4.1. In the cells of a $5 \times 5$ square, Vasya wants to arrange 25 different natural numbers such that any two numbers in adjacent cells (sharing a side) are not coprime. What is the smallest value that the largest number in such a table can take?
4.1. Solution. We will prove that there is a number $\geqslant 33$. Indeed, suppose all numbers are no more than 32. Note that if there is a prime $p$ on the board, then there are at least two numbers next to it, one of which is no less than $3 p$. Then the prime numbers 11, 13, $17, 19, 23, 29, 31$, as well as, obviously, the number 1 cannot appear in this table. But then it turns out that there are no more than $32-8=24$ different numbers in the table, which is not true. Therefore, there is a number not less than 33. Now let's provide an example.

| 11 | 33 | 3 | 21 | 7 |
| :---: | :---: | :---: | :---: | :---: |
| 22 | 30 | 9 | 12 | 14 |
| 2 | 4 | 24 | 27 | 6 |
| 8 | 16 | 20 | 18 | 15 |
| 26 | 28 | 32 | 10 | 5 |