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Starting from the number $1$ we write down a sequence of numbers where the next number in the sequence is obtained from the previous one either by doubling it or rearranging its digits (not allowing the first digit of the rearranged number to be $0$) for instance a sequence might begin $$1,2,4,8,16,61,122,212,424,\ldots$$ Is it possible to make a sequence that ends in $1000000000$ and a sequence that ends in $9876543210$. Please show me how and if there is any working.

Empiricist
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0327
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  • 1000000000 shouldn't be hard, if you have just one non-zero digit, could it have been rearranged from any number? EDIT: oh, ok, maybe not that obvious. – Ennar Aug 16 '15 at 09:21
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    Hint: $1000 = 8\cdot 125$. – Daniel Fischer Aug 16 '15 at 09:28
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    You can try to work backwards: halve the number successively a few times. Rearrange. Repeat. If you're lucky you can eventually rearrange it into a power of $2$. At least I think that that method is slightly more constructive. – Arthur Aug 16 '15 at 10:17
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    @DanielFischer Combine that with 8⋅625 and you will have covered all exponents except from three possibilities. 10 and 100 are impossible to reach. 100000 is possible if you go via 256, 265, 1060, 1600, 51200, 12500. – kasperd Aug 16 '15 at 13:31
  • I'm sorry but I'm new to higher mathematics. Why exactly are the numbers written with dollar signs? Is there some place where I could look this hole notation up? Never have seen \Idots either – BlueWizard Aug 16 '15 at 13:45
  • Oh okay. Basic LaTeX notation. Sorry that i haven't seen this myself :D Thanks fro your help – BlueWizard Aug 16 '15 at 13:57
  • Arguably, the question as worded does not prohibit $(00)1->100$ – March Ho Aug 16 '15 at 14:24
  • misread question – Neil W Aug 17 '15 at 06:13
  • I'm having a lot of fun trying to write a recursive function to find an answer to this question, I still need to add some memoization or tail call optimization to allow it to work with larger number but for now: http://jsfiddle.net/20Lmrun9/ shows that for the number 46 you could get there by 1*2=2 2*2=4 4*2=8 8*2=16 16*2=32 32->23 23*2=46, by the number 125 you are already dealing with too much recursion. – Jason Sperske Aug 17 '15 at 12:28
  • Here is a slight update http://jsfiddle.net/9emop3sa/3/ you can now pass it console.log(solve(1, 500, ' 1 ', true)); and get 1 2 4 8 16 32 64 128 256 512 125 250 500 – Jason Sperske Aug 17 '15 at 15:45
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    This has been turned into a challenge on CodeGolf.SE. :) – Martin Ender Aug 19 '15 at 08:38
  • So many interesting questions here - I'm not even sure of the rough order of growth of $f(n)=#{x\leq n:$ there exists a double-and-rearrange sequence reaching $x}$... – Steven Stadnicki Aug 19 '15 at 18:36
  • I got curious enough to ask my order-of-growth question: http://math.stackexchange.com/questions/1403089/ – Steven Stadnicki Aug 19 '15 at 19:29

1 Answers1

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We first note that we can attain $1000$ from $1$ by $$1,2,4,8,16,32,64,128,256,512,125,250,500,1000$$ Therefore $1000^n$ can be attained by performing the same operations as above with $1$ replaced by $1000^{n-1}$.

On the other hand, note that rearranging the digits does not change the remainder when divided by $3$, and $2^n \neq 0 \pmod{3}$. Therefore the sequence would not reach any multiple of $3$.

Empiricist
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