5. At a round table, 8 gnomes are sitting, each with three diamonds. Every minute, the gnomes simultaneously do the following: they divide all their diamonds into two piles (possibly one or both piles are 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 gnomes. One of them has seven diamonds. How many do the other two have?
Solution. Let's number the gnomes in order. Notice that gnomes with even numbers always share with gnomes with odd numbers, and vice versa. Since all distributions happen simultaneously, all the diamonds of the gnomes with even numbers will go to the 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.

Of these:

Proof that the even-numbered gnomes (and the odd-numbered ones) will always have a total of 12 diamonds - 3 points.

Proof that there will be a gnome with 5 diamonds - 1 point.

Proof that there will be a gnome with 12 diamonds - 1 point.