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Powers of 2 with all even digits

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Re: Powers of 2 with all even digits

#21

Post by andrewla »

andrewla wrote: post #13
lifthrasiir wrote: post #6

Earlier quoted context omitted.

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…

> whether 2^k mod 10 is odd 2^k mod 10 is never odd; it's the cycle (2, 4, 8, 6). Related here is the length of the cycles mod 2^k, https://oeis.org/A005054 . Interestingly, the number of all-even-digit elements in those cycles does not appear to be in the oeis, I get 4, 10, 25, 60, 150 as the first five terms. This does appear to get more efficient as k gets higher; for k=11 I get a cycle length of 39,062,500 with a…

I see my error here -- you can in fact eliminate half even for 2^k mod 10, because both 6 and 8 force a carry; so ending in a 2 or a 6 means that the next higher digit must be odd.

Re: Powers of 2 with all even digits

#22

Post by mtoner23 »

netsharc wrote: post #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).

why comment about your misunderstanding of the title?

Re: Powers of 2 with all even digits

#24

Post by 1970-01-01 »

heffer wrote: post #17
1970-01-01 wrote: post #14

[flagged]

The link is about 2^n not n^2.

You assumed this out of air.

"Powers of 2" means this:

Here are the powers of 2 from \( 2^{-11} \) to \( 2^{11} \) in a table format:

     | Power of 2   | Value              |
     |--------------|--------------------|
     | \( 2^{-11} \) | 0.00048828125      |
     | \( 2^{-10} \) | 0.0009765625       |
     | \( 2^{-9} \)  | 0.001953125        |
     | \( 2^{-8} \)  | 0.00390625         |
     | \( 2^{-7} \)  | 0.0078125          |
     | \( 2^{-6} \)  | 0.015625           |
     | \( 2^{-5} \)  | 0.03125            |
     | \( 2^{-4} \)  | 0.0625             |
     | \( 2^{-3} \)  | 0.125              |
     | \( 2^{-2} \)  | 0.25               |
     | \( 2^{-1} \)  | 0.5                |
     | \( 2^{0} \)   | 1                  |
     | \( 2^{1} \)   | 2                  |
     | \( 2^{2} \)   | 4                  |
     | \( 2^{3} \)   | 8                  |
     | \( 2^{4} \)   | 16                 |
     | \( 2^{5} \)   | 32                 |
     | \( 2^{6} \)   | 64                 |
     | \( 2^{7} \)   | 128                |
     | \( 2^{8} \)   | 256                |
     | \( 2^{9} \)   | 512                |
     | \( 2^{10} \)  | 1024               |
     | \( 2^{11} \)  | 2048               |
The evens include "0"

Re: Powers of 2 with all even digits

#25

Post by andrewla »

Sharlin wrote: post #20
andrewla wrote: 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.

Proofs of non-existence aren't usually straightforward.

I mean, clearly it isn't in this case. But given that the digits of 2^n are cyclical at each decimal position, it does feel like this should fall out of some sort of chinese remainder theorem manipulation.

Re: Powers of 2 with all even digits

#27

Post by thehappypm »

1970-01-01 wrote: post #24
heffer wrote: post #17

Earlier quoted context omitted.

The link is about 2^n not n^2.

You assumed this out of air. "Powers of 2" means this: Here are the powers of 2 from \( 2^{-11} \) to \( 2^{11} \) in a table format: | Power of 2 | Value | |--------------|--------------------| | \( 2^{-11} \) | 0.00048828125 | | \( 2^{-10} \) | 0.0009765625 | | \( 2^{-9} \) | 0.001953125 | | \( 2^{-8} \) | 0.00390625 | | \( 2^{-7} \) | 0.0078125 | | \( 2^{-6} \) | 0.015625 | | \( 2^{-5} \) | 0.03125 | | \( 2^{-4} \…

0 is not in the “Value” column

Re: Powers of 2 with all even digits

#29

Post by andrewla »

1970-01-01 wrote: post #24
heffer wrote: post #17

Earlier quoted context omitted.

The link is about 2^n not n^2.

You assumed this out of air. "Powers of 2" means this: Here are the powers of 2 from \( 2^{-11} \) to \( 2^{11} \) in a table format: | Power of 2 | Value | |--------------|--------------------| | \( 2^{-11} \) | 0.00048828125 | | \( 2^{-10} \) | 0.0009765625 | | \( 2^{-9} \) | 0.001953125 | | \( 2^{-8} \) | 0.00390625 | | \( 2^{-7} \) | 0.0078125 | | \( 2^{-6} \) | 0.015625 | | \( 2^{-5} \) | 0.03125 | | \( 2^{-4} \…

??

This does not clarify -- your initial post made a claim about 0^2, which (correctly) does not appear in this list.

Moreover it is trivial that there are no negative powers of 2 that have all even digits, since the trailing digit will always be 5. So the question reduces to whether there are powers of 2 greater than 2048 that have all even digits.

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