Timeline for Zero's in the decimal representation of powers of 3
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 10, 2014 at 22:41 | comment | added | Gerry Myerson | See also oeis.org/A238938 (powers of 2 without a zero in the decimal expansion). | |
| Oct 10, 2014 at 17:42 | comment | added | Anthony Quas | See also mathoverflow.net/questions/30357/… for a closely related question. | |
| Oct 10, 2014 at 17:40 | comment | added | Anthony Quas | So this is a version of a known open problem. The version I heard is about 7's in base 10 expansions of $2^n$. You can with a bit of ingenuity prove that this holds for almost all $n$, but all $n$ is probably really hard. If you believe the digits are random (as they "should" be), then you expect the conjecture to be true by Borel-Cantelli type arguments. | |
| Oct 10, 2014 at 12:45 | history | edited | user9072 |
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| Oct 10, 2014 at 12:39 | comment | added | Jeremy Rouse | Also, the fact that $\alpha = \log(3)/\log(10)$ is irrational, together with the fact that $\{ n \alpha \}$ is uniformly distributed mod $1$ means that for all positive integers $m$, there is an integer $n(m)$ so that the first $m$ digits of $3^{n(m)}$ are all nonzero. | |
| Oct 10, 2014 at 12:38 | comment | added | Douglas Zare | @James Cranch: The $5\times 10^{n-2}$ values hit mod $10^n$ are arbitrary initial sequences with one of the $500$ allowed final $4$ digits. Since $3^8 = 6561$, we can find a power of $3$ that ends in $1111...1116561$. For example, $3^{195508}$ ends in $...1116561.$ | |
| Oct 10, 2014 at 12:33 | comment | added | James Cranch | Perhaps I'm being stupid, but I don't see why that means it has to work early. | |
| Oct 10, 2014 at 12:28 | comment | added | Douglas Zare | @James Cranch: The multiplicative order of $3$ mod $10^n$ is $5\times 10^{n-2}$ for $n \ge 4$. I think this means it would have to work by $n=4$ but it doesn't. | |
| Oct 10, 2014 at 12:15 | comment | added | James Cranch | My advice is to try the last n decimal digits instead. The function is periodic modulo 10^n for any n. That means if you can find an n such that a long enough string have a zero among the last n digits, then you'll be done. | |
| Oct 10, 2014 at 12:13 | review | First posts | |||
| Oct 10, 2014 at 12:26 | |||||
| Oct 10, 2014 at 12:11 | history | asked | Omran Kouba | CC BY-SA 3.0 |