Timeline for Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integers $n$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 26, 2022 at 0:51 | vote | accept | Eric | ||
| May 25, 2022 at 15:54 | comment | added | Dattier | $\{\dfrac{2^n}{10^m} :(m, n) \in \mathbb N^2\} $ is dense on $\mathbb R^+$ | |
| May 25, 2022 at 14:08 | history | edited | LSpice | CC BY-SA 4.0 |
Slightly more TeX
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| May 25, 2022 at 14:00 | comment | added | spin | @Gro-Tsen: More generally, let $c > 0$ be any positive integer. Then there exist infinitely many powers of $2$ which have decimal expansion starting with $c$. | |
| May 25, 2022 at 13:03 | answer | added | JoshuaZ | timeline score: 7 | |
| May 25, 2022 at 10:06 | comment | added | Gro-Tsen | But saying anything beyond this is probably an open problem, as this older answer to a related problem suggests. | |
| May 25, 2022 at 9:56 | comment | added | Gro-Tsen | An easy remark: $2^{68}$ contains all decimal digits, and it does so in its last $17$ digits. Now $2^{4\times 5^{16}}$ is $1$ mod $5^{17}$ by Euler's theorem. So if $n$ is congruent to $68$ mod $4\times 5^{16}$, then $2^n$ ends in the same $17$ decimal digits as $2^{68}$ and hence, contains all decimal digits. This shows that there are arbitrarily large $n$ such that $2^n$ contains all decimal digits (this is, of course, weaker than showing that all sufficiently large $n$ are such). | |
| May 25, 2022 at 6:31 | comment | added | Eric | @JoséHdz.Stgo. oops, let's say it's a different kind of powerful then : ) | |
| May 25, 2022 at 6:14 | history | edited | Eric | CC BY-SA 4.0 |
n must be integer
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| May 25, 2022 at 5:20 | comment | added | José Hdz. Stgo. | Methinks the label "k-powerful" is already taken: en.wikipedia.org/wiki/Powerful_number | |
| May 25, 2022 at 4:32 | history | edited | Eric | CC BY-SA 4.0 |
added 17 characters in body
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| May 25, 2022 at 4:24 | history | asked | Eric | CC BY-SA 4.0 |