Problem 10.6. On an island, there are two tribes: knights and liars. Knights always tell the truth, while liars always lie. One day, 80 people sat around a round table, and each of them stated: "Among the 11 people sitting after me in a clockwise direction, there are at least 9 liars." How many knights are sitting at the round table? List all possible options.
Answer: 20.

Solution. First, we prove that among 12 consecutive people, there are no more than 3 knights. Suppose this is not the case. Consider the first knight in this group. Among the 11 people sitting after him in a clockwise direction, there are at least 3 knights, which contradicts the problem's condition.

Next, suppose there are 4 consecutive liars. Number all people clockwise from 1 to 80 so that these liars have numbers from 1 to 4.

Consider the first liar. From his statement, it follows that among people with numbers from 2 to 12, there are at least 3 knights, but from the above proof, we know there are no more than 3 knights. Since the first three of them are definitely liars, among people with numbers from 5 to 12, there are exactly 3 knights.

Now consider the group of people with numbers from 5 to 16. Among them, there are no more than 3 knights, but among people with numbers from 5 to 12, there are exactly 3 knights. Therefore, people with numbers from 13 to 16 are liars. Thus, we have found a new set of 4 consecutive liars, shifted by 12 people in a clockwise direction from the original set.

Continuing similar reasoning for the new set of 4 consecutive liars, we will get another such set, then another, and so on. Since the greatest common divisor (GCD) of 80 and 12 is 4, all people at the table will fall into such sets. But they cannot all be liars, which is a contradiction.

Thus, our assumption was incorrect, and in any set of 4 consecutive people, there is at least one knight. But we know that among 12 consecutive people, there are no more than 3 knights. Therefore, in any set of 4 consecutive people, there is exactly one knight, meaning there are exactly 20 knights at the table.

It remains to construct an example of such a seating arrangement of knights and liars. Let all people with numbers divisible by 4 be knights, and all others be liars. It is not difficult to verify that this example satisfies the problem's condition.

Another solution. Let's call the sorrow of a certain person the number of liars among 12 people, including himself and 11 others sitting consecutively after him in a clockwise direction. According to the problem, the sorrow of each knight is at least 9, and the sorrow of each liar is at most 9. We will prove that the sorrow of all people is exactly 9.

First, note that the sorrow of adjacent people cannot differ by more than 1. Indeed, when moving to the next person in a clockwise direction, one person is removed from their set of 12, and one is added. Therefore, the number of liars can decrease by no more than 1 and increase by no more than 1.

Now suppose that a certain knight has a sorrow greater than 9. Then the sorrow of the next person is no less: from their set of 12 people, this knight is removed, and an unknown person is added, i.e., the number of liars does not decrease. Therefore, a liar cannot follow such a knight, and we have found another knight with a sorrow greater than 9. Clearly, we can continue this process in a circle and arrive at a contradiction: all at the table cannot be knights.

Similarly, it can be proven that a liar cannot have a sorrow less than 9. Indeed, the sorrow of the next person after such a liar cannot be greater than the liar's; therefore, this must also be a liar with a sorrow less than 9; continuing this, we arrive at the conclusion that all at the table are liars, which is also impossible.

Now it is clear that the number of liars among any 12 consecutive people is equal to the sorrow of the first of them, i.e., 9. From this, it is easy to understand that two people at a distance of 12 (i.e., with 11 others between them) have the same type (knight or liar), since adding either of them to the 11 intermediate people should result in 9 liars. This means that the sequence of knights and liars is periodic with a period of 12; but since the GCD of 80 and 12 is 4, it is also periodic with a period of 4.

Dividing the entire table into identical sets of four, it is easy to see that in each of them there are 3 liars (otherwise, in a group of three adjacent sets of four, there would not be 9). Therefore, there is one knight in each, meaning there are 20 knights in total.

A periodic seating arrangement of knights with three liars in between, as is easy to verify, satisfies the condition.