In a movie theater, $80 \%$ of the adults are male, $40 \%$ of the male viewers are children, and $20 \%$ of the children are boys. What is the minimum number of people watching the movie?
I. solution. Let the number of adults be $x$, and the number of children be $y$. Thus, the number of men is $0.8 x$, and the number of boys is $0.2 y$. Clearly, $x, y$ are integers, and the number of boys and men must also be integers. Therefore, $0.8 x$ and $0.2 y$ must be integers; $0.8 x=\frac{4}{5} x$ is an integer if $x$ is divisible by 5.

The male audience consists of men and boys, and the male children are the boys. Thus, the number of male audience members is $0.8 x + 0.2 y$, and the number of boys is $0.2 y = (0.8 x + 0.2 y) \cdot 0.4$, from which $3 y = 8 x$ follows. The left side of the equation is divisible by 3, so the right side must also be divisible by 3. This is possible if $x$ is divisible by 3.

Therefore, we know that $x$ must be divisible by both 5 and 3. The smallest such number is 15. If $x = 15$, then $y = 40$. In this case, $0.2 y = 8$, which is an integer, and $0.8 x = 12$ is also an integer. Since $x + y = 15 + 40 = 55$, at least 55 people are watching the film.

II. solution. 1. If 20% of the children are boys, then there are four times as many girls as boys. Based on this, there must be at least 1 boy and 4 girls in the cinema.

2. There are one and a half times as many men as boys in the cinema (60% - 40%), so there must be at least 2 boys in the cinema.
3. If 80% of the adults are men, this means there are four times as many men as women in the cinema. Therefore, there must be at least 1 woman and 4 men in the audience.

Thus, the number of men must be a multiple of 4 (see point 3) and 3 (see point 2), and the smallest common multiple of these two numbers is 12.

Therefore, there must be at least 12 men, 3 women, 8 boys, and 32 girls, making a total of at least 55 audience members in the cinema. If the number of audience members differs from this, then the number of audience members is a multiple of 55.