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I recently saw a conjecture on a blog ( http://blog.tanyakhovanova.com/?p=311 ) which the author refers to as the 86 conjecture. The conjecture claims that all powers of 2 greater than $2^{86}$ have a zero in their base 10 representation.

At first glance this seems like a numerical curiosity, and I wasn't sure that this was an interesting or deep mathematical question. I wanted to ask the community if they knew of any existing work that deals with either this type of question or something similar.

I tried searching for 86 conjecture on Google and didn't get anything useful.

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    $\begingroup$ Decimal expansions of powers are pretty chaotic creatures. I spent a lot of time trying to bound the total number of zeros in powers of five. I couldn't even linearly bound them. But it is possible to show that any amount of zeros in a row eventually appears in the powers of five (section 10: arxiv.org/abs/1910.13829). For numbers coprime to $10$, showing any amount of zeros in a row appear in its powers is easy. Just note that $n$ is a unit in $(\mathbb{Z}/10^k\mathbb{Z})^\times$. $\endgroup$
    – user682219
    Nov 12, 2019 at 19:26

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Richard Guy's Unsolved Problems in Number Theory, Problem F24, mentions only that Dan Hoey has verified this conjecture for $2^n$ up to $n = 2,500,000,000$. Guy tends to be fairly complete in his references. Since he doesn't give any others I doubt there's much more out there.

Several other related questions on decimal representations of powers of integers appear in F24 as well, most of which also appear to be open problems. There are some sequences in the On-line Encyclopedia of Integer Sequences mentioned, too; those might be worth chasing down.

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  • $\begingroup$ Thanks for the reference. This problem really is in the mathematical wilderness. $\endgroup$
    – svenkatr
    Mar 8, 2011 at 2:57
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    $\begingroup$ The main sequence is A007377. $\endgroup$
    – Charles
    Mar 8, 2011 at 3:40
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Is there any new result? is that conjecture still open?

I ask because we have a related conjecture:

Every term in the sequence $49\times 2^n$ has an odd digit.

We tested the conjecture for $n\leq 37533295636$, you can see our calculations here

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