The square in Figure I is called special because:

- it is divided into 16 equal squares;
- in each row and in each column, the digits 1, 2, 3, and 4 appear;
- in each of the squares \(A, B, C\), and \(D\) (as in Figure II), the digits 1, 2, 3, and 4 appear.

| 4 | 2 | 1 | 3 |
| :--- | :--- | :--- | :--- |
| 1 | 3 | 2 | 4 |
| 3 | 1 | 4 | 2 |
| 2 | 4 | 3 | 1 |

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-08.jpg?height=222&width=200&top_left_y=246&top_left_x=1168)

a) Complete the square below so that it becomes special.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-08.jpg?height=192&width=203&top_left_y=635&top_left_x=1029)

b) Is it possible to complete the square below to obtain a special square? Why?

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-08.jpg?height=185&width=206&top_left_y=1001&top_left_x=1028)

c) Show all the ways to complete the square below to obtain a special square.

| 1 | 2 |  |  |
| :--- | :--- | :--- | :--- |
| 3 | 4 |  |  |
|  |  |  |  |
|  |  |  | 1 |

d) How many special squares exist?
Solution

a) The solution is presented in the figure below:

| $\mathbf{1}$ | 2 | $\mathbf{4}$ | $\mathbf{3}$ |
| :--- | :--- | :--- | :--- |
| 3 | 4 | 2 | 1 |
| $\mathbf{2}$ | $\mathbf{3}$ | 1 | $\mathbf{4}$ |
| $\mathbf{4}$ | $\mathbf{1}$ | $\mathbf{3}$ | 2 |

b) No. Since the small squares in the last column of square $D$ are filled with 1 and 2, the two small squares in the last column of square $B$ should be filled with 3 and 4. However, neither 3 nor 4 can appear in the second row, as they already appear in the second row of square $A$.

c) In square $D$, the 2 can appear in the same column as the 1 (as seen in the previous item). With a similar argument, it can be shown that the 3 cannot appear in the same row as the 1. Thus, we have the following possibilities for filling in square $D$:

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-33.jpg?height=114&width=574&top_left_y=2556&top_left_x=644)

In each of these cases, the special square can be filled in a unique way:

| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 3 | 4 | $\mathbf{1}$ | $\mathbf{2}$ |
| $\mathbf{4}$ | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ |
| $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | 1 |


| 1 | 2 | $\mathbf{4}$ | $\mathbf{3}$ |
| :--- | :--- | :--- | :--- |
| 3 | 4 | $\mathbf{1}$ | $\mathbf{2}$ |
| $\mathbf{2}$ | $\mathbf{1}$ | $\mathbf{3}$ | $\mathbf{4}$ |
| $\mathbf{4}$ | $\mathbf{3}$ | $\mathbf{2}$ | 1 |


| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 3 | 4 | $\mathbf{1}$ | $\mathbf{2}$ |
| $\mathbf{2}$ | $\mathbf{1}$ | $\mathbf{4}$ | 3 |
| $\mathbf{4}$ | $\mathbf{3}$ | $\mathbf{2}$ | 1 |

d) To fill in square $A$, we can place the 1 in 4 ways, the 2 in 3 ways, the 3 in 2 ways, and the 4 in 1 way. Therefore, it can be filled in $4 \times 3 \times 2 \times 1=24$ ways. For each of these choices, the number of ways to fill in the rest of the special square is the same. Therefore, to count the number of ways to complete the filling of the special square, we can assume that square $A$ is filled as in the previous item. To fill in square $C$, we can place the 1 in any of the 4 cells. Once the 1 is fixed, there are 3 ways to complete the square, as seen in the previous item. The total number of possibilities for filling is, therefore, $24 \times 4 \times 3=288$.