2. In each cell of a $3 \times 3$ square, a natural number is written. Each number is distinct and different from one. It is known that the number written in each cell is a divisor of the product of all numbers in the cells adjacent to it by a side. Find the maximum possible value of the number of prime numbers among those written.
Answer: Six. Solution. We will prove that it is impossible to use only two composite numbers. We will reason by contradiction. Suppose there are only two composite numbers and 7 primes. The product of several prime numbers cannot be divisible by a prime number different from them. Therefore, the composite numbers must be: one of the numbers $A, B, D$; one of the numbers $D, G, H$; one of the numbers $C, E, F, K$. In order for there to be only two composite numbers, one of them must be the number $D$, and the other must be the number $F$. But then the prime number $H$ is surrounded by three prime numbers $E, G, K$ - a contradiction. An example with three composite numbers can be constructed, for example, as shown on the right, where $A, B, C, G, H, K$

| $A$ | $B$ | $C$ |
| :--- | :--- | :--- |
| $A G$ | $B H$ | $C K$ |
| $G$ | $H$ | $K$ |

- are different prime numbers.