8.1. 72 vertices of a regular 3600-gon are painted red such that the painted vertices are the vertices of a regular 72-gon. In how many ways can 40 vertices of this 3600-gon be chosen so that they are the vertices of a regular 40-gon and none of them are red?
Answer: 81

Solution. Number the vertices in order $1,2, \ldots, 3600$, so that the painted vertices are those with numbers divisible by $3600 / 72=50$. 40 vertices form the vertices of a regular 40-gon if their numbers give the same remainder when divided by $3600 / 40=90$. There are 90 ways to choose a regular 40-gon. Consider one of these variants, in which the vertex numbers have the form $a+90 k$, where $a \in$ $\{1,2, \ldots, 90\}$. In $90-9=81$ cases, $a$ is not divisible by 10; in such cases, $a+90 k$ is not divisible by 10, and thus not divisible by 50, making these variants suitable. If, however, $a=10 m$, then for some $k \in\{1,2,3,4,5\}$, the number $m+9 k$ will be divisible by 5, and thus, $a+90 k=10(m+9 k)$ will be divisible by 50. Therefore, variants where $a$ is divisible by 10 are not suitable.