1. In how many ways can a $3 \times 3$ square table be filled with the numbers $2,2,3,3,3,4,4,4,4$ so that the sum of the numbers in each $2 \times 2$ square of this table is equal to 14?
ANOTHER SOLUTION. Just like in the previous solution, we will show that the number 14 can be obtained from the given numbers as a sum of four numbers only as $4+4+4+2$ or $4+4+3+3$. It follows from this that in any $2 \times 2$ square that contains the number 2 (or 3), no other 2 can be present. In the central square (which is part of four $2 \times 2$ squares), there cannot be a 2 (there would be no room for the threes), nor a 3 (there would be no room for the twos). Therefore, the number 4 must be in the central square.

From the previous solution, we know that in the square adjacent to the central square, there cannot be a 2, because both $2 \times 2$ squares containing this 2 would already contain only fours, and we do not have enough of them. Therefore, the central square must contain a 4, and one of the corners must contain a 2. This determines the numbers in the corresponding $2 \times 2$ square of the table:

| 2 |  |  |
| :--- | :--- | :--- |
|  | 4 |  |
|  |  |  |$\rightarrow$| 2 | 4 |  |
| :--- | :--- | :--- |
| 4 | 4 |  |
|  |  |  |

We have five unfilled squares left, which must contain the numbers $2,3,3,3,4$. The number 3 must appear in at least one of the two $2 \times 2$ squares that share two squares with the already filled square (and this must happen twice). Let's choose one of them (choosing the other one would result in the same process and a symmetrical situation). The other numbers can then be written in only one way, because the 2 cannot be in the same $2 \times 2$ square as the 3:

| 2 | 4 | 3 |
| :--- | :--- | :--- |
| 4 | 4 | 3 |
|  |  |  |$\longrightarrow$| 2 | 4 | 3 |
| :--- | :--- | :--- |
| 4 | 4 | 3 |
| 2 |  |  |$\rightarrow$| 2 | 4 | 3 |
| :--- | :--- | :--- |
| 4 | 4 | 3 |
| 2 | 4 |  |$\rightarrow$| 2 | 4 | 3 |
| :--- | :--- | :--- |
| 4 | 4 | 3 |
| 2 | 4 | 3 |

By rotating by a multiple of $90^{\circ}$, we get three more distinct fillings. It is easy to see that these are all the possibilities that can be obtained by choosing a different initial corner with a 2 and choosing an adjacent square with two threes. The table can therefore be filled in four ways.