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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.

Ramand
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  • This is an interesting question. What have you tried already? – daOnlyBG Jan 11 '15 at 20:40
  • Have any work to show us what you have done? – Julian Rachman Jan 11 '15 at 20:40
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    For two powers of $2$ to have the same number of digits, one of them must be $2,4,$ or $8$ times the other. Also they must both have the same remainder when divided by $9$. This should be enough for you to show a contradiction. – TonyK Jan 11 '15 at 20:44
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    @TonyK This is true unless we are allowed to rearrange so that $0$s go to the front. – Peter Woolfitt Jan 11 '15 at 20:45
  • Yes, the real question is whether you are allowed to put the $0$ digits in the front. The digit sum works if you are not allowed to move zero digits to the front. – Thomas Andrews Jan 11 '15 at 20:50
  • @TZakrevskiy: Yes, I didn't realise that. But a power of $2$ can never equal $3$ (or $6$) mod $9$,so I think my argument survives. – TonyK Jan 11 '15 at 20:57
  • @TonyK already found it and deleted the comment=) – TZakrevskiy Jan 11 '15 at 20:58
  • If we are allowed to put leading zeroes in the rearrangement, then the factor has to be of the form $64^k$, $k\in \Bbb N$ – TZakrevskiy Jan 11 '15 at 21:21

1 Answers1

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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 $a,b$

Gridley Quayle
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  • Good answer. basically $2^n,2^{n+1},2^{n+2},2^{n+3}$ are all different mod $9$, but if we could get one by rearranging the other they would be the same mod 9. – Asinomás Jan 12 '15 at 01:13