Timeline for Calculation of modular multiplicative inverse of A mod B when A > B
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 15, 2020 at 18:53 | history | suggested | JB-Franco | CC BY-SA 4.0 | improved readability |
| Jun 15, 2020 at 18:24 | review | Suggested edits | |||
| S Jun 15, 2020 at 18:53 | |||||
| Aug 16, 2018 at 10:11 | vote | accept | Danetta | ||
| Aug 16, 2018 at 10:11 | vote | accept | Danetta | ||
| Aug 16, 2018 at 10:11 | |||||
| Aug 16, 2018 at 10:11 | vote | accept | Danetta | ||
| Aug 16, 2018 at 10:11 | |||||
| Jun 15, 2018 at 19:24 | comment | added | fleablood | "A lot of online calculators that I tried are also said that a has to be bigger than be," Calculators don't do mathematics. Calculators do calculations. | |
| Jun 15, 2018 at 19:16 | comment | added | fleablood | $-1\equiv 2 \mod 3$. So $-1$ is equivalent to $2$. It doesn't make the slightest difference which one you use. (We don't call these things "equivalences" for nothing, you know...) | |
| Jun 15, 2018 at 19:08 | answer | added | fleablood | timeline score: 0 | |
| Jun 15, 2018 at 18:48 | comment | added | fleablood | "We have X=4, Y=-1 for A=3,B=11, and X=4 is valid inverse, while -1 is indeed not. " Huh. $X \equiv A^{-1}\mod B$ because $X*A = 4*3 \equiv 1 \mod 11$. An $Y\equiv B^{-1} \mod A$ because $Y*B=(-1)*11 \equiv 1 \mod 3$. They are both valid inverses. What is your issue? | |
| Jun 15, 2018 at 18:23 | answer | added | Joffan | timeline score: 1 | |
| Jun 15, 2018 at 16:17 | history | edited | Bernard | CC BY-SA 4.0 | edited body |
| Jun 15, 2018 at 16:05 | review | First posts | |||
| Jun 15, 2018 at 16:11 | |||||
| Jun 15, 2018 at 16:00 | history | asked | Danetta | CC BY-SA 4.0 |