3. In each cell of a $2 \times 2$ table, a number was written, and all the numbers are different. It turned out that the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second column. What can the sum of all four numbers in the table be? Find all possible options and explain why there are no others.
# Answer: 0.

## Solution:

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

Let's denote the numbers in the table as shown on the left. According to the condition, $a+b=c+d$, $ac=bd$. Then $a=c+d-b$, substitute this into the product equality: $(c+d-b)c=bd$. Expand the brackets and move everything to the left: $c^2 + cd - bc - bd = 0$. Factorize: $(c+d)(c-b)=0$. Since $c-b \neq 0$ (all numbers are different), then $c+d=0$. Therefore, the sum in the first row is also 0, and accordingly, the sum in the entire table is also 0.

## Criteria:

Correct answer without correct justification - 1 point.

If it is not justified why $c$ can be canceled out - deduct 2 points.