Problem 1146

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 $\boxed{4}$.