8. (5 points) On the Island of Eye, there live knights who always tell the truth, and liars who always lie. In addition, all residents of the island are either blue-eyed or brown-eyed. One day, 100 residents of the island met, and each one said to each other one of two phrases: "You are a liar" or "You are blue-eyed," and the phrase "You are a liar" was more than half. How many brown-eyed residents could there be on the island? Find all possible options and prove that there are no others.
Answer: The answer is from 46 to 54.

Solution: Note that at the meeting, there could not have been brown-eyed knights and blue-eyed liars, because no knight can say the given phrase, and no liar can say it either. Therefore, the islanders at the meeting are divided into blue-eyed knights and brown-eyed liars. Thus, the phrase "You are a liar" was said by people with different eye colors to each other, while the phrase "You have blue eyes" was said by people with the same eye color.

Let the number of blue-eyed knights be $50+k$, and the number of brown-eyed liars be $50-k$. We can introduce such notation because the sum of these numbers is 100. Here, $k$ is an integer, but not necessarily a natural number.

Then, the total number of phrases "You are a liar" is $2(50+k)(50-k)=2 \cdot 50^{2}-2 k^{2}$. We know that the number of these phrases is more than half, that is, more than $100 \cdot 99 / 2=4950$. Therefore, $5000-2 k^{2}>4950$, from which $2 k^{2}<50$, that is, $|k|<5$. This means that the number of liars differs from 50 by no more than 4, that is, it is in the range from 46 to 54, and all situations are clearly possible.

## 8th grade

2nd variant