7.4 Can the cells of a $5 \times 8$ table be filled with the numbers 1 and 3 so that the sum of the numbers in each row and each column is divisible by 7?
Solution: Suppose it was possible to arrange the numbers 1 and 3 in the table such that their sum in each of the 5 rows and in each of the 8 columns is divisible by 7. The sum in any row is not less than $8 \cdot 1=8$ and not more than $8 \cdot 3=24$. The numbers in the range from 8 to 24 that are divisible by 7 are 14 and 21. Note that the sum of the numbers in a row is even - it is the sum of eight odd numbers 1 and 3. Therefore, the sum in each row is 14. Similarly, the sum of the numbers in each column lies in the range from 5 to 15, is odd, and is divisible by 7, i.e., this sum is 7. Let's calculate the sum $\mathrm{S}$ of all the numbers in the table:

1) by rows: $S=14 \cdot 5=70 ; 2)$ by columns: $S=7 \cdot 8=56$. The contradiction obtained indicates that the assumption about the possibility of the required arrangement of numbers in the condition of the problem is incorrect. It is impossible to arrange the numbers in this way.

Answer: impossible.