Given 10 numbers - one one and nine zeros. It is allowed to choose two numbers and replace each of them with their arithmetic mean.

What is the smallest number that can end up in the place of the one?

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We will prove by induction on $k$ the following statement: if at some moment there are exactly $k$ non-zero numbers, then each of these numbers is not less than $2^{1-k}$.

Base. For $k=1$, this is true, since after the first non-trivial operation, at least two non-zero numbers will appear, and the number of non-zero numbers will not decrease thereafter.

Inductive Step. Suppose at some moment there are $k+1$ non-zero numbers. Consider the sequence of operations that resulted in these numbers. In this sequence of operations, find the last moment when there were exactly $k$ non-zero numbers. By the inductive hypothesis, each of these numbers is not less than $2^{1-k}$. In the next operation, one zero and one non-zero number were replaced by their average. After this, each of these numbers became not less than $2^{-k}$. In subsequent operations involving only non-zero numbers, the minimum of non-zero numbers does not decrease.

Since we have a total of 10 numbers, each of them is not less than $2^{-9}=1 / 512$.

On the other hand, it is not difficult to provide an example of operations that result in $1 / 512$ from a unit. Replace the unit and one of the zeros with $1 / 2$ and $1 / 2$, then replace $1 / 2$ and another zero with $1 / 4$ and $1 / 4$, and so on; in the end, we get $1 / 512$ from the unit.

## Answer

$1 / 512$.