Seven students took a math exam. Every problem was solved at at most $3$ students. For every pair of students, there is at least one problem which was solved by both of these two students. What is the minimal number of problem this exam contains?
The answer is $\boxed{7}$.
Let $n$ be the number of problems in this exam and $m$ be the number of pairs $(P, SS)$ where $P$ denotes a problem and $SS$ is a pair of students who both solved $P$.
On one hand, it should hold that $m > 21$. This is because there are $\binom{7}{2}=21$ possible pairs of students and at least one $(P, SS)$ should exist for each pair.
On the other hand, it should be true that $m\le 3n$. This is because for each problem $P$, there are at most $\binom{3}{2}=3$ possible pairs of $(P, SS)$ exits. Otherwise, there will be more than $3$ students who have solved this problem.
Combining both facts lead to $n\ge 7$. And indeed one solution exists for $n=7$. Label these problems as $1$, $2$, $3$, $\cdots$, $7$. Then the solution is shown below where each set below lists the problem one student has solved. $$\{1,2,3\},\{1,4,7\}, \{1,5,6\}, \{2,5,7\}, \{2,4,6\},\{3,4,5\},\{3,6,7\}$$
This solution is generated from the diagram below:
