Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students?
Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students?

| Solution | Criteria |
| :---: | :---: |
| Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm{a}_{4}, \mathrm{a}_{5}$ be the number of problems solved by the students. Note that no more than one student could have solved more than 4 problems, otherwise there would be two students who solved more than 9 problems in total. Without loss of generality, we can assume that $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$ are no more than 4. Then $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\mathrm{a}_{4}+\mathrm{a}_{5} \leq 4+4+4+\left(\mathrm{a}_{4}+\mathrm{a}_{5}\right) \leq 4+4+4+9=21$. Let's provide an example: $4,4,4,4,5$. In this example, all conditions of the problem are met and the sum of the numbers is 21. | ![](https://cdn.mathpix.com/cropped/2024_05_06_6614d7e4ddcc88834716g-1.jpg?height=483&width=798&top_left_y=1285&top_left_x=1121) |