1. On an $8 \times 8$ chessboard, tokens are placed according to the following rule. Initially, the board is empty. A move consists of placing a token on any free square. With this move, exactly one of the tokens that ends up adjacent to it is removed from the board (if there is such an adjacent token). What is the maximum number of tokens that can be placed on the board, considering the specified rule? Adjacent squares are those that are nearest horizontally, vertically, and diagonally. Solution.

The number of tokens on the board can be increased by one only when a new token is placed on a completely free square, i.e., a square all of whose adjacent squares are free of tokens.
![](https://cdn.mathpix.com/cropped/2024_05_06_cca78509636089d39cb9g-1.jpg?height=748&width=1678&top_left_y=1022&top_left_x=201)

Let's highlight one horizontal row A1-H1, see the figure. We can fill this row with tokens from left to right. To do this, we will place the tokens in sequence, as shown in the figure. The numbers in circles indicate the sequence of placing the tokens. A crossed-out number means removing a token when placing an adjacent one. When filling the bottom row, each subsequent token removes the previous one from the board, so placing a token with an odd number in the row A1-H1 removes the previous even-numbered token from the row A2-H2. To place a token on the H1 field, we first need to place token number 14 on the H3 field, then token number 15 on the H2 field, removing the previous one. Finally, we can place token number 16 on the corner field H1, removing token number 15 from the board.

Next, we can fill all the horizontal rows up to A6-H6 in a similar manner. Rows A7-F7 and A8-F8 will have to be filled differently, as there are not enough free squares above them. The order of filling the two vertical squares A7 and A8 is shown in the second figure. To fill them, we need to start twice from the C8 field. This way, we will reach the F7 and F8 fields. We cannot move further to the right, as there is no free field except for the H8 field. We can place a token on this field and, if we wish to fill the G8 field, remove the token from the H8 field. Thus, we have shown how to fill the board with 61 tokens. There are other sequences of placements and other final arrangements of 61 tokens.

![](https://cdn.mathpix.com/cropped/2024_05_06_cca78509636089d39cb9g-2.jpg?height=951&width=1033&top_left_y=324&top_left_x=523)

It remains to prove that it is impossible to place 62 tokens. Suppose 61 tokens are placed on the board. The three free squares do not allow us to select one that does not have any adjacent occupied squares. The minimum number of sides of a free square that can adjoin another free square is two, in the corner of the board. Therefore, the minimum necessary number of free neighbors is three, taking into account the diagonally adjacent square, see the figure.
Answer: No more than 61 chips can be placed.