The number $2^{29}$ has a $9$-digit decimal representation that contains all but one of the $10$ (decimal) digits. Determine which digit is missing
1. **Determine the modulo 9 of \(2^{29}\):**
   \[
   2^{29} \equiv (2^6)^4 \cdot 2^5 \pmod{9}
   \]
   First, calculate \(2^6 \mod 9\):
   \[
   2^6 = 64 \quad \text{and} \quad 64 \div 9 = 7 \quad \text{remainder} \quad 1 \quad \Rightarrow \quad 64 \equiv 1 \pmod{9}
   \]
   Therefore,
   \[
   (2^6)^4 \equiv 1^4 \equiv 1 \pmod{9}
   \]
   Next, calculate \(2^5 \mod 9\):
   \[
   2^5 = 32 \quad \text{and} \quad 32 \div 9 = 3 \quad \text{remainder} \quad 5 \quad \Rightarrow \quad 32 \equiv 5 \pmod{9}
   \]
   Combining these results:
   \[
   2^{29} \equiv 1 \cdot 5 \equiv 5 \pmod{9}
   \]

2. **Sum of digits of \(2^{29}\):**
   Since \(2^{29}\) has a 9-digit decimal representation containing all but one of the 10 decimal digits, the sum of all digits from 0 to 9 is:
   \[
   0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
   \]
   Let \(m\) be the missing digit. The sum of the digits of \(2^{29}\) is:
   \[
   45 - m
   \]

3. **Modulo 9 condition:**
   The sum of the digits of a number is congruent to the number itself modulo 9. Therefore:
   \[
   45 - m \equiv 2^{29} \equiv 5 \pmod{9}
   \]
   Solving for \(m\):
   \[
   45 - m \equiv 5 \pmod{9}
   \]
   Subtract 5 from both sides:
   \[
   45 - m - 5 \equiv 0 \pmod{9}
   \]
   Simplify:
   \[
   40 - m \equiv 0 \pmod{9}
   \]
   Therefore:
   \[
   40 \equiv m \pmod{9}
   \]
   Since \(40 \div 9 = 4\) remainder \(4\):
   \[
   40 \equiv 4 \pmod{9}
   \]
   Thus, the missing digit \(m\) is:
   \[
   m = 4
   \]

The final answer is \(\boxed{4}\).