9. In a $3 \times 3$ table, all natural numbers from 1 to 9 are written. It is known that: the product of the numbers in the top row is 84; the product of the numbers in the left column is 16; the sum of the numbers in the middle row is equal to the sum of the numbers in the bottom row; the product of the numbers on one of the two diagonals is odd. Which number is written in the cell on the right of the middle row?
(A) 5
(B) 4
(C) 9
(D) 6
(E) 3
Problem 9. The correct answer is (B).

To obtain the product 84 in the top row, the three numbers must be 2, 6, and 7, or 3, 4, and 7. To obtain the product 16 in the left column, the three numbers must be $1, 2, 8$. Therefore, the top row must necessarily be formed by 2, 6, 7, and the number 2 must be in the top-left cell. In the 4 unoccupied cells, the numbers 3, 4, 5, 9 must therefore be placed. To ensure that the middle row and the bottom row have equal sums, given that one of them must start with the number 1 and the other with the number 8, it is necessary that one of them be formed by 1, 5, 9 and the other by 8, 3, 4. To have a diagonal where the product is odd, since there is a 2 in the top-left cell (so the product on that diagonal will be even), the product on the other diagonal must be odd. The only possibility is that the bottom row is formed by 1, 5, 9 (with 1 on the left), the middle row is formed by $8, 3, 4$ (in this order from left to right), and the top row is formed by 2, 6, 7 (in this order from left to right). The cell on the right of the middle row will therefore contain the number 4.

| 2 | 6 | 7 |
| :--- | :--- | :--- |
| 8 | 3 | 4 |
| 1 | 5 | 9 |


| 2 | 6 | 7 |
| :--- | :--- | :--- |
| 8 | 3 | 4 |
| 1 | 9 | 5 |