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

Solution 1:

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.

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