6. Eight dwarfs are sitting around a round table, each with three diamonds. The dwarfs' chairs are numbered in order from 1 to 8. Every minute, the dwarfs simultaneously do the following: they divide all their diamonds into two piles (one or both piles may be empty), then give one pile to their left neighbor and the other to their right neighbor. At some point, all the diamonds are gathered by three dwarfs. One of the dwarfs ended up with 7 diamonds. How many diamonds did each of the others have?
Solution. Note that gnomes with even numbers always share with gnomes with odd numbers, and vice versa. Since all distributions happen simultaneously, all diamonds of gnomes with even numbers will go to gnomes with odd numbers and vice versa. That is, the gnomes with even numbers will always have a total of 12 diamonds, and the gnomes with odd numbers will also have 12. Therefore, the gnome with 7 diamonds has a partner with the same parity and he will have 5 diamonds. Then the remaining third can only have 12 diamonds.

Criteria. Full solution - 6 points.

1 point - just the correct answer;

2 points - solution by example.