## Solution.

One rook covers its own square and attacks 7 additional squares in its row and column.

The second rook covers its own square and attacks 6 additional squares in its row and column (not counting those already attacked by the first rook).

Continuing this way, the sixth rook covers its own square and 2 additional squares in its row and column. In total, there are $15+13+11+9+7+5=60$ covered or attacked squares.

Thus, if we place six rooks, only 4 free squares remain, and we cannot place six kings on them.

In the figure, one valid arrangement of five kings (軑) and five rooks (單) is shown, so the answer is 10.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-12.jpg?height=568&width=580&top_left_y=1578&top_left_x=664)

Note: In the proof that it is not possible to place six kings and six rooks, it is sufficient to comment that after placing six rooks, 6 rows and 6 columns are blocked; therefore, only 4 free squares remain.

## COUNTY COMPETITION IN MATHEMATICS

3rd grade - high school - A variant

## February 28, 2019.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.
## Task A-2.5.

On a board of dimensions $8 \times 8$, kings and rooks are placed such that no figure is under attack. A king attacks the adjacent fields (eight of them, unless the king is on the edge of the board), and a rook attacks all the fields in the row and column it is in. What is the maximum number of figures on the board if the number of rooks is equal to the number of kings?