A week-old set's three-element subsets need to be colored such that if the intersection of two subsets is empty, then their colors differ. How many colors do we need at least?
Solution. 1. Three colors are sufficient for the desired coloring. Let the 7-element set be $\{1,2,3,4,5,6,7\}$. The color of the 3-element subsets containing 1 is red, those containing 2 but not 1 are blue, and the rest are yellow. Then any two red subsets contain the element 1, so they cannot be disjoint. Similarly, any two blue subsets have the element 2 in common. Consider two yellow subsets; these do not contain 1 or 2, so their elements can be at most five: $3,4,5,6$ or 7. Two such 3-element subsets cannot be disjoint, as this would require $3+3=6$ distinct elements.

2. We will show that the coloring cannot be achieved with two colors. Suppose it were, and consider the sets $H_{1}=\{1,2,3\}, H_{2}=\{4,5,6\}, H_{3}=\{1,2,7\}, H_{4}=\{3,5,6\}, H_{5}=\{2,4,7\}, H_{6}=\{1,3,6\}, H_{7}=\{4,5,7\}$. Any two adjacent sets in this order are disjoint, so if, for example, $H_{1}$ is red, then $H_{2}$ is blue, $H_{3}$ is red, $H_{4}$ is blue, $H_{5}$ is red, $H_{6}$ is blue, and $H_{7}$ is again red. However, $H_{7}$ and $H_{1}$ are also disjoint, so they cannot both be red, a contradiction.

Note. Péter Énekes and Balázs Zsakó mention the following generalization of the problem: Let $n$ be a positive integer. We color the $n$-element subsets of a $2n+1$ element set such that any two disjoint subsets have different colors. This can be achieved with at least three colors. The first part of the provided solution remains unchanged in form, while the second part can be adapted with some modifications to prove this more general statement.