<s>[INST] 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?
 [/INST] 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 
$24^2 = 9.64 = 576 \\
\Rightarrow 3.24^2 = 1,728\\
24^3 = 24.576 = 3.8.576 = 3.4,608 = 13,824 \\
\Rightarrow 1024^3 = 10^9 + 72.10^6 + 1,728.10^3 + 13,824 \\
= 10^9 + 73.10^6 + 741.10^3 + 824 = 1,073,741,824 \\
\displaystyle \Rightarrow  10^{29} = \frac{1,073,741,824}{2} = 536,870,912$
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