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?
 A: That $2^{29}$ has exactly nine distinct digits is a coincidence, and a fairly appropriate one. One (a very good one) mathematical proof that $81619^2$ contains only two distinct digits, is calculating it : $81619^2 = 6661661161$. Even though sometimes these do yield patterns that are interesting, my personal take is that this is a one-off, a coincidence and a nice triviality to put in the back page of your notebook.
What we can do , is show that it has nine digits, and find the first few digits on both left and right. On the right, we would note that $2^{20} \equiv 1 \mod 100$ (checking remainders mod $25$), so the last two digits match with the first two digits of $512$, which give the last two digits as $12$. 
The first digit would come from the fact that $\log_{10}(2^{29}) = 29 \log_{10}2$. Noting that $\log_{10}(2) \approx 0.30103$ (for mathematical proof, use Taylor series) we get that $\log_{10}(2^{29}) \approx 10^{8.73}$, so from here you get that the number has $9$ digits, and the first digit is $10^{0.73}$, which again using a couple of Taylor series $e^{0.73 \ln 10}$ you can see to be $5$. Quite a bit of effort, but at least this method of deriving the digits is different from multiplying $2$ again and again.

Knowing that it has exactly $9$ distinct digits, we can actually calculate the missing digit using the fact that the sum of digits must be between $36$ and $45$. This can be located using the remainder modulo $9$.
Note that $2^3 = 8 \equiv -1 \pmod 9$. Therefore, $2^{27} \equiv (2^3)^{9} \equiv -1 \pmod 9$. Therefore, $2^{29} \equiv -4 \equiv 5 \pmod{9}$.
Therefore, the sum of digits of $2^{29}$ must leave remainder $5$ upon division by $9$, and be between $36$ and $45$. Thus, it is $41$. Now $45-41 = \color{blue}{4}$ is the missing digit.
A: 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.
A: We know that $2^{10}=1024$, so we can calculate the value of $2^{29}$ by hand as follows
$2^{30} = 1024^3 = (10^3 + 24)^3 = 10^9 + 3.24.10^6 + 3.24^2.10^3 + 24^3$
Now $24=3.8$, so 
$24^2 = 9.64 = 576 \\
\Rightarrow 3.24^2 = 1,728\\
24^3 = 24.576 = 3.8.576 = 3.4,608 = 13,824 \\
\Rightarrow 1024^3 = 10^9 + 72.10^6 + 1,728.10^3 + 13,824 \\
= 10^9 + 73.10^6 + 741.10^3 + 824 = 1,073,741,824 \\
\displaystyle \Rightarrow  10^{29} = \frac{1,073,741,824}{2} = 536,870,912$
