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Are there positive integers $n$ and $m$ for which reversing the base-10 digits of $2^n$ yields $7^m$?


I've answered this question for powers of 2 and powers of 3 in the negative. Permutations of the digits of a number divisible by 3 yields another number divisible by 3 in base-10. This arises from the fact that the sum of base-10 digits of a number divisible by 3 is itself divisible by 3.

Thus since $2^n$ is not divisible by 3, reversing the digits can't yield another number divisible by 3, and hence no natural number power of 2 when reversed will be a natural power of 3.


I'm currently trying to put limitations on n and m by considering modular arithmetic. Suggestions for techniques in the comments would also be appreciated.

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  • $\begingroup$ If the number of digits is even, adding $2^n$ to $7^m$ would give a multiple of 11. By letting 7=-4 modulo 11, some things can be said about $m$ and $n$. $\endgroup$ – Marco Aug 25 '18 at 3:20
  • $\begingroup$ Also going on with the same idea of remainders modulo 3, you can say $n$ must be even since $2^n=7^m=1$ modulo 3. $\endgroup$ – Marco Aug 25 '18 at 3:22
  • $\begingroup$ Related: math.stackexchange.com/questions/1653079/… $\endgroup$ – Misha Lavrov Aug 25 '18 at 3:55
  • $\begingroup$ $1024=2^{10}$ and $2401=7^4$ are close to being reversals of each other.... $\endgroup$ – Gerry Myerson Aug 25 '18 at 4:23
  • $\begingroup$ This is also related $\endgroup$ – mbjoe Aug 25 '18 at 18:29
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Some basic observations:

$n$ must be even: one has $2^n=7^m=1 \pmod{3}$ and so $n$ is even.

$n-m$ must be divisible by 3: one has $2^n=7^m=(-2)^m \pmod{9}$ and so $2^{n-m}=(-1)^m \pmod{9}$ and so $n-m$ is divisible by 3.

$n-2m$ is divisible by 10 as the following two paragraph show.

If the number of digits is even, then $2^n+7^m$ is a multiple of 11. Therefore, $2^n+(-4)^m =0 \pmod{11}$. Since $n$ is even, we let $n=2l$, and so $4^l+(-4)^m=0 \pmod{11}$. It follows that $m$ is odd and $l-m$ is divisible by 5.

If the number of digits is odd, then $2^n=7^m \pmod{11}$. Again we have $4^l=(-4)^m \pmod{11}$ and so $4^{l-m}=(-1)^m \pmod{11}$, which implies that $m$ is even and $l-m$ is divisible by 5.

The number of digits of $2^n$ is given by the inequality $10^s<2^n<10^{s+1}$. So $s<n \log_{10} 2<s+1$. In particular $|n\log 2 - m \log 7|<1$.

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