1. In a $3 \times 4$ table, 12 numbers are arranged such that all seven sums of these numbers in the rows and columns of the table are distinct. What is the maximum number of numbers in this table that can be zero?
Answer: 8.

Solution. Let's call the index of the table the total number of its zero rows and columns. According to the condition, the index does not exceed 1. Let $n$ be a non-zero element of the table. Then in the same row or in the same column with $n$ there is another non-zero number (otherwise the sums in the row and column containing $n$ are equal to $n$). Therefore, non-zero elements occur in pairs. There are at least two such pairs, otherwise the index of the table is not less than 4. If there are exactly two pairs, then they do not intersect, otherwise the index of the table is 3. Thus, the table contains at least 4 non-zero numbers, and the number of zeros does not exceed 8. An example of a table with 8 zeros is given below.

| 0 | 0 | 0 | 0 |
| :--- | :--- | :--- | :--- |
| 1 | 2 | 0 | 0 |
| 0 | 0 | 4 | 5 |