Problem 2. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all possible options and prove that there are no others.
Answer: 30.

Solution. Consider any four consecutive people. If there were at least 3 knights among them, the leftmost knight would definitely lie, which is impossible. If there were at least 3 liars, the leftmost liar would definitely tell the truth, which is also impossible. Therefore, in every group of four consecutive people, there are exactly 2 knights and 2 liars. By dividing all the people into 15 such groups, we get that there are $15 \cdot 2 = 30$ knights.

Another solution. Notice that there cannot be only knights or only liars sitting at the table. Then there must be a pair of knight-liar sitting next to each other in that exact order. Consider the 4 cases of who can sit to the right of this pair.

Case 1. Two knights sit to the right.

Then we get the sequence

$$
\text { knight-liar-knight-knight, }
$$

which contradicts the condition, as the first knight lied.

Case 2. Two liars sit to the right.

Then we get the sequence

$$
\text { knight-liar-liar-liar. }
$$

Whoever sits immediately after this sequence, the first liar will tell the truth, which contradicts the condition.

Case 3. A knight and a liar sit to the right in that exact order.

Then we get the sequence

$$
\text { knight-liar-knight-liar. }
$$

Consider the first liar. He lies, so among the next three, there must be at least two knights. From this, we uniquely get

knight-liar-knight-liar-knight

Now consider the second knight. He tells the truth, so the sixth person in the sequence will be a liar:

knight-liar-knight-liar-knight-liar.

Continuing this reasoning, we get an alternating arrangement where there are an equal number of knights and liars.

Case 4. A liar and a knight sit to the right in that exact order. Then we get the sequence

$$
\text { knight-liar-liar-knight. }
$$

Consider the first liar. He lies, so among the next three, there must be at least two knights. From this, we uniquely get

knight-liar-liar-knight-knight

Now consider the second knight. He tells the truth, so the sixth and seventh people in the sequence will be liars

knight-liar-liar-knight-knight-liar-liar.

Continuing this reasoning, we get an arrangement where there are an equal number of knights and liars.

## Criteria

The highest applicable criterion is used:

7 6. Any complete solution to the problem.

7 6. The problem is correctly solved for the case where both knights and liars are present at the table.

4 6. Both possible cases of arranging knights and liars are considered in the solution, but there is no justification for why there are no other arrangements.

2 6. One of the two possible cases of arranging knights and liars is considered in the solution, but there is no justification for why there are no other arrangements.

1 6. Only the correct answer is present.