3. A $3 \times 3$ table was filled with prime numbers. It turned out that the sum of the numbers in any two adjacent cells is a prime number. What is the maximum number of different numbers that can be in the table? Provide an example and prove that there cannot be more different numbers.

Fill the $3 \times 3$ table with prime numbers such that the sum of the numbers in any two adjacent cells is a prime number. Determine the maximum number of different prime numbers that can be used and provide an example. Prove that it is not possible to use more different prime numbers.
3. Answer: 6.

Example:

| 5 | 2 | 29 |  |
| :---: | :---: | :---: | :---: |
| 2 | 3 | 2 |  |
| 11 | 2 | 17 |  |
|  | -2 |  | 0 |

![](https://cdn.mathpix.com/cropped/2024_05_06_7965a576f88598e4f309g-2.jpg?height=281&width=280&top_left_y=1059&top_left_x=886)

We will prove that it cannot be more. Divide the cells of the table into 4 pairs of adjacent cells and one central cell. In each pair of adjacent cells, the sum of the numbers is a prime number. But the sum of two odd primes is an even number greater than two and, therefore, composite. Thus, in each pair, there must be a number 2. Therefore, there must be at least four 2s. The different numbers can be the five remaining and the 2, meaning there are no more than 6 different numbers.

## Grading Criteria.

Only the correct example is provided: 3 points.

Proved that there are at least four 2s, 6 different numbers, but no example: 4 points.

It can also be proved by coloring the board in a checkerboard pattern and explaining that on each color, the numbers have the same parity.