3. Find all $3 \times 3$ square tables of natural numbers in which the product of all numbers in each row, each column, and both diagonals is the same, and for which the sum of the four numbers in their corner fields is a single-digit number.

School - written part of the 1st round, category B is held

## on Tuesday, January 27, 1998

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the start of the competition.
3. Let $a, b, c, d$ be the single-digit numbers in the corner fields of the desired table (Fig. 1) and $e$ the number in its central field. Given the $\mathrm{k}$ symmetry (flipping along one of the diagonals or the central column or row does not change the properties of the table), we can assume that $a \leqq d, b \leqq c$ and $a+d \leqq b+c$, and since $a+b+c+d \leqq 9$, under the given assumptions, $a+d \leqq 4$ and $b+c \leqq 5$. From the equality $a e d = b e c$ it follows that $a d = b c$, so it suffices to examine the following five possibilities:

![](https://cdn.mathpix.com/cropped/2024_04_17_68223bd8f0f7e94aebddg-3.jpg?height=226&width=214&top_left_y=1275&top_left_x=1635)

Fig. 1

$$
\begin{array}{l|lllll}
a & 1 & 1 & 1 & 2 & 2 \\
d & 1 & 2 & 3 & 2 & 2 \\
b & 1 & 1 & 1 & 1 & 2 \\
c & 1 & 2 & 3 & 4 & 2
\end{array}
$$

In each of these five cases, we can express the other numbers in the table using the "middle" number $e$ in the same way, by using the equalities of the products of the numbers in both diagonals, both outer rows, and both outer columns. The tables then look like this:

| 1 | $e$ | 1 |
| :--- | :--- | :--- |
| $e$ | $e$ | $e$ |
| 1 | $e$ | 1 |


| 1 | $2 e$ | 1 |
| :---: | :---: | :---: |
| $e$ | $e$ | $e$ |
| 2 | $\frac{1}{2} e$ | 2 |


| 1 | $3 e$ | 1 |
| :---: | :---: | :---: |
| $e$ | $e$ | $e$ |
| 3 | $\frac{1}{3} e$ | 3 |


| 2 | $2 e$ | 1 |
| :---: | :---: | :---: |
| $\frac{1}{2} e$ | $e$ | $2 e$ |
| 4 | $\frac{1}{2} e$ | 2 |


| 2 | $e$ | 2 |
| :--- | :--- | :--- |
| $e$ | $e$ | $e$ |
| 2 | $e$ | 2 |

If we now compare the mentioned products with the product of the numbers in the second row (or in the second column), we get in each of the mentioned cases the same equation

$$
e^{3} = (a d) e, \quad \text{where successively } a d = 1, 2, 3, 4, 4
$$

This equation has solutions in natural numbers only for $a d \in \{1, 4\}$, and these correspond to the three tables in Fig. 2. From the last table, we get three more by the mentioned symmetries, but as we can easily see, each of them is obtained by rotating the given table by $90^{\circ}$.

| 1 | 1 | 1 |
| :--- | :--- | :--- |
| 1 | 1 | 1 |
| 1 | 1 | 1 |


| 2 | 2 | 2 |
| :--- | :--- | :--- |
| 2 | 2 | 2 |
| 2 | 2 | 2 |


| 2 | 4 | 1 |
| :--- | :--- | :--- |
| 1 | 2 | 4 |
| 4 | 1 | 2 |

Fig. 2

For the correct solution, award 6 points, 1 point for finding the first two tables, 2 points for finding the third type of table, and the remaining 1 to 2 points according to the completeness of the reasoning that excludes the existence of other solutions. Do not deduct points for omitting possible symmetric solutions.