Problem 9.1. In a $3 \times 3$ table, the digits from 1 to 9 are arranged (each digit is written in exactly one cell). The digits 1, 2, 3, 4 are placed as shown in the figure. It is also known that the sum of the digits in the cells adjacent to the digit 5 is 9. Find the sum of the digits in the cells adjacent to the digit 6.

| 1 |  | 3 |
| :--- | :--- | :--- |
|  |  |  |
| 2 |  | 4 |
Answer: 29.

Solution. Note that the smallest possible sum of the digits around the digit 5 is exactly $1+2+6=9$, and this is only possible when it is below the digit 1 and above the digit 2, with the digit 6 in the center.

Indeed, if 5 is in the center, then the sum of the surrounding digits is $6+7+8+$ $9>9$. And if 5 is not in the center (and the digit in the center is at least 6) and not below 1, then the sum of the surrounding digits is at least $1+3+6>9$.

| 1 |  | 3 |
| :--- | :--- | :--- |
| 5 | 6 |  |
| 2 |  | 4 |

Since 6 is in the center, the sum of the surrounding digits is $5+7+8+9=29$.