Problem 2. 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

![](https://cdn.mathpix.com/cropped/2024_05_06_9cb48d57b1f9bbd6792cg-03.jpg?height=210&width=214&top_left_y=1029&top_left_x=1062)
less than the sum of the other digits.

b) Prove that the problem has a unique solution. [6 points] (A.V.ІІapovalov)
Answer. a) See the diagram.

Solution. b) If the sum of the digits in the two upper circles is 7 times less than the sum of the remaining digits, then it is 8 times less than the sum of all five digits. Reasoning similarly, we get that

![](https://cdn.mathpix.com/cropped/2024_05_06_9cb48d57b1f9bbd6792cg-03.jpg?height=73&width=186&top_left_y=1407&top_left_x=1080)
the sum of the digits in the two left circles is 6 times less than the sum of all five digits. Therefore, the sum of all digits is divisible by both 6 and 8. 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 upper digits is $24: 8=3$, and the sum of the two left digits is $24: 6=4$. It is easy to see that the digits in the three circles on the left and above can be arranged in only two ways (see the diagram), 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_9cb48d57b1f9bbd6792cg-04.jpg?height=440&width=372&top_left_y=137&top_left_x=904)