Find all natural numbers of the form $2^n$ whose all digits are even

Asked May 12 '16 at 16:49

Active May 12 '16 at 16:54

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Find all natural numbers of the form $2^n$ whose all digits are even. For example: $2, 4, 8, 64, 2048$ (I believe they are the only such numbers).

For $n \geq 11$, so far, I can prove that the last digit of such number must be $4$ or $8$. Moreover, the last $2$ digits must be $24,64,08,48$ or $88$. Then I don't know how to procede next. I also found that such number must be of the form $2^{4k+2}$ or $2^{4k+3}$, for $k \geq 3$.

edited May 12 '16 at 16:54

asked May 12 '16 at 16:49

SiXUlm

2,436

1

See http://oeis.org/A068994 -- it looks like it's an open problem. – Barry Cipra May 12 '16 at 16:54
@BarryCipra: Thanks! (Sadly) this is what I'm looking for – SiXUlm May 12 '16 at 16:56
Possible duplicate: http://math.stackexchange.com/questions/116026/is-2048-the-highest-power-of-2-with-all-even-digits-base-ten – N.S.JOHN May 13 '16 at 12:17
@N.S.JOHN: yes, you are right. – SiXUlm May 13 '16 at 16:00

Find all natural numbers of the form $2^n$ whose all digits are even

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