503. Three workers need to make 80 identical parts. Together, they make 20 parts per hour. The first worker started the job alone. He made 20 parts, spending more than 3 hours on their production. The remaining work was done by the second and third workers together. The entire job took 8 hours. How many hours would it take the first worker to make all 80 parts?
Let's denote the productivity of the first, second, and third workers as \( x, y, \) and \( z \) parts per hour, respectively. Based on the conditions of the problem, we obtain a system of two equations with three unknowns:

\[
\left\{\begin{array}{l}
x+y+z=20 \\
\frac{20}{x}+\frac{60}{y+z}=8
\end{array}\right.
\]

To find \( x \), we express \( y+z \) in terms of \( x \) from the first equation: \( y+z=20-x \), and then substitute this expression into the second equation:

\[
\frac{20}{x}+\frac{60}{20-x}=8
\]

This equation can be reduced to a quadratic equation \( x^{2}-15 x+50=0 \). The roots of this equation are \( x_{1}=10 \) and \( x_{2}=5 \). Do both fit? According to the condition, the first worker spent more than 3 hours on 20 parts. But \( \frac{20}{10}=2 \), so only \( x=5 \) is valid. Then he will manufacture all 80 parts in \( \frac{80}{5}=16 \) hours.

Answer: 16 hours.