9.4. After the bankruptcy of the company "Horns and Hooves," 17 horns, 2 hooves, and one weight remained. All this wealth was divided equally by weight between Pankovskiy and Balaganov, with the weight entirely going to Balaganov. The horns and hooves were not cut into pieces. Each horn is heavier than each hoof and lighter than the weight by the same amount. How many horns and hooves did Pankovskiy get? Provide all possible options and prove that there are no others.
Solution: Let one hoof weigh $k$, and one horn weigh $k+\delta$ (all weights in the same units of measurement, for example, in puds). Then, according to the condition, the weight of the weight (the iron piece) is $k+2\delta$, and the total weight of the divided property is $20k+19\delta$. Each person received $10k+9.5\delta$. Balaganov took the weight, and with horns and hooves, he gathered $9k+7.5\delta$. This weight is less than the weight of 9 horns but more than the weight of 7 horns and 2 hooves. Therefore, Balaganov took fewer than 9 but more than 7 horns, which means 8.8 horns weigh $8k+8\delta$, and for hooves, Balaganov has $k-0.5\delta$ left, which is less than the weight of one hoof. This means all hooves and all other horns went to Panikovsky. Incidentally, we established that $k-0.5\delta=0$, meaning the horn is 3 times heavier than the hoof, and the weight is 5 times heavier than the hoof.

Answer: 9 horns and 2 hooves.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and justified answer with the derived relationship between the weights of the horn, hoof, and weight | 7 points |
| Proven that all hooves went to Panikovsky | 2 points |
| Proven that Balaganov could only have 9 horns | 2 points |
| The problem is solved for specifically chosen weights of horns, hooves, and the weight | 0 points |
| Correct answer without justification | 0 points |