![](https://cdn.mathpix.com/cropped/2024_05_06_57b0dc32154003794737g-13.jpg?height=232&width=2041&top_left_y=1250&top_left_x=0)

Auto: :Shapovalov A.B.

a) Fill in each circle with a non-zero digit so that the sum of the digits in the two top circles is 7 times less than the sum of the other digits, and the sum of the digits in the two left circles is 5 times less than the sum of the other digits.

b) Prove that the problem has a unique solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_57b0dc32154003794737g-13.jpg?height=226&width=232&top_left_y=1716&top_left_x=913)
b) From the condition, it is clear that the sum of all digits is divisible by both 8 and 6. The smallest such natural number is 24. The next number is 48, but the sum of all five digits cannot exceed \(5 \cdot 9 = 45\).

Thus, the sum of all digits is 24, the sum of the two top digits is 3, and the sum of the two left digits is 4. It is easy to see that the digits in the three circles on the left and top can be arranged in only two ways (see the figure), and in the first case, the sum of the two remaining digits is 19, which is impossible, while in the second case, it is 18, which is possible only if both are nines.
![](https://cdn.mathpix.com/cropped/2024_05_06_57b0dc32154003794737g-13.jpg?height=224&width=574&top_left_y=2342&top_left_x=752)

## Answer

a) See the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_57b0dc32154003794737g-14.jpg?height=229&width=1237&top_left_y=323&top_left_x=-1)

Difficulty: $3+$

Grades: $9,10,11$

Author: R Rubanov I.S.

On a plane, all vertices of a regular $n$-gon and its center are marked. Then, the contour of this $n$-gon is drawn, and the center is connected to all vertices; as a result, the $n$-gon is divided into $n$ triangles. Vasya writes a number in each marked point (the numbers can be equal). In each triangle of the division, he writes in any order the three numbers at its vertices; after that, he erases the numbers in the marked points. For which $n$ can Petya always restore the number in each marked point from the triples of numbers written in the triangles?

## Solution

Let $A_{1} A_{2} \ldots A_{n}$ be the given polygon, and $S$ be its center.

We will show that if $n$ is even, there exist two different arrangements such that the corresponding triples will be the same. Then Petya will not be able to restore the original numbers. Place the number 1 in $S$, and in the vertices of the $n$-gon, place ones and twos alternately. By shifting the numbers one step clockwise, the triples $\{1,1,2\}$ will not change.

For odd $n$, Petya will be able to restore all numbers. Note that the number from vertex $A_{i}$ is written in two triples, while the number from $S$ is written in $n$ triples. Therefore, only the number from $S$ will appear in all triples an odd number of times, and Petya can determine it. After that, for each pair of adjacent vertices of the $n$-gon, he knows which two numbers are written in them. It remains to restore the numbers in the vertices from this data. By listing the numbers from all pairs of adjacent vertices, we obtain a "doubled" set of all numbers in the vertices. Comparing it with the set of numbers in pairs $\left(A_{1}, A_{2}\right), \left(A_{3}, A_{4}\right), \ldots, \left(A_{n-2}, A_{n-1}\right)$, we can find the number in vertex $A_{n}$. Similarly, the numbers in the other vertices can be found.

## Answer

For odd $n$.