Shapovalov A.V.

A monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines?
Let's put the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed no more than one tangerine to each monkey, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. Here's how we can satisfy 45 monkeys:

5 monkeys eat a pear, a banana, and a tangerine;

15 monkeys - a pear, a peach, and a tangerine;

25 monkeys - a peach, a banana, and a tangerine.

## Answer

45. 

## [ Mathematical Logic (Miscellaneous). $\quad]$  Problem 65111 Topics: [ Examples and Counterexamples. Constructions]  Author: Poilipsky O.K.  Around a round table sit 2015 people, each of whom is either a knight or a liar. Knights always tell the truth, liars always lie. Each person was given one card, with a number written on each card; all the numbers on the cards are different. After looking at their neighbors' cards, each person said: "My number is greater than that of each of my two neighbors." Then $k$ of those present said: "My number is less than that of each of my two neighbors." For what maximum $k$ could this have happened?

## Solution

Let $A$ and $B$ be the people who received the cards with the largest and smallest numbers, respectively. Since both of them said the first phrase, $A$ is a knight, and $B$ is a liar. Therefore, neither of them could have said the second phrase. Consequently, $k \leq 2013$.

A situation where the remaining 2013 people could say the second phrase is possible. Suppose the people sitting at the table received (clockwise) cards with the numbers 1, 2, 3, ..., 2015; with the card numbered 2015 going to a knight, and the rest to liars. Then the first phrase can be said by everyone, and the second by everyone except those with cards numbered 1 and 2015.

## Answer

At $k=2013$.