Pemer V.B.

It is known that the difference between the largest and smallest of the numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{9}, x_{10}$ is 1. What is the a) largest; b) smallest possible difference between the largest and smallest of the 10 numbers $x_{1}, 1 / 2\left(x_{1}+\right.$ $\left.x_{2}\right), 1 / 3\left(x_{1}+x_{2}+x_{3}\right), \ldots, 1 / 10\left(x_{1}+x_{2}+\ldots+x_{10}\right)$?

c) What would the answer be if there are not 10, but $n$ numbers?
Let $y_{k}$ denote the number $1 / k\left(x_{1}+x_{2}+\ldots+x_{k}\right)$. If we add a certain number $a$ to all $x_{i}$, then instead of the numbers $y_{i}$, we will get the numbers $y_{i}+a$. The maximum differences for the numbers $y_{i}$ and for the numbers $y_{i}+a$ are the same. Therefore, from the set $\left\{X_{i}\right\}$, using a suitable choice of $a$, we can transition to a set $\left\{X_{i}\right\}$ where the smallest $X_{i}$ are zero, and the largest are one. In the future, we will only consider such sets. Similarly, if we replace the numbers $x_{i}$ with $1-x_{i}$, then $y_{i}$ will be replaced with $1-y_{i}$. Therefore, from the set $\left\{x_{i}\right\}$, we can transition to the set $\{1 - x_{i}\}$: the maximum differences between the numbers $y_{i}$ and the numbers $1-y_{i}$ are the same.

Let $y_{k}$ be the smallest, and $y_{m}$ be the largest of the numbers $\left\{y_{i}\right\}$. If $k < m$, then $\Delta_{1}=y_{k}-y_{k-1}=\frac{1}{k-1}\left(1-y_{k}\right), \Delta_{2}=y_{m-1}-y_{m}=1 /{ }^{1} \cdot y_{m-1}$, and if $y_{k} \leq y_{m-1}$, then $\Delta_{2} \geq 1 / m \cdot y_{k}$. The largest of the numbers $\Delta_{1}, \Delta_{2}$ is not less than $1 / k+m-1$, since $1 \leq(k-1) \Delta_{1}+m \Delta_{2} \leq(k+m-1) M$, where $M=\max \left\{\Delta_{1}\right.$, $\left.\Delta_{2}\right\}$. If $y_{k}>y_{m-1}$, then consider the difference again

$\Delta_{3}=y_{k}-y_{m}=y_{k}-y_{m-1}+1 / m \cdot y_{m-1} . m \Delta_{3}+(k-1) \Delta_{1}=(m-1)\left(y_{k}-y_{m-1}\right)+1>1$, from which $\max \left\{\Delta_{1}, \Delta_{3}\right\} \geq 1 / k+m-1$. Thus, we have proved that the difference is not less than $1 / k+m-1$. Therefore, for an arbitrary set, it is not less than $1 / 2(n-1)$ ( $m \leq n, k \leq n-1$). Such a difference can be realized: for example, consider the set $x_{1}=x_{2}=\ldots=x_{n-2}=1 / 2, x_{n-1}=1, x_{n}=0$. Thus, the smallest difference is $1 / 2(n-1)$.

## Answer

a) 0, 0.9; b) $\frac{1}{18}$; c) ${ }^{n-1 / n}, \frac{1}{2(n-1)}$.