Problem 1288

1. Assume the group is 6-great but not 7-great.
- Consider a graph $G$ where vertices represent people and edges represent friendships.
- The group is 6-great if for any choice of 6 vertices, there exists a cycle containing these vertices.
- Assume there are 7 vertices in $G$ for which no cycle contains all of them.

2. Choose 7 vertices and analyze the subgraph.
- Let $v$ be one of these 7 vertices.
- Since the group is 6-great, there is a cycle containing the remaining 6 vertices.

3. **Analyze the degree of $v$.**
- If more than 3 of the dashed edges (edges connecting $v$ to the other 6 vertices) are part of $G$, then there are two neighboring dashed edges, forming a cycle with all 7 vertices, which contradicts our assumption.
- Therefore, $\deg(v) \leq 3$.

4. **Consider a cycle containing 6 vertices including $v$.**
- Let $w$ be a neighbor of $v$ in this cycle.
- Since there is a cycle through the 6 vertices excluding $w$, $\deg(v) \geq 3$.

5. Combine the results to find a contradiction.
- From the previous steps, we have $\deg(v) = 3$.
- Summing the degrees of the 7 vertices, we get $21$, which contradicts the handshaking lemma (which states that the sum of the degrees of all vertices in a graph is even).

6. Conclude that the assumption is false.
- Therefore, there is no group of people that is 6-great but not 7-great.

$\blacksquare$