1. In a $3 \times 3$ table, 9 numbers are arranged such that the six products of these numbers in the rows and columns of the table are all different. What is the maximum number of these numbers that can equal one?
Answer: 5.

Solution. Let's call the index of the table the total number of its rows and columns consisting of ones. According to the condition, the index does not exceed 1. Let $n$ be an element of the table different from 1. Then in one row or one column with $n$ there is another number not equal to 1 (otherwise the products in the row and column containing $n$ would be equal to $n$). Therefore, non-unit elements occur in pairs. There are at least two such pairs, otherwise the index of the table would be no less than 3. If there are exactly two pairs, then they do not intersect, otherwise the index of the table is 2. Thus, the table contains at least 4 numbers different from 1, and the number of ones does not exceed 5. An example of a table with 5 ones is given below.

| 1 | 1 | 1 |
| :--- | :--- | :--- |
| 1 | 2 | 3 |
| 5 | 7 | 1 |