All digits of $2^n$ are even if and only if $n=1,2,3,6,11$.
For example, $2^1=2,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,\ldots,2^{11}=2048,2^{12}=4096$.
Do you know a proof of this fact or some related results ? Or do you have a counter example ?
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11$\begingroup$ math.stackexchange.com/questions/116026/…. $\endgroup$– Josiah ParkCommented Oct 16, 2019 at 1:59
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$\begingroup$ Based on the comment of the OP in the linked question about finite sequences appearing in large enough powers of 2, maybe a (quite hard too) way to tackle the problem would be to prove that the number $0.1248163264...$ obtained writing the successive powers of 2 is universal (if such a terminology exists, in French we say "nombre univers"). $\endgroup$– Sylvain JULIENCommented Oct 16, 2019 at 6:19
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$\begingroup$ Ok, wikipedia just taught me it is called a "disjunctive sequence" in English. $\endgroup$– Sylvain JULIENCommented Oct 16, 2019 at 6:46
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2$\begingroup$ If all the digits of $2^n$ are even, then all of the digits of $2^{n-1}$ are no more than $4$... $\endgroup$– Jeff StromCommented Oct 16, 2019 at 13:40
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1$\begingroup$ I'm voting to close this question as off-topic because it is a duplicate of a question on math.stackexchange.com $\endgroup$– Timothy ChowCommented Oct 16, 2019 at 20:24
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