1. The automatic line for processing body parts initially consisted of several identical machines. The line processed 38880 parts daily. After the production was modernized, all the machines on the line were replaced with more productive but also identical ones, and their number increased by 3. The automatic line began to process 44800 parts per day. How many parts did each machine process daily initially?
Solution: Let $x$ be the number of machines before modernization, $y$ be the productivity of each machine, i.e., the number of parts processed per day, and $z$ be the productivity of the new machines. Then we have $x y = 38880 = 2^{5} \cdot 3^{5} \cdot 5, (x+3) z = 44800 = 2^{8} \cdot 5^{2} \cdot 7, x > 1, y \frac{38880 \cdot 3}{5920} = \frac{2^{5} \cdot 3^{6} \cdot 5}{2^{5} \cdot 5 \cdot 37} = \frac{3^{6}}{37} = 19 \frac{26}{37}, \quad x \geq 20$. Since $(x+3) z = 2^{8} \cdot 5^{2} \cdot 7$, $x$ is not divisible by 3, and $x = 2^{\alpha} \cdot 5^{\beta}$, where $\alpha \in \{0,1,2,3,4,5\}, \beta \in \{0,1\}$.

1) $\beta=0, x=2^{\alpha}, \alpha \in \{5\}$. When $\alpha=5$ we have $x=32, x+3=35, y=3^{5} \cdot 5=1215, z=2^{8} \cdot 5=1280$
2) $\beta=1, x=2^{\alpha} \cdot 5, \alpha \in \{2,3,4,5\}$.

When $\alpha=2$ we have $x=20, x+3=23$, which is impossible.

When $\alpha=3$ we have $x=40, x+3=43$, which is impossible.

When $\alpha=4$ we have $x=80, x+3=83$, which is impossible.

When $\alpha=5$ we have $x=160, x+3=163$, which is impossible.

Answer: 1215.