon division by 8. Thus all 100 terms of the sequence have the same remainder modulo 8.

Suppose, by way of contradiction, that not all of the x_{i} are equal and let _{i} among all integer sequences satisfying the average condition. Assume the x_{i} form a sequence with smallest difference between terms equal to

Thus, when all the x_{i}'s are integers, the x_{i}'s must be equal. Therefore, when all the x_{i}'s are rational, the x_{i}'s must be equal.

But now we can claim that all the x_{i}'s are equal EVEN IF THE x_{i}'s ARE REAL! For, the average condition defines a rational homogeneous linear system of equations which the x_{i} 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 x_{i}'s!

### Evolution of a Solution

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

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