[ Tournaments and tournament tables ] Counting in two ways
[ Examples and counterexamples. Constructions ]

In the competition, 10 figure skaters participate. The competition is judged by three judges in the following way: each judge independently distributes places (from first to tenth) among the skaters, after which the winner is the skater with the smallest sum of places. What is the maximum value that this sum can take for the winner (the winner is unique)?
Evaluation. Since each of the three judges distributed a set of places from first to tenth, the sum of the places awarded by all judges to all participants in the competition is $3 \cdot (1 + 2 + \ldots + 10) = 165$.

On the other hand, if the winner received a sum of no less than 16, then all the others received a sum of no less than 17, and the considered total sum is no less than $16 + 9 \cdot 17 = 169$. Contradiction. Therefore, the sum of the places of the winner is no more than 15.

Example. See the table:

| Judges | Places |  |  |  |  |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| I | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| II | 10 | 8 | 9 | 3 | 5 | 1 | 2 | 7 | 4 | 6 |
| III | 6 | 7 | 4 | 9 | 5 | 10 | 8 | 2 | 3 | 1 |
| Sum of places | 17 | 17 | 16 | 16 | 15 | 17 | 17 | 17 | 16 | 17 |

## Answer

15. 

![](https://cdn.mathpix.com/cropped/2024_05_06_57b0dc32154003794737g-01.jpg?height=222&width=2041&top_left_y=2513&top_left_x=0)

On each cell of a chessboard, there is a checker, white on one side and black on the other. In one move, you can choose any checker and flip all the checkers standing on the same vertical and all the checkers standing on the same horizontal.
a) Come up with a way to flip exactly one checker on a $6 \times 6$ board, arbitrarily filled with checkers.

b) Can you achieve that all checkers on a $5 \times 6$ board become white if exactly half of the checkers were initially black?

## Solution

a) Perform the described operation with the given checker and with all checkers in the same row and column. Then the given checker will change color 11 times and, therefore, will be flipped, while all other checkers on the board will be flipped an even number of times and, thus, will not change color.

b) Note that with each execution of the described operation, the color of an even number of checkers changes. For further details, see problem 35111.

## Answer

b) No.