2. A $3 \times 3$ table is filled with distinct natural numbers such that in each row and column, the sum of the outer numbers equals the number written between them. Determine the smallest number that can be written in the center of the table.

(Tomáš Jurík)
Solution. Let's denote the numbers in the corners of the table as $a, b, c, d$ (from left to right, top to bottom). These four numbers uniquely determine all other numbers in the table, as we can sequentially calculate the numbers between them and finally the number $a+b+c+d$ in the center of the table.

| $a$ |  | $b$ |
| :---: | :---: | :---: |
|  |  |  |
| $c$ | $d$ |  |
| $a+c$ |  | $b+d$ |
| $c$ | $c+d$ | $d$ |
| $a$ | $a+b$ | $b$ |
| $a+c$ | $a+b+c+d$ | $b+d$ |
| $c$ | $c+d$ | $d$ |

For the values $a=1, b=3, c=6$ and $d=2$, we get the table of different numbers

| 1 | 4 | 3 |
| :---: | :---: | :---: |
| 7 | 12 | 5 |
| 6 | 8 | 2 |

with the number $a+b+c+d=12$ in the center.

Now we will show that the number in the center of the table cannot be smaller than 12. A number smaller than 12 can be obtained as the sum of four different natural numbers in only two ways: as $1+2+3+4$ or $1+2+3+5$.

In both cases, the numbers 1, 2, and 3 are among the numbers written in the corners of the table. Let's try to see if they can be written in the corners of the table in such a way that we can fill the entire table as required.

The numbers 1 and 2 cannot be written in the same row or column, because then the number 3 would be written between them as their sum, and we need to have the number 3 in one of the corners of the table. Therefore, the numbers 1 and 2 must be in opposite corners of the table. For the number 3, we only have two possible corner positions in the table, and no matter which one we choose, we will have to write the number 4 between 3 and 1 and the number 5 between 3 and 2, so we will not be able to have either the number 4 or the number 5 in the last corner. Therefore, we cannot write the numbers $1,2,3,4$ or $1,2,3,5$ in the corners of the table, and thus the number $a+b+c+d$ in the center of the table, satisfying the conditions of the problem, always has a value of at least 12.