Problem 8.1. Gleb placed the numbers $1,2,7,8,9,13,14$ at the vertices and the center of a regular hexagon such that in any of the 6 equilateral triangles, the sum of the numbers at the vertices is divisible by 3. What number could Gleb have written in the center? It is sufficient to provide one suitable example.

![](https://cdn.mathpix.com/cropped/2024_05_06_ed8f792abd651132e9d3g-1.jpg?height=294&width=325&top_left_y=521&top_left_x=564)
Answer: 9.

Solution. Let's add the sums of the numbers in all six triangles; denote this value by $X$. On the one hand, $X$ is divisible by 3, since each term was divisible by 3 according to the condition. On the other hand, each number at the vertex of the hexagon entered $X$ twice (from two triangles), and the central number entered it 6 times. Subtracting from $X$ twice the sum of all numbers, which is $2 \cdot 54$, we get the quadrupled central number.

Divisibility by 3 is preserved in this process. This means that the number in the center is a multiple of 3. We have only one such number, which is 9.

The other numbers can be arranged around the perimeter of the hexagon so that their remainders when divided by 3 alternate; for example, in the order $1,2,7,8,13,14$.