Timeline for Solving Linear Congruences.
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2017 at 11:47 | vote | accept | laura | ||
| Mar 23, 2017 at 11:41 | answer | added | B. Goddard | timeline score: 1 | |
| Mar 23, 2017 at 7:43 | comment | added | kingW3 | You get the same solution as in your book why do you think you're wrong? | |
| Mar 23, 2017 at 7:22 | answer | added | Dr. Sonnhard Graubner | timeline score: 0 | |
| Mar 23, 2017 at 7:20 | comment | added | Mike | @laura You mean like $20=2(7)+6$? $x\equiv20\pmod7$ means $x=7k+20$ for some integer $k$. It's a little unclear what you do and do not understand. | |
| Mar 23, 2017 at 7:11 | comment | added | laura | @Mike it would be better if you help me out in solving the same ! | |
| Mar 23, 2017 at 6:56 | comment | added | Mike | It looks like you did pretty much the same thing. You both multiplied by the inverse of 3. The only thing I didn't see you do was actually find the smallest positive integer congruent to 20. | |
| Mar 23, 2017 at 6:37 | comment | added | laura | @EricClapton there is no harm in learning new concepts ! :) | |
| Mar 23, 2017 at 6:36 | comment | added | Yes | I guess your instructor just intended to prevent negative integers. So she chose to multiply the congruence by $5$. | |
| Mar 23, 2017 at 6:34 | comment | added | laura | @JackD'Aurizio sir, i know this method.i want to solve this question using the above mentioned method! | |
| Mar 23, 2017 at 6:33 | comment | added | Jack D'Aurizio | If you write your equation as $3x\equiv -3\pmod{7}$ the solution is clearly $x\equiv -1\equiv 6\pmod{7}$, there is no need to compute any explicit inverse. | |
| Mar 23, 2017 at 6:29 | history | asked | laura | CC BY-SA 3.0 |