Problem 10.4. The price of a merchandize dropped from March to April by $x \%$, and went up from April to May by $y \%$. It turned out that in the period from March to May the prize dropped by $(y-x) \%$. Find $x$ and $y$ if they are positive integers (the prize is positive for the whole period).

Ivailo Kortezov
10.4. It follows from the condition of the problem that

$$
\frac{100-x}{100} \cdot \frac{100+y}{100}=\frac{100-y+x}{100}
$$

Hence $(100-x)(100+y)=100(100-y+x)$ which can be written as $(200-$ $x)(200+y)=200^{2}$. Further, $100<200-x<200$, i. e. $200-x$ equals 125 or 160 .

Therefore the solutions are $x_{1}=75, y_{1}=120$ and $x_{2}=40, y_{2}=50$, both of which fulfill the condition.