Problem 3. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex?
Answer: 61.

Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, continuing this sequence in a cycle will include all numbers of the form $x_{3 k+1}$. Similar sequences can be constructed starting from the numbers $x_{2}$ and $x_{3}$. Thus, the numbers at the vertices that have the same remainder when divided by 3 are equal.

Since each set of nine consecutive numbers contains three sets of vertices with different remainders when divided by 3, the sum of the numbers in each such set is 100. Note that 19 gives a remainder of 1 when divided by 3, 20 gives a remainder of 2, and 2019 gives a remainder of 0, so the number at vertex 2019 is $100 - 19 - 20 = 61$.

## Criteria

## The largest suitable criterion is used:

## 7 p. Any complete solution to the problem.

4 p. It is proven that numbers that are 2 apart are equal.

1 p. It is noted that numbers that are 8 apart are equal.