2026/040) The number $2^{29}$ has exactly 9 distinct digits. Find the missing digit.

Let us work mod 9.

We have $2^3 \equiv -1 \pmod 9$

Hence $2^{27} \equiv (-1)^9  \equiv -1 \pmod 9$ 

Hence $2^{29} \equiv (-1) * 4  \equiv -4 \pmod 9$

If we have all the digits(once) that is 10 digits  then sum of digit is 45 so it is divisible by 9 or 0 mod 9 

So removing 4 we shall have -4 mod 9. 

So  missing digit is 4