Problem 6.3. In the cells of a $4 \times 4$ table, the numbers $1,2,3,4$ are arranged such that

- each number appears in each row and each column;
- in all four parts shown in the figure, the sums of the numbers are equal.

Determine in which cells the twos are located based on the two numbers in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_b46fbd582cc3a82460aeg-14.jpg?height=305&width=301&top_left_y=683&top_left_x=576)

Construct the correspondence

- In row $A$, the two is located
- In row $B$, the two is located
- In row $C$, the two is located
- In row $D$, the two is located
- in column number 1.
- in column number 2.
- in column number 3.
- in column number 4.
Answer: In row $A$, the two is in column 2, in row $B-1$, in row $C-4$, in row $D-3$.

Solution. First, let's find what the sum of the numbers in each of the parts into which the board is divided is. In each column, the numbers from 1 to 4 appear once. Their sum is $1+2+3+4=10$, and the sum of all numbers on the board is $10 \cdot 4=40$. Since the sums of numbers in all parts are equal, they are equal to $40: 4=10$.

Let's look at the lower left corner of three cells. One of the numbers is a four, so the sum of the other two should be equal to $10-4=6$. At the same time, these two numbers cannot be fours, as a four is already present in the same line with them. The only possible option is two threes. We get that in cells $C 1$ and $D 2$ there are threes.

In the upper right corner of three cells, the sum of the three numbers is also 10. Such a sum can be obtained either as $4+3+3$, or as $4+4+2$. In the first case, in cell $A 4$ there should be a four, and in the second case, a two.

In cells $C 3$ and $C 4$ there are a two and a four in some order, in cells $D 3$ and $D 4$ there are a one and a two. Then the number in cell $B 3$ is equal to $10-2-4-1-2=1$.

In cells $A 4$ and $B 4$ there are either a four and a three, or a four and a two. There is a four in any case, so there cannot be a four in $C 4$, which means that in $C 4$ there is a two, and in $C 3$ there is a four.

In the rightmost column, there is already a two, so in cell $D 4$ there is a one, and in cell $D 3$ there is a two.

Let's consider the rightmost column again. There is already a one and a two, so in the two remaining upper cells there are a three and a four, and in cell $A 4$ there is definitely a four, and in cell $B 4$ there is a three. We can also immediately determine that in cell $A 3$ there is a three.

In the top row, there is already a four. In the second column, a two and a four are missing. It turns out that in cell $A 2$ there is a two, and in cell $B 2$ there is a four.

The remaining empty upper cells of the first column are filled without difficulty: in cell $A 1$ there is a one, and in cell $B 1$ there is a two.

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

Thus, the twos are in cells $A 2, B 1, C 4, D 3$.