For any integer $n\ge 2$ denote by $A_n$ the set of solutions of the equation
\[x=\left\lfloor\frac{x}{2}\right\rfloor+\left\lfloor\frac{x}{3}\right\rfloor+\cdots+\left\lfloor\frac{x}{n}\right\rfloor .\]
a) Determine the set $A_2\cup A_3$.
b) Prove that the set $A=\bigcup_{n\ge 2}A_n$ is finite and find $\max A$.

[i]Dan Nedeianu & Mihai Baluna[/i]
For any integer \( n \ge 2 \), denote by \( A_n \) the set of solutions of the equation
\[ x = \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor + \cdots + \left\lfloor \frac{x}{n} \right\rfloor. \]

### Part (a): Determine the set \( A_2 \cup A_3 \).

1. **Case \( n = 2 \):**
   \[ x = \left\lfloor \frac{x}{2} \right\rfloor. \]
   Let \( x = 2k \). Then,
   \[ 2k = \left\lfloor k \right\rfloor = k. \]
   This implies \( k = 0 \), so \( x = 0 \).

   Let \( x = 2k + 1 \). Then,
   \[ 2k + 1 = \left\lfloor k + \frac{1}{2} \right\rfloor = k. \]
   This implies \( k = -1 \), so \( x = -1 \).

   Therefore, \( A_2 = \{-1, 0\} \).

2. **Case \( n = 3 \):**
   \[ x = \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor. \]
   Let \( x = 6k \). Then,
   \[ 6k = \left\lfloor 3k \right\rfloor + \left\lfloor 2k \right\rfloor = 3k + 2k = 5k. \]
   This implies \( k = 0 \), so \( x = 0 \).

   Let \( x = 6k + 1 \). Then,
   \[ 6k + 1 = \left\lfloor 3k + \frac{1}{2} \right\rfloor + \left\lfloor 2k + \frac{1}{3} \right\rfloor = 3k + 2k = 5k. \]
   This implies \( k = -1 \), so \( x = -5 \).

   Let \( x = 6k + 2 \). Then,
   \[ 6k + 2 = \left\lfloor 3k + 1 \right\rfloor + \left\lfloor 2k + \frac{2}{3} \right\rfloor = 3k + 2k = 5k + 1. \]
   This implies \( k = -1 \), so \( x = -4 \).

   Let \( x = 6k + 3 \). Then,
   \[ 6k + 3 = \left\lfloor 3k + 1.5 \right\rfloor + \left\lfloor 2k + 1 \right\rfloor = 3k + 2k + 1 = 5k + 2. \]
   This implies \( k = -1 \), so \( x = -3 \).

   Let \( x = 6k + 4 \). Then,
   \[ 6k + 4 = \left\lfloor 3k + 2 \right\rfloor + \left\lfloor 2k + 1.33 \right\rfloor = 3k + 2k + 1 = 5k + 3. \]
   This implies \( k = -1 \), so \( x = -2 \).

   Let \( x = 6k + 5 \). Then,
   \[ 6k + 5 = \left\lfloor 3k + 2.5 \right\rfloor + \left\lfloor 2k + 1.67 \right\rfloor = 3k + 2k + 2 = 5k + 3. \]
   This implies \( k = -2 \), so \( x = -7 \).

   Therefore, \( A_3 = \{-7, -5, -4, -3, -2, 0\} \).

Combining the sets:
\[ A_2 \cup A_3 = \{-7, -5, -4, -3, -2, -1, 0\}. \]

### Part (b): Prove that the set \( A = \bigcup_{n \ge 2} A_n \) is finite and find \( \max A \).

1. **Finiteness of \( A \):**
   Since \( \left\lfloor \frac{x}{k} \right\rfloor \le \frac{x}{k} \) and \( \left\lfloor \frac{x}{k} \right\rfloor > \frac{x}{k} - 1 \), we have:
   \[ x = \sum_{k=2}^n \left\lfloor \frac{x}{k} \right\rfloor \le x \sum_{k=2}^n \frac{1}{k}. \]
   \[ x = \sum_{k=2}^n \left\lfloor \frac{x}{k} \right\rfloor > x \sum_{k=2}^n \left( \frac{1}{k} - \frac{1}{x} \right) = x \sum_{k=2}^n \frac{1}{k} - (n-1). \]

   Therefore,
   \[ x \sum_{k=2}^n \frac{1}{k} - (n-1) < x \le x \sum_{k=2}^n \frac{1}{k}. \]
   This implies:
   \[ 0 \le x \left( \sum_{k=2}^n \frac{1}{k} - 1 \right) < n-1. \]

   For \( n \ge 4 \), \( \sum_{k=2}^n \frac{1}{k} - 1 > 0 \), so:
   \[ 0 \le \frac{x}{n} < \frac{1 - \frac{1}{n}}{\sum_{k=2}^n \frac{1}{k} - 1}. \]

   For large \( n \) (e.g., \( n \ge 10 \)), we get:
   \[ 0 \le \frac{x}{n} < 1. \]
   Hence, \( x \in A_{n-1} \) for large \( n \), implying \( A \) is finite.

2. **Finding \( \max A \):**
   We need to check \( A_2, A_3, \ldots, A_9 \) to find the maximum value.

   - For \( n = 4 \):
     \[ x = \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor + \left\lfloor \frac{x}{4} \right\rfloor. \]
     The maximum solution is \( x = 23 \).

   - For \( n = 5 \):
     \[ 0 \le x < 15. \]

   - For \( n = 6 \):
     \[ 0 \le x < 12. \]

   - For \( n = 7 \):
     \[ 0 \le x < 11. \]

   - For \( n = 8 \):
     \[ 0 \le x < 10. \]

   - For \( n = 9 \):
     \[ 0 \le x < 10. \]

   - For \( n \ge 10 \):
     \[ 0 \le x < 10. \]

   Therefore, the maximum value in \( A \) is \( 23 \).

\(\blacksquare\)

The final answer is \( \boxed{ \{-7, -5, -4, -3, -2, -1, 0\} } \) for part (a) and \(23\) for part (b).