Problem 2. (4 points) In a box, there are red, blue, yellow, and white balls, a total of 111. If 100 balls are drawn from the box with closed eyes, among them there will definitely be four balls, each two of which are of different colors. How many balls at least need to be drawn from the box (with closed eyes) to be able to assert with certainty that among them there are three balls, each two of which are of different colors?
Solution: If the balls of any color are no more than 11, then the remaining balls are at least $111-11=100$, and among them, there are no 4 balls of different colors. Therefore, the balls of each color are at least 12.

However, if 88 balls are taken, among them there will certainly be three balls of different colors (since $88=111-12 \cdot 2+1$, i.e., $12 \cdot 2-1$ balls are not taken, so at most one color is not taken at all).

We will show that taking 87 balls is not sufficient. Suppose there are 12 red balls, 12 blue balls, 12 green balls, and 75 white balls. In this case, the condition of the problem is satisfied: if we take 100 balls, we will not take more than 11 balls of one color, and therefore we will have taken a ball of each color. But if we take all the green and white balls, a total of 87 balls, there will be no three balls of different colors among them.