Problem - 2796
Show that, among randomly selected $11$ numbers from $1$, $2$, $3$, $\cdots$, $19$, $20$, one of them must be a multiple of another.
Dividing the $20$ numbers into the following $10$ groups:
- $\{1, 2, 4, 8, 16\}$
- $\{3, 6, 12\}$
- $\{5, 10, 20\}$
- $\{7, 14\}$
- $\{9, 18\}$
- $\{11\}$
- $\{13\}$
- $\{15\}$
- $\{17\}$
- $\{19\}$
Then by the pigeonhole principle, $2$ must be in the same group if $11$ numbers are chosen. Then by the construction of these groups, these two numbers must meet the requirement.