Authors: Kazzczzna T.V., Frankinn B.R., Shapovalov A.V.

Out of 100 members of the Council of Two Tribes, some are elves, and the rest are dwarves. Each member wrote down two numbers: the number of elves in the Council and the number of dwarves in the Council. Each member correctly counted their own tribe but made an error of exactly 2 when counting the other tribe. In the numbers written down, one digit appeared no less than 222 times. How many elves and how many dwarves could there be in the Council? If there are multiple solutions, provide one of them.

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Participants were required to find just one example. Let's show how to find them all.

Let the digit $A$ be written at least 222 times. Someone wrote $A$ at least three times (otherwise, the number of $A$s in the answers would not exceed $100 \cdot 2 = 200$). Therefore, Vasya wrote $A A$ and $B A$. Since the sum of these numbers ends in 2 or 8, $A$ can be 1, 6, 4, or 9. There are the following 4 cases:

1. If $A A=11$, then $B A=102-11=91$;
2. If $A A=66$, then $B A=102-66=36$;
3. If $A A=44$, then $B A=98-44=54$;
4. If $A A=99$, then $B A=98-99<0$ - does not fit.

In the case of $A A=11, B A=91$, the number of gnomes and elves (in some order) is either 11 and 89, or 9 and 91. In the first case, 1 is named no more than three times by each of the 11 people and at most once by each of the 89, which is less than 200 times in total. In the second case, 1 can be named three times by each of the 91 people and not at all by each of the 9, which is a maximum of $91 \cdot 3 = 273$ times. This case fits and it gives two options: 9 elves and 91 gnomes, or 9 gnomes and 91 elves.

Similar reasoning shows that in case 2, the council could have 66 elves and 34 gnomes (or vice versa, 66 gnomes and 34 elves), and case 3 does not fit.

## Answer

Possible options:

- 66 elves, 34 gnomes (or vice versa - 66 gnomes, 34 elves);
- 9 elves, 91 gnomes (or vice versa - 9 gnomes, 91 elves).

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