# Problem 3. (3 points)

30 people are standing in a row, each of them is either a knight, who always tells the truth, or a liar, who always lies. They were numbered from left to right, after which each person with an odd number said: "All people with higher numbers than mine are liars," and each person with an even number said: "All people with lower numbers than mine are liars."

How many liars could there be? If there are multiple correct answers, list them in ascending order separated by semicolons.
# Answer: 28

## Solution:

We will prove that under the given conditions, there must be exactly two knights, and all others are liars.

Consider the people with odd numbers. If a person with an odd number $n$ is telling the truth, then the person with the odd number $n+2$ must also be telling the truth, since all people with numbers greater than his are liars. On the other hand, if the person with number $n$ is telling the truth, the person with number $n+2$, according to his statement, is also a liar. This leads to a contradiction. This means that the only person with an odd number who can be telling the truth is the one with the largest odd number, which is the second-to-last person.

Similarly, it can be proven that the only person with an even number who can be a knight is the person numbered 2.

At the same time, since the first and the last are definitely liars, the second and the second-to-last are telling the truth.

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