Problem 11.7. An archipelago consists of $N \geqslant 7$ islands. Any two islands are connected by no more than one bridge. It is known that from each island, no more than 5 bridges lead, and among any 7 islands, there are definitely two connected by a bridge. What is the largest value that $N$ can take?
Answer: 36.

Solution. There can be 36 islands in the archipelago, for example, as follows: they form 6 groups of 6 islands each, and two islands are connected by a bridge if and only if they are in the same group. It is clear that from each island, exactly 5 bridges lead out, and among any 7 islands, there are necessarily two that are in the same group, and they are connected to each other by a bridge.
![](https://cdn.mathpix.com/cropped/2024_05_06_8af0c885427e3e323cf9g-44.jpg?height=150&width=1166&top_left_y=1224&top_left_x=143)

Suppose there can be $N \geqslant 37$ islands. Choose any of the islands, call it $A$, it is connected to no more than five other islands. Temporarily forget about these no more than 6 islands, and consider the remaining at least $N-6$ islands (among which none are connected to $A$). Choose any of these islands, call it $B$, it is connected to no more than five other islands. Temporarily forget about these no more than 6 islands, and consider the remaining at least $N-12$ islands (among which none are connected to $A$ and $B$), and so on. Forgetting the sixth island $F$ and the no more than 5 islands connected to it, at least $N-36 \geqslant 1$ islands will remain (among which none are connected to $A, B, C, D, E, F$). Choose any of these islands, call it $G$. Then among the islands $A, B, C, D, E, F, G$, there are no islands connected by a bridge. Contradiction.