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In this note the famous mathematical physicists Freeman Dyson gives an example of a true statement that is impossible to prove. Or so he states. The statement is as follow:

Numbers that are exact powers of two are 2, 4, 8, 16, 32, 64, 128 and so on. Numbers that are exact powers of five are 5, 25, 125, 625 and so on. Given any number such as 131072 (which happens to be a power of two), the reverse of it is 270131, with the same digits taken in the opposite order. Now my statement is: it never happens that the reverse of a power of two is a power of five.

He gives some arguments to support his claim and it indeed seems to be probable that his statement is correct. However, another part of a story is a possibly to prove this statement. So here is my question: Can anyone prove Dyson's statement, or at least give some thought along what lines this statement can be proved.

Artem
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    $2^0=1$ backwards is $1=5^0$ :) –  Apr 15 '14 at 02:17
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    Thank you, Captain Obvious @user142299. – 2012ssohn Apr 15 '14 at 02:18
  • On a more serious note, I highly doubt there is such a number. That power of $2$ in question would have to start with $5260$, $5213$, $5265$, or $5218$. – 2012ssohn Apr 15 '14 at 02:19
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    @2012ssohn Also on a more serious note, $5260135901548373507240989882880128665550339802823173859498280903068732154297080822113666536277588451226982968856178217713019432250183803863127814770651880849955223671128444598191663757884322717271293251735781376$ is a power of two. –  Apr 15 '14 at 02:24
  • Did you happen to check whether or not the reverse is a power of $5$? – MPW Apr 15 '14 at 02:25
  • @MPW No it is not. I have checked up to (and not including) $2^{5000}$. –  Apr 15 '14 at 02:28
  • Using a computer, that is... :) –  Apr 15 '14 at 02:29
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    Have now checked up to $2^{20,000}$ no success so far. –  Apr 15 '14 at 02:37
  • seems somewhat similar to a conjecture of Erdős and Graham that the base 3 expansion of $2^n$ avoids the digit “2” for infinitely many $n$, see also some analysis by RJLipton – vzn Sep 10 '14 at 15:54
  • Related – leo Sep 10 '14 at 16:37
  • A cute (albeit likely useless) way of restating this: Is there a power of 2 whose reciprocal, when reflected across the decimal point, is also a power of 2? (All this really amounts to is dividing the power of 5 by as many powers of 10, and therefore sliding its digits past the decimal point.) – Semiclassical Sep 11 '14 at 22:31
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    Probably not acceptable, and with apologies in advance, but $2^5$ is a power of $2$, and written backwards it is, arguably, $5^2$, a power of $5$. – paw88789 Sep 13 '14 at 21:41
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    I have checked up to $n=80,000$. I found that the highest power of 5 present is at $r(2^{7148})$ with $5^6$ dividing it. But what was really cool was that: $5^5|r(2^{55555})$. I mean, wow. That's a lot of fives. – amcalde Sep 16 '14 at 22:10
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    It seems like there might be some weird way to get from the fact that any power of 2 represented in base 2 simply has a single 1 for a digit and all 0s everywhere else, and similarly for any power of 5 in base 5. That means a palindrome number of a power in its own base will also have a single 1 as its representation. I feel like this almost makes intuitive sense that in base 10 they wouldn't then be palindrome numbers, but really it's just a hunch that could very well just be wrong. – Kainui Sep 18 '14 at 09:15
  • For what it's worth, I can confirm that there are no solutions where the exponent of the power of $2$ is less than $524288$. – Dylan Oct 28 '16 at 09:00

1 Answers1

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Just a partial answer. We can formalize the problem like the following. We are looking for an $(n,k) \in \mathbb{N}^2$ pair, which satisfies the following equality:

$$r(2^n)=5^k,$$

where $r : \mathbb{N} \rightarrow \mathbb{N}$ function gives the reverse of a number.

We can get $r$ like this:

$$r(n)=\sum_{\ell=0}^{m-1}10^{m-1-\ell}\left(\left\lfloor\frac{n}{10^\ell}\right\rfloor-10\left\lfloor\frac1{10}\left\lfloor\frac{n}{10^\ell}\right\rfloor\right\rfloor\right),$$

where $n$ is the number we want to reverse, and $m$ is the digits of $n$.

We know that $m$ is

$$m=\lfloor\log_{10}n\rfloor+1.$$

So using this we get for $r$ the following.

$$r(n)=\sum_{\ell=0}^{\lfloor\log_{10}n\rfloor}10^{\lfloor\log_{10}n\rfloor-\ell}\left(\left\lfloor\frac{n}{10^\ell}\right\rfloor-10\left\lfloor\frac1{10}\left\lfloor\frac{n}{10^\ell}\right\rfloor\right\rfloor\right),$$

where $n$ is the number we want to reverse.

We know that $r(2^n)=5^k$ is the same as

$$k = \log_5 r(2^n) = \log_5 \left( \sum_{\ell=0}^{\lfloor\log_{10}2^n\rfloor}10^{\lfloor\log_{10}2^n\rfloor-\ell}\left(\left\lfloor\frac{2^n}{10^\ell}\right\rfloor-10\left\lfloor\frac1{10}\left\lfloor\frac{2^n}{10^\ell}\right\rfloor\right\rfloor\right) \right).$$

From here we could use this identity, but I don't know how to prove that there exist such an $n$ that $k$ is an integer, or not.

Another approach to "brute force" a solution. I have written the following Maple program.

with(MmaTranslator[Mma]);
with(ListTools);

revnum := proc(n)
 FromDigits(Reverse(RealDigits(n)[1]));
end proc;

check := proc(a,b)
 local i;
 for i from a to b do 
  if type(eval(log[5](revnum(2^i))),integer) then 
   print([i,eval(log[5](revnum(2^i)))]);
  fi
 od;
 print("Done.");
end proc;

With check function we could test any interval of $n$ that $k$ is an integer or not. With check(1,15000) I tested upto $2^{15000}$ without any positive result.

That is why I believe the statement is true.

I think the article is very weird, Skewes' number is a good counterexample for this approach. We can find such parametrization, and create related problem, which has solution. $r(11^n)=11^n$ is true for $n=1,2,3,4$, since $11, 121, 1331,14641$ are palindromes. You can read about the generalization of this example in this paper.

If you like "almost solutions" here are some.

  • $(n,k)=(896, 385.9994262)$
  • $(n,k)=(1269, 547.0014642)$
  • $(n,k)=(1698, 732.0013312)$
  • $(n,k)=(1712, 737.9991581)$
  • $(n,k)=(2243, 967.0008879)$
  • $(n,k)=(3347, 1442.001656)$
  • $(n,k)=(3893, 1676.000497)$
  • $(n,k)=(4125, 1775.999455)$
  • $(n,k)=(4451, 1917.000274)$
  • $(n,k)=(4670, 2010.999274)$
  • $(n,k)=(4838, 2083.999410)$
Community
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