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Powers of 2 with all even digits

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Re: Powers of 2 with all even digits

#5
post #4

No additional terms up to 2^(10^10). - Michael S. Branicky, Apr 16 2023 How did he do this?

yeah that's weird - its kind of a pointless comment without an included algorithm or something

There’s probably a smart way to rule out a lot of cases so you only have to check a relatively small number of candidates. It would be good to know what it is.

Re: Powers of 2 with all even digits

#6

No additional terms up to 2^(10^10). - Michael S. Branicky, Apr 16 2023 How did he do this?

As noted in 3) in the Shepherd's comment, 2^k has no odd digits when 2^k mod 10^n for all integer n have no odd digits as well. So many k would be filtered by checking whether 2^k mod 100 has an odd digit, then another portion of the remainder will get filtered with 2^k mod 1000, 2^k mod 10000 and so on. (EDITED: Thanks to andrewla!) All of them would be periodic, so first few steps can be made into a lookup table to filter almost every k.

Re: Powers of 2 with all even digits

#7
This is remarkable! I always find it fascinating that simple to express properties lack a proof. This is a very simple thing to evaluate and seems like it should be straightforward to establish that 2048 is the highest such power.

Re: Powers of 2 with all even digits

#9
Somehow I missed the title and wondered what the fuck was going on...

2, 4, 8, 64, 2048 are powers of 2 (i.e. 2^n), and they don't contain odd numbers (e.g. 16, 128, 1024 contain 1 so are not in this list, same with 4096 containing 9).

Re: Powers of 2 with all even digits

#10
post #7

This is remarkable! I always find it fascinating that simple to express properties lack a proof. This is a very simple thing to evaluate and seems like it should be straightforward to establish that 2048 is the highest such power.

why should it be straightforward to establish that?
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