Problem 7. (11 points) On a 10 x 10 board, some cells are occupied by a single beetle. Once a minute, the beetles jump simultaneously, with each one jumping to an adjacent (by side) cell. Each beetle jumps in one of the four possible directions, maintaining its direction as long as possible, and then jumps in the opposite direction. Dog Sharik observed the beetles for an entire hour and never saw two beetles in the same cell. What is the maximum number of beetles that can jump on the board?

## Answer. 40.
Solution. Evaluation. On one vertical, there can be no more than two vertically jumping fleas (otherwise, the fleas located on the same-colored fields in a chessboard coloring would meet). The same is true for the horizontals. Therefore, on 20 horizontals and verticals, there can be no more than 40 fleas.

Example. It is clear that fleas from different-colored fields cannot meet. Therefore, it is sufficient to indicate only 20 "white" fleas (the arrangement of the "black" fleas is symmetrical). This is shown in the drawing on the left. The arrow indicates the initial position of the flea and the direction of its movement.

In the middle drawing, one "vertical" flea B and all the prohibited positions of the horizontal fleas (i.e., those that can end up in the same field as B) are indicated. As we see, the prohibited arrows lie only in the fields of the diagonals passing through B and in the diagonal rectangle defined by the smaller diagonal. The situation is essentially the same for the other vertical fleas from the first drawing; all the prohibited fields from them are shown in the drawing on the right. We see that the horizontal fleas from the first drawing are in one-sided prohibited fields and are in the opposite direction of the prohibited one.
![](https://cdn.mathpix.com/cropped/2024_06_03_376d2fa4074cfcd15764g-3.jpg?height=526&width=1564&top_left_y=1483&top_left_x=262)