My question is in the title:

Is it possible to find $n≥4$ such that $2^n$ is a palindromic number (in base $10$)?

A palindromic number is a number which is the same, independently from which side we read it (forwards and backwards), for example $121, 484, 55755$.

My guess is "no". I know that a palindromic number $x$ with even length (i.e. the number of digits $\max\{n : 10^n \mid x\}$ is even) is a multiple of $11$: see here or here or here. In particular, a power of $2$ with an even length can't be a palindromic number.

However, I don't know what to do with the case of an odd length. For instance, if $x=abcdcba$, then $abccba$ is a multiple of $11$, but I don't see how this can help.

Here is a related question. On MathOverflow, related questions are: (1) and (2). Maybe also this thread (since $(1)$ is focused on binary expansion).

I tried with Mathematica and there is no palindromic power of $2$ with exponent $n<10000$:

 palindromeQ[n_] := IntegerDigits[n] === Reverse@IntegerDigits[n];
 For[i = 1, i < 10000, i++, If[PalindromeQ[2^i], Print[i]] ]

Finally, I think that the answer will be the same if we replace $2$ by any integer $n>1$ which is not a multiple of $11$. I don't know how I could prove (even for the case of even length…) that $11^n$ is not a palindromic number for $n≥5$.

Any hint will be helpful. Thank you!

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    $\begingroup$ The general claim is false...$2201^3=10662526601$. No idea about powers of $2$. $\endgroup$
    – lulu
    Feb 13 '16 at 11:40
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    $\begingroup$ Maybe you could use $11^{n} = (10+1)^{n} = \sum_{k=0}^{n} {n \choose k} 10^{k}$ for your last problem. $\endgroup$
    – Vincent
    Feb 13 '16 at 11:41
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    $\begingroup$ @lulu, the question was about exponent at least 4. This also rules out $26^2=676$. $\endgroup$ Feb 13 '16 at 12:17
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    $\begingroup$ @Downvoter : could you justify the downvote? $\endgroup$
    – Watson
    Sep 12 '16 at 18:59

According to the Wikipedia page on palindromic numbers:

G. J. Simmons conjectured [that] there are no palindromes of form $n^k$ for $k > 4$ (and $n > 1$).

The cited reference is:

Murray S. Klamkin (1990), Problems in applied mathematics: selections from SIAM review, p. 577

  • 1
    $\begingroup$ Thank you! I didn't looked closely enough at this Wikipedia page. However, the page 520 in that book is only a bibliography for geometry problems. On page 577, the problem 70-8 (the only one I found on palindromic numbers in that book) asks : "Does there exist any number greater than $1$ and not of the form $10 \cdots 01$ whose fourth power is a palindrome?". I have the 1990 edition… I don't know why I don't find precisely this conjecture in this reference. $\endgroup$
    – Watson
    Feb 13 '16 at 12:41
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    $\begingroup$ On that page, another reference is given : G. J. Simmons, Palindromic powers, Journal of Recreational Mathematics, 3 (No. 2, 1970), 93-98. $\endgroup$
    – Watson
    Feb 13 '16 at 12:42
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    $\begingroup$ I found it! I love the OEIS! This is question 5 on page 98. $\endgroup$
    – Watson
    Feb 13 '16 at 12:46
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    $\begingroup$ (Sorry to post so many comments…). Just out of curiosity, we can read on this old Wikipedia page: "M. Markovič's conjecture: If $n^4>0$ is a palindrome, then $n$ begins and ends with digit $1$, all other digits of $n$ being $0$. OEIS A186080. This conjecture has been tested by Simmons up to $10^{14}$". See also OEIS A056810. $\endgroup$
    – Watson
    Feb 13 '16 at 12:51
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    $\begingroup$ Here is a version of Klamkin's book, available online. I let you guess what happens for page 577… $\endgroup$
    – Watson
    Feb 13 '16 at 13:07

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