Timeline for How many integers divide a number that involves just three non-zero digits?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 23, 2014 at 14:33 | vote | accept | Vesselin Dimitrov | ||
| Aug 20, 2014 at 13:57 | comment | added | Vesselin Dimitrov | Thank you! I hadn't taken into account this incompatibility for powers of $2$ mod distinct prime powers. | |
| Aug 20, 2014 at 10:53 | comment | added | Jan-Christoph Schlage-Puchta | As far as I know it is an unsolved problem whether there exist infinitely many primes with $\mathrm{ord}^\times_p(2)$ significantly bigger than $\sqrt{p}$. Also glueing together different prime factors is more difficult than usual: If $p, q$ are prime numbers with $(p-1, q-1)$ large, then solutions of $2^{e_1}+2^{e_2}+2^{e_3}\equiv0\pmod{p}$ and $\pmod{q}$ might exist, while a solution $\pmod{pq}$ need not exist. As far as I know this incompatibility of powers of 2 was first exploited by Erd\H os and van der Corput n their work on integers of the form $p+2^a$. | |
| Aug 18, 2014 at 18:53 | comment | added | Vesselin Dimitrov | Very nice, thank you very much! Do you know if (for every $\epsilon > 0$) there should be a positive density of integers all of whose prime factors satisfy $\mathrm{ord}_p^{\times}{2} > p^{1-\epsilon}$? If this is true for some $\epsilon < 1/4$, then the argument with Weil's bound for the Fermat curve will show that also a positive density of the integers have a multiple with digit sum $3$; this is because all mod $p$ points on the Fermat curve of exponent $(p-1)/r$ lift to $p$-adic points. Or is almost every $n$ divisible by some prime for which $2$ has order $< p^{1-\epsilon}$? | |
| Aug 17, 2014 at 9:56 | comment | added | Jan-Christoph Schlage-Puchta | The first half of my answer didn't make much sense. I deleted it and expanded the second half. | |
| Aug 17, 2014 at 9:55 | history | edited | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |
As Dimitrov already noted, the first half of my answer did not make much sense. I deleted it and improved the second part. I just hope I got the computations right.
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| Aug 12, 2014 at 18:05 | comment | added | Vesselin Dimitrov | Thank you for your answer, but please note that that question I asked was whether the density of such $a$ [dividing a number with bounded digit sum], was positive, or zero. It is indeed obvious that this density cannot be $1$, and I had also noted, in the text of the question, the observation you make in your first paragraph. | |
| Aug 12, 2014 at 12:22 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |