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As the title says, I want to prove that there is a natural number $k$ such that $2^k$ is starting with $999$. Can you help me please ?

Watson
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Claudiu Bbn
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  • See http://math.stackexchange.com/questions/544214, http://math.stackexchange.com/questions/1261980, http://math.stackexchange.com/questions/13131 – Watson Sep 08 '16 at 13:07
  • Related: http://math.stackexchange.com/questions/131281 – Watson Sep 08 '16 at 13:16
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    I think this is not a duplicate, since we do not need to prove that any prefix is achievable, but just the prefix $999$, that is very specific. In fact, we may also find the least power of two with such property. – Jack D'Aurizio Sep 08 '16 at 13:29
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    Due to my previous comment, I am voting for reopening. – Jack D'Aurizio Sep 08 '16 at 13:46
  • @JackD'Aurizio I answered on the other question but it's about a general proof of existence . Finally , I don't know if it was a good idea ... –  Sep 08 '16 at 15:04
  • @igael, is there any solution without using log ? i know it can be solved with Dirichlet's box principle , but i don't know the actual solution – Claudiu Bbn Sep 08 '16 at 17:36
  • @ClaudiuBbn: very interesting, but I don't see for the moment why we could assume that all the results are possible ( even if cannot find an exception on the 50000 1st powers ) –  Sep 08 '16 at 18:02
  • @ClaudiuBbn : to prove that $k \log_{10}(2)$ can be extremely close to an integer, you can use http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset/272713#272713 (using the pigeonhole's principle). See also http://math.stackexchange.com/questions/136665 – Watson Sep 08 '16 at 18:09

1 Answers1

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If $\log_{10}(2^k) = k \log_{10}(2)$ is extremely close to an integer, but less than such integer, we are happy. To get some working values for $k$, it is enough to compute the continued fraction of $$\frac{\log 2}{\log 10}=[0; 3, 3, 9, 2, 2, 4, 6, 2, 1, 1, 3, 1, 18,\ldots ]. $$

By considering the expansion of $[0; 3, 3, 9, 2, 2, 4, 6, 2, 1, 1, 3]$ we get: $$ 254370\cdot\log_{10}(2) = 76572.999997\ldots $$ hence the number $\color{red}{2^{254370}}$ starts with the digits $999$ as wanted.
The same happens with $\color{red}{2^{13301}}$ that is associated with the continued fraction $[0; 3, 3, 9, 2, 2, 4, 6]$.

The least power of two with the wanted property should be $\color{red}{2^{2621}}$ that is associated with the continued fraction $[0, 3, 3, 9, 2, 2, 5]$.

Jack D'Aurizio
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  • I got $2^{2621}$ . And $4757$, $6893$ , $9029$ , $11165$ too –  Sep 08 '16 at 13:29
  • @igael: yes, by playing a bit with the given continued fraction we should get that $2^{2621}$ is the minimal example. – Jack D'Aurizio Sep 08 '16 at 13:31
  • I used a trivial method : the script truncated the last digit to keep only 7 in the loop and deep tested the selected powers ( edit : I tried your method , nice idea :) But may we find something less computational ? I have a doubt ) –  Sep 08 '16 at 13:34
  • @igael: yes, we are following two slightly different ways for providing good rational approximations of $\log_{10}(2)$, but the outcomes have to be the same. – Jack D'Aurizio Sep 08 '16 at 13:37