Is there a power of 2 that, written backward, is a power of 5?

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.

• $2^0=1$ backwards is $1=5^0$ :) – user142299 Apr 15 '14 at 2:17
• Thank you, Captain Obvious @user142299. – 2012ssohn Apr 15 '14 at 2:18
• @2012ssohn Also on a more serious note, $5260135901548373507240989882880128665550339802823173859498280903068732154297080822113666536277588451226982968856178217713019432250183803863127814770651880849955223671128444598191663757884322717271293251735781376$ is a power of two. – user142299 Apr 15 '14 at 2:24
• Have now checked up to $2^{20,000}$ no success so far. – user142299 Apr 15 '14 at 2:37
• 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

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.

• $(n,k)=(896, 385.9994262)$
• $(n,k)=(1269, 547.0014642)$
• $(n,k)=(1698, 732.0013312)$
• $(n,k)=(1712, 737.9991581)$
• $(n,k)=(2243, 967.0008879)$
• $(n,k)=(3347, 1442.001656)$
• $(n,k)=(3893, 1676.000497)$
• $(n,k)=(4125, 1775.999455)$
• $(n,k)=(4451, 1917.000274)$
• $(n,k)=(4670, 2010.999274)$
• $(n,k)=(4838, 2083.999410)$
• I tested upto $2^{25000}$ without finding a solution. – user153012 Sep 17 '14 at 15:16