## 10. Triomino

A triomino tile of shape $\square$ is placed on a board of dimensions $5 \times 5$ such that each covers three cells, and they do not overlap with each other. One cell remains uncovered. What is the sum of the numbers on all the cells of the board that can remain uncovered?

## Result: $\quad 234$
Solution. Eight triomino tiles of the given shape cover 24 squares, leaving exactly one uncovered. We will show that only the squares marked in the image can remain uncovered.

![](https://cdn.mathpix.com/cropped/2024_05_30_e6d161e17f72e23e8b9eg-6.jpg?height=337&width=372&top_left_y=2413&top_left_x=842)

A single triomino tile cannot cover two of them, so 8 triomino tiles cannot cover all 9 squares. This means that one of these squares will always remain uncovered.

This also means that in any arrangement of eight triominoes on the board, all other squares will be covered.

We still need to check if each of the nine marked squares can remain uncovered.
![](https://cdn.mathpix.com/cropped/2024_05_30_e6d161e17f72e23e8b9eg-7.jpg?height=344&width=1216&top_left_y=614&top_left_x=387)

The tiles can also be rotated, so these examples show that it is possible to arrange the triominoes such that the uncovered square is the center square (number 26), any of the four middle squares along the edge of the board (number 37), or any of the four corner squares (number 15).

Therefore, any of the 9 marked squares in the first image can remain uncovered (and only these squares).

The sum of the numbers on them is \(26 + 4 \cdot 37 + 4 \cdot 15 = 234\).