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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.

user196574
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1 Answers1

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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 $|n\log 2 - m \log 7|<1$.

Marco
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