22.3. Given a segment $O A$. From the end of the segment $A$, 5 segments $A B_{1}, A B_{2}, A B_{3}, A B_{4}, A B_{5}$ extend. From each point $B_{i}$, five new segments can extend, or no new segments, and so on. Can the number of free ends of the constructed segments be equal to $1001$?
22.3. Answer: yes, it can. When five segments are drawn from the end of a segment, 5 new free ends appear and one old end disappears. As a result, the number of free ends increases by 4. Therefore, if sets of five segments are drawn $k$ times, the number of free ends is $4 k+1$. For $k=250$, we get the required number of free ends.