TITLE: Problem from Iran Olympiad?
QUESTION [5 upvotes]: Does there exist a positive integer that is a power of $2$ and we get another power of $2$ by swapping its digits? Justify your answer.
I gussed the answer is no.
Let $\overline{a_n ,...,a_1 ,a_0}$ be in the form $2^k$ and $\sigma\in S_n$ be sch that $\overline{a_{\sigma(n)},...,a_{\sigma_(0)}}$ also be in the form of $2^m$. I couldnt to find any contradition. it is clear that $\vert m-k\vert$ must be 1 or 3 or 2.

REPLY [4 votes]: Let $a=2^x$ and $b=2^y$ where $x,y$ are distinct and $b$ is the rearrangement of $a$.
We know that $|x-y|<4$ otherwise $a$ and $b$ would have a different number of digits. Therefore $|x-y|=1,2,3$ 
Successive powers of 2 are congruent in $ mod(9)$ to $1,2,4,8,7,5,1,2,4...$. This implies that powers of 2 differing by factors of $2^1,2^2,2^3$ cannot be congruent.
Yet if $a,b$ have the same digits then $b=a \; mod(9)$ which means $|x-y|\neq 1,2,3$.
Due to the contradiction we can therefore conclude that there are no such $a,b$