Anthroov A.

What is the maximum number of white and black pawns that can be placed on a $9 \times 9$ checkerboard (a pawn, regardless of its color, can be placed on any cell of the board) so that no pawn attacks any other pawn (including those of the same color)? A white pawn attacks the two diagonally adjacent cells in the next higher row, while a black pawn attacks the two diagonally adjacent cells in the next lower row (see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_de7bb834a1e75b584fa6g-02.jpg?height=314&width=320&top_left_y=1292&top_left_x=869)
An example with 56 pawns is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_de7bb834a1e75b584fa6g-02.jpg?height=674&width=686&top_left_y=1844&top_left_x=686)

Evaluation. Suppose in a rectangle of three rows and two columns, there are at least five pawns. Then on three cells of one color, there are three pawns, and a pawn from the central row can capture one of the two remaining pawns. Therefore, this is impossible, and in any $3 \times 2$ rectangle, there are no more than 4 pawns.

Assume we managed to place 57 pawns. In the first 8 columns (which can be divided into 12 $3 \times 2$ rectangles), there are no more than 48 pawns, and in the ninth column, there are therefore 9 pawns. However, in the eighth column, there are no more than two pawns (otherwise, there would be a pawn not in the first or last row that could capture a pawn in the ninth column). Thus, in the eighth and ninth columns together, there are no more than 11 pawns, in the columns from the second to the seventh - no more than 36 pawns (which can be divided into 9 $3 \times 2$ rectangles), and in the first column - no more than nine. In total, no more than

$9 + 36 + 11 = 56$ pawns. Contradiction.

## Answer

56.

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