25. All fishermen are divided into ordinary and honest ones. An honest fisherman exaggerates the weight of the fish he caught exactly by 2 times, while an ordinary fisherman exaggerates it by an integer greater than six times (these coefficients can be different for different ordinary fishermen). 10 fishermen caught a total of 120 kg of fish together. Each claimed that they caught exactly 60 kg of fish. How many of them were ordinary fishermen? Find all answers and prove that there are no others.
25. Each honest fisherman caught exactly 30 kg, while each ordinary one caught less than $60 / 6=10$ kg of fish. Therefore, there could not have been zero honest fishermen (10 ordinary fishermen would have caught less than 100 kg). Similarly, there could not have been exactly one honest fisherman (9 ordinary fishermen would have caught less than 90 kg, and together with the honest one, less than 120 kg).

Furthermore, if there were four honest fishermen, the ordinary fishermen would not have caught anything, and could not have claimed that they caught 60 kg. Moreover, there could not have been more than four honest fishermen.

Thus, there could have been only two or three honest fishermen. Both these scenarios are possible. For example, 2 honest fishermen each caught 30 kg and 8 ordinary fishermen each caught $60 / 8=15 / 2$ kg, or 3 honest fishermen each caught 30 kg and 7 ordinary fishermen each caught $60 / 14=30 / 7$ kg.