Prove $2^{29}$ has exactly 9 distinct digits There is a problem that say: The number $2^{29}$ has exactly 9 distinct digits. Which digit is missing?
It is an Olympiad problem and I see it solved by using remainder modulo 9.
$2^{29}=536870912$  interesting!
My Question: Can we find any mathematically way to show $2^{29}$ has exactly 9 distinct digits?

You can write a mathematical statement that would serve as the basis for a proof, but it would involve tremendously more arithmetical calculation than simply calculating $2^{29}$ and examining the result.
Let $a_1=(2^{29}\bmod 10), a_2=\Bigl (\Bigl \lfloor \frac{2^{29}}{10}\Bigr \rfloor \bmod 10\Bigr ),\dots, a_k=\Bigl (\Bigl \lfloor \frac{2^{29}}{10^{k-1}}\Bigr \rfloor \bmod 10\Bigr ),\dots, a_9=\Bigl (\Bigl \lfloor \frac{2^{29}}{10^{8}}\Bigr \rfloor \bmod 10\Bigr )$
Now you just need to show that $a_i=a_j \iff i=j$
Much better you should perform the exponentiation and examine the result.
