A set of $8$ problems was prepared for an examination. Each student was given $3$ of them. No two students received more than one common problem. What is the largest possible number of students?
1. **Generalization and Problem Setup:**
   We are given a set of \( n \) problems and each student is given \( m \) problems such that no two students share more than one common problem. We need to find the largest possible number of students, denoted as \( S \).

2. **Deriving the Upper Bound:**
   Suppose a particular problem is given to \( x \) students. Each of these \( x \) students will receive \( m-1 \) additional problems. Therefore, the total number of problems required is at least \( x(m-1) + 1 \).

   This gives us the inequality:
   \[
   n \geq x(m-1) + 1
   \]
   Solving for \( x \), we get:
   \[
   x \leq \left\lfloor \frac{n-1}{m-1} \right\rfloor
   \]

3. **Counting the Assignments:**
   Each problem can be assigned to at most \( \left\lfloor \frac{n-1}{m-1} \right\rfloor \) students. Therefore, the total number of problem assignments is at most:
   \[
   n \left\lfloor \frac{n-1}{m-1} \right\rfloor
   \]

4. **Calculating the Number of Students:**
   Since each student is assigned \( m \) problems, the total number of students \( S \) is:
   \[
   S \leq \left\lfloor \frac{n \left\lfloor \frac{n-1}{m-1} \right\rfloor}{m} \right\rfloor
   \]

5. **Applying the Formula to the Given Problem:**
   For the given problem, \( n = 8 \) and \( m = 3 \):
   \[
   \left\lfloor \frac{8-1}{3-1} \right\rfloor = \left\lfloor \frac{7}{2} \right\rfloor = 3
   \]
   Therefore, the total number of problem assignments is:
   \[
   8 \times 3 = 24
   \]
   And the number of students is:
   \[
   S \leq \left\lfloor \frac{24}{3} \right\rfloor = 8
   \]

The final answer is \( \boxed{ 8 } \).