algebra precalculus - Finding the missing digit in a an Olympiad question.

The question is problem N0. $5$

The question states that

$2^{29}$ is a $9$ digit number with distinct digits. Which digit is missing?

$(a.) \space0 \quad\quad (b.) \space3 \quad\quad (c.)\space4 \quad\quad (d)\space5 \quad\quad (e.) \space 7$

The answer provided is a brute method and does not look very appealing. I tried to solve this without directly computing the number. Can anyone help me ??

My Attempt :

It is obvious to see that the number must include $0$ and $9$ , otherwise it would be divisible by $9$ which cannot be true .

Using basic modular arithmetic , it also easy to show that

$$2 ^ {29} \equiv 2 \mod 10$$

Hence the number also contains $2$.

Similarly to our first observation , we may also conclude that the number must also contain $3$ and $6$ , otherwise it would be divisible by $3$ , which is again not possible.

Hence the only missing digits could be $(1,4,5,7,8)$.

But I cannot further deduce any new constraints . Is it possible to further reduce the number of cases ?

Answer

Following the suggestion by Lulu, we have that

$$2^{29}\equiv 5 \pmod 9$$

wich is compatible with $4$ missing indeed

$$\left(\sum_{i=0}^9 i\right) \, -4=45-4=41 \equiv 5 \mod 9$$