5. For which $n$ can a complete graph with $n$ vertices be drawn on a plane so that each arc has no more than one common interior point with other arcs and no three arcs have a common interior point? (A graph is called complete if any two of its vertices are connected.)
5. Answer: $n<7$.

It is easy to construct a picture for $n=6$ and obtain an example for smaller $n$ by erasing edges. Similarly, an example for $n+k$ implies the existence of an example for $n$, so it is sufficient to show that the complete graph on 7 vertices does not work.

Assume the contrary. Consider a 7-vertex graph satisfying the conditions of the problem. It has 21 edges and $X$ points of intersection of edges. Let $O$ be a point of intersection of arcs $A C$ and $B D$.

Consider the quadrilateral formed by arcs $A B, B C, C D$, and $D A$. Its sides do not intersect, because if an edge enters the triangle $A B O$ intersecting $A B$, it will have to exit the triangle through $A B, A O$, or $B O$ and intersect one of its sides a second time. Except for the case where the edge enters $A B O$ through $A B$ and ends at vertex $A$ or vertex $B$. But then a small movement can eliminate this intersection.

We can assume that we are drawing the graph with the minimum number of intersection points, and each intersection corresponds to 4 edges, on which there can be no intersection points. Since each such edge without an intersection point can be counted at most twice, the number of such edges does not exceed $2 X$. The number of edges with intersection points is exactly $2 X$, so $4 X \leqslant 21$ and thus $X \leqslant 5$.

Consider the graph whose vertices are the vertices of the original graph plus the points of intersection of edges. This graph is planar. It has $V=7+X$ vertices, $E=21+2 X$ edges, and $F \leqslant 2 / 3 E$ faces, so by Euler's formula $2=F-E+V \leqslant V-E / 3=$ $7+X-(7+2 X / 3)=X / 3$. Therefore, $6 \leqslant X$. But we established above that $X \leqslant 5$, leading to a contradiction.