Problem 7.3. Numbers were placed in the cells of a square such that the sums of the numbers in each vertical, horizontal, and each diagonal of three cells are equal. Then some numbers were hidden. What is the sum of the numbers in the two shaded cells?

| 16 |  |  |
| :--- | :--- | :--- |
|  |  | 10 |
| 8 |  | 12 |
Answer: 34.

Solution. Let $a$ be the number in the top right cell, and $b$ be the number in the central cell. The sum of the diagonal containing the number 8 and the sum of the numbers in the right column have a common term $a$, so $8+b=10+12$, from which $b=14$. The sums of the three numbers along both diagonals have a common number $b$, so $16+12=8+a$, from which $a=20$. Therefore, $a+b=20+14=34$.

Remark. The other numbers can also be placed to verify that the problem's condition is correct:

| 16 | 6 | 20 |
| :---: | :---: | :---: |
| 18 | 14 | 10 |
| 8 | 22 | 12 |