8.5. A round table was sat at by 10 people - liars and knights. Liars always lie, and knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, each one said: “I have more coins than my right neighbor.” What is the maximum number of knights that could have been sitting at the table?
Answer: 6.

Solution: After the coins are passed, each person sitting at the table can have 0, 1, or 2 coins. Note that 3 knights cannot sit in a row. Indeed, let knights $A, B, C$ sit next to each other, with $B$ sitting to the right of $A$, $C$ to the right of $B$, and $D$ to the right of $C$. If $A$ has $x$ coins, $B$ has $y$ coins, $C$ has $z$ coins, and $D$ has $t$ coins, then the inequality $x > y > z > t$ would hold, which is impossible for the numbers 0, 1, 2. Therefore, among any 3 people sitting in a row, there is a liar. Choose any liar sitting at the table (such a person exists because there are 3 people sitting in a row), and divide the remaining people into 3 groups of 3 people each sitting next to each other. In each of these groups, there is a liar. Thus, there are at least $1 + 3 = 4$ liars at the table, and therefore no more than 6 knights. We will show that 6 knights could be sitting at the table. Suppose they are seated as follows: -K-K-L-L-K-K-L-K-K-L-. And the knights sitting next to each other exchange coins, while the liars give their coins to the people sitting to their right. Then the number of coins each person has will be -K(2)-K(1)-L(0)-L(1)-K(2)-K(1)-L(0)-K(2)-K(1)-L(0)-. And the knights will tell the truth, while the liars will lie.

Comment: It is proven that no more than 6 knights can sit at the table - 5 points. It is proven that 6 knights can sit at the table - 2 points.