What digit must be put in place of the "$?$" in the number $888...88?999...99$ (where the $8$ and $9$ are each written $50$ times) in order that the resulting number is divisible by $7$? 

(M . I. Gusarov)
1. **Understanding the problem**: We need to find the digit \( x \) that makes the number \( 888\ldots88x999\ldots99 \) (where there are 50 eights and 50 nines) divisible by 7.

2. **Using properties of divisibility by 7**: We know that \( 7 \mid 111111 \implies 1111\ldots1 \) (50 times) \( \equiv 4 \mod 7 \). This is because \( 1111\ldots1 \) (50 times) can be written as \( \frac{10^{50} - 1}{9} \), and we need to find its equivalence modulo 7.

3. **Simplifying the number**: The number can be broken down as follows:
   \[
   888\ldots88x999\ldots99 = 8 \times 10^{51} + 8 \times 10^{50} + \ldots + 8 \times 10 + 8 \times 1 + x \times 10^{50} + 9 \times 10^{49} + 9 \times 10^{48} + \ldots + 9 \times 10 + 9 \times 1
   \]

4. **Modulo 7 calculations**:
   - \( 10 \equiv 3 \mod 7 \)
   - \( 10^2 \equiv 3^2 = 9 \equiv 2 \mod 7 \)
   - \( 10^3 \equiv 3 \times 2 = 6 \equiv -1 \mod 7 \)
   - \( 10^4 \equiv 3 \times (-1) = -3 \equiv 4 \mod 7 \)
   - \( 10^5 \equiv 3 \times 4 = 12 \equiv 5 \mod 7 \)
   - \( 10^6 \equiv 3 \times 5 = 15 \equiv 1 \mod 7 \)
   - This cycle repeats every 6 powers.

5. **Summing the contributions**:
   - For \( 8 \times 10^{51} \equiv 8 \times 10^3 \equiv 8 \times (-1) \equiv -8 \equiv -1 \mod 7 \)
   - For \( 8 \times 10^{50} \equiv 8 \times 10^2 \equiv 8 \times 2 \equiv 16 \equiv 2 \mod 7 \)
   - Continuing this pattern, we sum the contributions of all 50 eights and the digit \( x \).

6. **Combining the results**:
   - The sum of the contributions of the 50 eights:
     \[
     8 \times (10^{51} + 10^{50} + \ldots + 10^1 + 10^0) \equiv 8 \times (1 + 3 + 2 + (-1) + 4 + 5) \times 8 \equiv 8 \times 14 \equiv 8 \times 0 \equiv 0 \mod 7
     \]
   - The sum of the contributions of the 50 nines:
     \[
     9 \times (10^{49} + 10^{48} + \ldots + 10^0) \equiv 9 \times (1 + 3 + 2 + (-1) + 4 + 5) \times 9 \equiv 9 \times 14 \equiv 9 \times 0 \equiv 0 \mod 7
     \]
   - The contribution of \( x \):
     \[
     x \times 10^{50} \equiv x \times 2 \mod 7
     \]

7. **Setting up the equation**:
   \[
   0 + x \times 2 \equiv 0 \mod 7 \implies 2x \equiv 0 \mod 7 \implies x \equiv 0 \mod 7
   \]

8. **Conclusion**: The digit \( x \) must be 0 to make the number divisible by 7.

The final answer is \( \boxed{0} \).