We divide a square into 3 equal wide rows and columns. For each of the resulting 9 cells, write one of the first 30 natural numbers - each cell with a different number - in such a way that the product of the 3 numbers in each row and column is 270. (We do not consider two arrangements different if they can be transformed into each other by swapping rows, swapping columns, or reflecting the square across one of its diagonals.)
Each field must be filled with a divisor of 270 that is not greater than 30; these are as follows:

$$
30, \quad 27, \quad 18, \quad 15, \quad 10, \quad 9, \quad 6, \quad 5, \quad 3, \quad 2, \quad 1
$$

We need to use 9 of these numbers. From our numbers, 270 can be represented as a product of three factors in the following forms:

| $A:$ | 30 | $\cdot$ | 9 | $\cdot$ | 1, |
| ---: | ---: | ---: | ---: | ---: | ---: |
| $B:$ | 27 | $\cdot$ | 10 | $\cdot$ | 1, |
| $C:$ | 27 | $\cdot$ | 5 | $\cdot$ | 2, |
| $D:$ | 18 | $\cdot$ | 15 | $\cdot$ | 1, |
| $E:$ | 18 | $\cdot$ | 5 | $\cdot$ | 3, |
| $F:$ | 15 | $\cdot$ | 9 | $\cdot$ | 2, |
| $G:$ | 15 | $\cdot$ | 6 | $\cdot$ | 3, |
| $H:$ | 10 | $\cdot$ | 9 | $\cdot$ | 3, |
| $J:$ | 9 | $\cdot$ | 6 | $\cdot$ | 5. |


|  | $C$ | $H$ | $D$ |
| :---: | :---: | :---: | :---: |
| $B$ | 27 | 10 | 1 |
| $E$ | 5 | 3 | 18 |
| $F$ | 2 | 9 | 15 |

We need to use six of these for the rows and columns.

Every number written belongs to a product representation in the row and column, so the representation $A$ cannot be used because the factor 30 does not appear elsewhere. From the remaining representations, we need to select three for the 3 rows and three for the columns, where all numbers are different. Two such sets can be formed: the triples $B, E, F$ and $C, D, H$. They have no common letters, and they are composed of the same numbers (the factor 6 does not appear in either), so the desired filling is definitely possible. (If two factors appear together in one of the triples in a representation, they cannot appear together in the other, since two factors determine the third factor in the representation.) By arranging the representations $B, E, F$ in the rows and writing the factors of the first row in the order 27, 10, 1 - in the subsequent columns, the representations $C, H, D$ containing the second occurrence of these factors must appear; the common factor of the corresponding row and column representation goes into the remaining fields. - There is no other solution.

András Talyigás, Budapest, Móricz Zs. g. II. o. t.)

Remarks. 1. Several people determined that the number of non-essentially different solutions that can be formed from the found arrangement, including the one above, is $72=2 \cdot 6 \cdot 6$, as there are 2 ways to choose the triple for the rows and 6-6 ways to arrange the triples internally.

2. We can ask which other number must be deleted after deleting 30 from the (1) numbers, since only nine numbers can appear. The product of the remaining 10 numbers is $270^{3} \cdot 6$, so only the 6 can be deleted, but this does not make the formation of the letter triples redundant.