27. (BLR) Ten points such that no three of them lie on a line are marked in the plane. Each pair of points is connected with a segment. Each of these segments is painted with one of $k$ colors in such a way that for any $k$ of
the ten points, there are $k$ segments each joining two of them with no two being painted the same color. Determine all integers $k, 1 \leq k \leq 10$, for which this is possible.
27. Since this is essentially a graph problem, we call the points and segments vertices and edges of the graph. We first prove that the task is impossible if $k \leq 4$.
Cases $k \leq 2$ are trivial. If $k=3$, then among the edges from a vertex $A$ there are two of the same color, say $A B$ and $A C$, so we don't have all the three colors among the edges joining $A, B, C$.
Now let $k=4$, and assume that there is a desired coloring. Consider the edges incident with a vertex $A$. At least three of them have the same color. say blue. Suppose that four of them, $A B, A C, A D, A E$, are blue. There is a blue edge, say $B C$, among the ones joining $B, C, D, E$. Then four of the edges joining $A, B, C, D$ are blue, and we cannot complete the coloring. So, exactly three edges from $A$ are blue: $A B, A C, A D$. Also, of the edges connecting any three of the 6 vertices other than $A, B, C, D$, one is blue (because the edges joining them with $A$ are not so). By a classical result, there is a blue triangle $E F G$ with vertices among these six. Now one of $E B, E C, E D$ must be blue as well, because none of $B C, B D, C D$ is. Let it be $E B$. Then four of the edges joining $B, E, F, G$ are blue, which is impossible.
For $k=5$ the task is possible. Label the vertices $0,1, \ldots, 9$. For each color. we divide the vertices into four groups and paint in this color every edge joining two from the same group, as shown below. Then among any 5 vertices, 2 must belong to the same group, and the edge connecting them has the considered color.
\begin{tabular}{lllll} 
yellow: & 011220 & 366993 & 57 & 48 \\
red: & 233442 & 588115 & 79 & 60 \\
blue: & 455664 & 700337 & 91 & 82 \\
green: & 677886 & 922559 & 13 & 04 \\
orange: & 899008 & 144771 & 35 & 26.
\end{tabular}

A desired coloring can be made for $k \geq 6$ as well. Paint the edge ij in the $(i+j)$ th color for $i<j \leq 8$, and in the $2 i$ th color if $j=9$ (the addition being modulo 9 ). We can ignore the edges painted with the extra colors. Then the edges of one color appear as five disjoint segments, so that any complete $k$-graph for $k \geq 5$ contains one of them.