9-6-1. Six pirates - a captain and five members of his crew - are sitting around a campfire facing the center. They need to divide a treasure: 180 gold coins. The captain proposes a way to divide the treasure (i.e., how many coins each pirate should receive: each pirate will receive a non-negative integer number of coins; different pirates may receive different numbers of coins). After this, the other five pirates vote on the captain's proposal. A pirate will vote "for" only if they receive more coins than each of their two neighbors. The proposal is accepted if at least three out of the five crew members vote "for."

What is the maximum number of coins the captain can get with this method of dividing the treasure?
Answer: 59.

Solution variant 1. Let's number the pirates clockwise, starting from the captain: First, Second, ..., Fifth. The main observation in this problem is:

two pirates sitting next to each other cannot both vote "for": one of them will not vote "for" because the captain offers no more to him than to his neighbor.

Therefore, for three to vote "for," they must be the First, Third, and Fifth - the proof of this fact is detailed in the comment. Then, if the captain keeps $x$ coins for himself, he must offer at least $x+1$ coins to the First and Fifth, and at least 1 coin to the Third, so they vote "for." Thus, the total number of coins should be at least $3x+3$, from which $3x+3 \leqslant 180, x \leqslant 59$.

Note that so far we have only proven that the captain cannot get more than 59 coins. From our reasoning, it is easy to construct an example where he can get exactly 59: it is sufficient to offer 60 coins to the First and Fifth, one coin to the Third, and nothing to the Second and Fourth.

Comment. The intuitive statement about the First, Third, and Fifth is rigorously proven through partitioning into pairs. From the pair First-Second, no more than one will vote "for," just as from the pair Third-Fourth. Therefore, the Fifth must definitely vote "for." Similarly, the First must vote "for." But then the Second and Fourth do not vote "for," so the Third votes "for."