9.5. On a checkerboard of size $8 \times 8$, 8 checkerboard ships of size $1 \times 3$ are placed such that no two cells occupied by different ships share any points. One shot is allowed to pierce all 8 cells of one row or one column. What is the minimum number of shots needed to guarantee hitting at least one ship?
Answer: 2 shots.

Solution. We will make 2 shots as shown in Fig. 5. Suppose we did not hit any ship. Then there are no ships in area 1. In each of areas 2 and 3, there is no more than 1 ship. Therefore, there are at least 6 ships in area 4.

Area 4 is a $5 \times 5$ square. Then in this area, horizontally placed ships cannot lie in adjacent rows, and vertically placed ships cannot lie in adjacent columns. Therefore, in this area, there are 3 "vertical" and 3 "horizontal" ships, and one of them lies

![](https://cdn.mathpix.com/cropped/2024_05_06_754265878a272b0ed3b4g-2.jpg?height=426&width=417&top_left_y=1055&top_left_x=1439)

Fig. 5 in the central row, and another in the central column of the area. But then both of these ships contain the central cell of the area. Contradiction. Therefore, at least one ship is hit.

We will show that if only one shot is made, it is possible not to hit any ship. Suppose the shot was made in some row. Note that in one row, two ships can be placed. Then if the shot was made in a row with an odd number, two ships could stand in rows with numbers $2, 4, 6$, and 8. If the shot was made in a row with an even number, two ships could stand in rows with numbers $1, 3, 5$, and 7.

Remark. It can be shown that even 5 ships cannot be placed in area 4.

Comment. Only shown how to wound a ship with 2 shots - 4 points.

Only proven that 1 shot may not be enough - 2 points.

A correct example of two shots is given, but it is not explained why at least one ship will be hit - 1 point (instead of 4 points for the first criterion).