Consider $M . \mathbf{B}$.

On some cells of a $10 \times 10$ board, there is a flea. Once a minute, the fleas jump simultaneously, and each one jumps to an adjacent cell (along a side). A flea jumps strictly in one of the four directions parallel to the sides of the board, maintains its direction as long as possible, and otherwise changes it to the opposite. Dog Barboss observed the fleas for an hour and never saw two of them sitting on the same cell. What is the maximum number of fleas that could have been jumping on the board?
Evaluation. On one vertical line, there can be no more than two fleas jumping along the vertical (otherwise, fleas in cells of the same color will meet). The same is true for the horizontals. In total, on 20 horizontals and verticals, there can be no more than 40 fleas.

Example. It is clear that fleas from cells of different colors will not meet. Therefore, it is sufficient to indicate only 20 "white" fleas (the placement of "black" fleas can be obtained, for example, by symmetry relative to the middle line).

On the left diagram, one "vertical" flea $B$ and all the forbidden positions of "horizontal" fleas (those from which a "horizontal" flea can end up on the same cell as $B$) are drawn. As we see, the forbidden cells form two rectangles built on the diagonals passing through $B$. On the right
|diagram, 10 "vertical" fleas and all the positions forbidden by them for "horizontal" fleas are placed. We see that in each horizontal line, there is at least one white cell that is not forbidden, where a "horizontal" flea can be placed.

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-05.jpg?height=658&width=1495&top_left_y=137&top_left_x=287)

## Answer

40 fleas.

Submit a comment