3. Alice wrote several positive integers. Sasha rewrote these numbers and added one integer, smaller than all of Alice's numbers. Each of them found the sum and the product of their numbers and divided the sum by the product. Sasha got a number 5 times smaller than Alice. What number could he have added?
Answer: 6 or -20.

Solution: Let Alice's sum be $a$, Alice's product be $b$, and the added number be $c$. Then the condition can be written as: $\frac{a}{b}=\frac{a+c}{b c} \cdot 5$, or $a c=5 a+5 c$. Factoring, we get: $(a-5)(c-5)=25$. The factor $a-5$ is a divisor of 25 and can take the values $\pm 1, \pm 5, \pm 25$. Then $c-5$ takes the values $\pm 25, \pm 5, \pm 1$. The pairs $(a, c)$ take the values $(6,30),(4,-20),(10,10),(0,0),(30,6),(-20,4)$. We immediately discard the impossible pairs where $c$ is not less than $a$. The remaining pairs $(a ; c)$ are: $(30 ; 6),(4 ;-20)$. For each of these pairs, there exists a set of numbers that satisfy the conditions. For $c=6: 15,15,6$. Indeed, $\frac{30}{225}=\frac{30+6}{225 \cdot 6} \cdot 5=\frac{2}{15}$. For $c=-20: 2,2,-20$. Indeed, $\frac{4}{4}=\frac{4-20}{4(-20)} \cdot 5=1$.

Comment: Correct solution - 20 points. Points are distributed as follows: 10 points for each found value; if the values are found by trial and error, and it is not proven that there are no other values in the given range - 8 points per value. All values are found, but negative values are discarded - 18 points. There is progress - 5 points. The solution is started, but the progress is insignificant - 2 points. The correct answer is given without explanation - 0 points.