3-ча 5. 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 \(1001\)?

By a free end of a segment, we mean a point that belongs to only one segment (except for the point \(O)\).
Solution 5. 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 the 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.

\section*{10th Grade}