Problem 4.2. Vika wrote the numbers from 1 to 9 in the cells of a $3 \times 3$ table such that the sum of any two numbers in adjacent cells is less than 12. The hooligan Andrey erased all the even numbers: $2, 4, 6,$ and $8$.

Help Vika restore where each number was.

| $A$ | $\mathbf{1}$ | $\mathbf{9}$ |
| :--- | :--- | :--- |
| $\mathbf{3}$ | $\mathbf{5}$ | $D$ |
| $B$ | $C$ | $\mathbf{7}$ |

Construct the correspondence

- In square $A$
- In square $B$
- In square $C$
- In square $D$
- the number 2 stands.
- the number 4 stands.
- the number 6 stands.
- the number 8 stands.
Answer: $A=8, B=6, C=4, D=2$.

Solution. Since the sum of the numbers 9 and $D$ is less than 12, $D$ can only be 2. Since the sum of the numbers 7 and $C \neq 2$ is less than 12, $C$ can only be 4. Since the sum of the numbers $C=4$ and $B$ is less than 12, and $B$ is 6 or 8, $B$ can only be 6. Then the remaining number $A$ is 8. It is easy to see that the obtained arrangement of numbers in the table satisfies all the conditions of the problem.

| 8 | 1 | 9 |
| :--- | :--- | :--- |
| 3 | 5 | 2 |
| 6 | 4 | 7 |