5. On the table, there are 7 clean sheets of paper. Every minute, Vasya chooses any 4 of them and draws one star on each of the chosen sheets. Vasya wants to end up with a different number of stars on each sheet (that is, there should be no two sheets with the same number of stars drawn on them). What is the minimum total number of stars he will have to draw to achieve this?
Answer: 28

Solution: Estimation. We will prove that to achieve the desired result, the number of moves must be at least 7. If 5 or fewer moves are made, then on each sheet there are no more than 5 stars, meaning there are only 6 different variants of the number of stars (from 0 to 5), while there are 7 sheets. By the Pigeonhole Principle, there will be at least two sheets with the same number of stars, which contradicts the condition. Suppose 6 moves are made, then the maximum number of stars on one sheet is 6. However, different numbers of stars on all sheets can only be achieved with the unique set of values: \(0, 1, 2, 3, 4, 5, 6\), which sum up to 21 stars, while by assumption, there are \(6 \cdot 4 = 24\) stars. Therefore, the number of moves must indeed be at least 7.

Example: We will show that 7 moves are sufficient.

The first 4 moves select sheets №1, №2, №3, №4.

Fifth move: sheets №1, №2, №3, №5.

Sixth move: sheets №1, №2, №5, №6.

Seventh move: sheets №1, №5, №6, №7.

In total, 7 stars are drawn on sheet №1, 6 stars on sheet №2, 5 stars on sheet №3, and so on, down to 1 star on sheet №7.

Comment: It is proven that 5 or fewer moves are insufficient - 2 points. It is proven that 6 moves are insufficient - 2 points. An example is provided of how to draw 28 stars (it is sufficient to describe each of the 7 moves) - 3 points. Points are added according to the listed criteria. Only the answer - 1 point.