Timeline for A strange property about modulus
Current License: CC BY-SA 4.0
28 events
| when toggle format | what | by | license | comment | |
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| Sep 26 at 17:43 | vote | accept | Dattier | ||
| Sep 26 at 16:16 | answer | added | Max Alekseyev | timeline score: 11 | |
| Sep 25 at 19:58 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Sep 25 at 15:30 | comment | added | Dave Benson | @LSpice Ha! too right. My observation was that the $n$th primorial number is the first time the function returns a number as low as $11-n$, up until $n=10$, but then the tenth primorial number is the order of $11$ mod $A$. This is probably a clue to how the numbers were constructed. | |
| Sep 25 at 15:13 | comment | added | LSpice | @DaveBenson, ha, non-mathematical auto-correct is more familiar with ‘primordial’ than ‘primorial’. 😄 | |
| Sep 25 at 15:11 | history | edited | LSpice | CC BY-SA 4.0 |
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| Sep 25 at 14:35 | review | Close votes | |||
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| Sep 25 at 13:58 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Sep 25 at 13:53 | comment | added | Max Alekseyev | @DavidLoeffler: $A,B\ll 11^{223092870}.$ | |
| Sep 25 at 13:50 | comment | added | David Loeffler | How exactly does this example qualify as "relatively small size"? | |
| Sep 25 at 13:27 | comment | added | Max Alekseyev | @AndrejBauer: In other words, OP claims that the base-11 expansion of the fraction $B/A$ contains only nonzero digits. If the claim is correct, it is a nontrivial construction due to the large period 223092870 of $B/A$ in base 11 with relatively small size of $A,B$. But the question itself sounds like a challenge to the readers to reveal the construction method that OP used. | |
| Sep 25 at 7:17 | comment | added | Andrej Bauer | Can someone explain to the rest of us why this is interesting? | |
| Sep 25 at 6:56 | history | edited | Dattier |
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| Sep 23 at 7:08 | review | Low quality posts | |||
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| Sep 22 at 14:00 | comment | added | Michael Hardy | @DaveBenson $\qquad\uparrow \qquad$ | |
| Sep 22 at 13:58 | comment | added | Michael Hardy |
When you type B\mod A it is rendered as $B\mod A$ and when you type B\bmod A it is rendered as $B\bmod A.$ Presumably the "b" stands for "binary". \bmod is to be used when ${\bmod}$ is used as a binary operation symbol.
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| Sep 22 at 13:55 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Sep 22 at 11:03 | comment | added | Dattier | By trial and error. | |
| Sep 22 at 10:44 | comment | added | Daniel Weber | How were these values of $A, B$ found? | |
| Sep 22 at 10:29 | comment | added | Dave Benson | Does this have to do with the sequence of primordial numbers (A002110)? I notice that $(B\times 11^{\rm primordial} \mod A) \mod 11$ is an interesting sequence. | |
| Sep 22 at 9:31 | history | edited | Dattier |
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| Sep 22 at 3:02 | comment | added | Robert Israel | @FedorPetrov The order of $11$ mod $A$ should be $223092870$, I think. | |
| Sep 21 at 18:55 | comment | added | Fedor Petrov | How many powers of 11 modulo A are there? | |
| Sep 21 at 15:34 | comment | added | Dattier | the interesting fact is that $(11^n\times B \mod A) \mod 11$, does not take all values between 0 and 10. | |
| Sep 21 at 15:28 | comment | added | Carlo Beenakker | you might need really large $n$, up to $n=10^6$ also 1,2,3 do not appear | |
| Sep 21 at 15:06 | comment | added | Dattier | I have not found any counter-examples. | |
| Sep 21 at 14:58 | comment | added | Carlo Beenakker | intriguing; will you share a proof for statement 1? | |
| Sep 21 at 13:17 | history | asked | Dattier | CC BY-SA 4.0 |