Authors: [Ribamko A.V., Zhenedarov R.G., Tokarev S.i.

On each cell of a $5 \times 5$ board, there is one coin, all coins look the same. Among them, exactly 2 coins are counterfeit, they weigh the same and are lighter than the genuine ones, which also weigh the same. The counterfeit coins lie in cells that share exactly one common vertex. Can one guarantee to find a) 13 genuine coins; b) 15 genuine coins; c) 17 genuine coins in one weighing on a balance scale without weights?
We will color the board and the coins on it in a checkerboard pattern (corner cells are black). Note that both counterfeit coins are of the same color.

a) Method 1. Place 16 coins on the scale, which lie on the edge of the board: 8 black on one pan, 8 white on the other. There are three possible cases.

1) The scale is in balance. Since the counterfeit coins could only be on one pan, they are not among the weighed coins, meaning all 16 weighed coins are genuine.
2) The "black" pan outweighs. Then one counterfeit coin is on the "white" pan. Therefore, all 13 black coins are genuine.
3) The "white" pan outweighs. Then one counterfeit coin is on the "black" pan. Therefore, the central coin is genuine. And all 12 white coins are also genuine.

Generalization. Set aside any black coin A and all its diagonal neighbors. From the remaining black coins, place at least 7 on one pan and the same number of white coins on the other. If the scale is in balance, all the coins on the scale are genuine (at least 14). If the black coins outweigh, then all the black coins are genuine, and if the white coins outweigh, then all the white coins and A are genuine.

Method 2. Place 4 black coins, adjacent to the central one, on the pans: the left ones on the left pan, the right ones on the right pan. If the scale is in balance, all black coins are genuine (since two counterfeit black coins cannot be on different pans or both outside the pans). If one pan outweighs, then the two coins on that pan and all white coins are genuine.

b) Method 1. Weigh the black coins against the white ones on the left diagram.
![](https://cdn.mathpix.com/cropped/2024_05_06_caf98da37e6d9925b347g-39.jpg?height=430&width=1358&top_left_y=2441&top_left_x=36)

In the case of balance, 15 coins in the lower three rows are genuine. If the black coins are heavier, the unweighed counterfeit coin can only be in the squares marked with crosses in the central diagram. Therefore, we have found $25-5-5=15$ genuine coins. If the white coins are heavier, the situation is shown in the right diagram with the same result.

Method 2. Weigh the black coins against the white ones on the left diagram.
![](https://cdn.mathpix.com/cropped/2024_05_06_caf98da37e6d9925b347g-40.jpg?height=414&width=1332&top_left_y=273&top_left_x=40)

In the case of balance, all 16 weighed coins are genuine. If the black coins are heavier, the unweighed counterfeit coin can only be in two squares marked with crosses in the central diagram. Thus, we have found $25-8-2=15$ genuine coins. If the white coins are heavier, we get the same result (right diagram).

Method 3. From the fifth row, place two white coins on the left pan, two black coins on the right pan, and set aside one coin. Place the remaining white coins on the right pan and the black coins on the left pan. If the scale is in balance, all coins from the first three rows are genuine. If one pan outweighs, then all 12 coins on that pan and all coins from the fifth row are genuine (a total of 15 coins).

c) A priori, any of the 25 coins is suspicious (could be counterfeit). Weighing can have 3 outcomes, so at least one of them will leave at least 9 suspicious coins, meaning no more than $25-9=16$ genuine coins will be found.

## Answer

a) possible; b) possible; c) not possible.