Timeline for Solving $3x\equiv 4\pmod 7$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2014 at 1:56 | vote | accept | Tehmas | ||
| Dec 7, 2014 at 20:26 | comment | added | Fmonkey2001 | Don't forget to choose my answer as the answer you liked! :D I'd appreciate it! | |
| Dec 7, 2014 at 14:43 | comment | added | Tehmas | Thanks Fmonkey2001. :) Your method is truly simpler and thanks for replying to other questions as well. | |
| Dec 7, 2014 at 13:48 | comment | added | Fmonkey2001 | It wasn't useless, it was just a different method to solving the same problem. You'd get the same answer if you did it your way. Just like @N. F. Taussig did you could always substitute the answer of $6$ back into the original equation to see that $x=6 \rightarrow 18\equiv4\pmod{7}$ | |
| Dec 7, 2014 at 13:39 | comment | added | Tehmas | Ahan. But you didn't even use the inverse(-2). Was it a useless step? @_@ | |
| Dec 7, 2014 at 13:36 | comment | added | Fmonkey2001 | yes you can always add multiples of $7$ (in this particular example) since $7\equiv0\pmod{7}$ so what you're doing is really just adding $0$ to the equation. More generally $a\equiv0\pmod{a}$ (just to make sure I don't confuse you and make you add $7$ to all mod equations lol | |
| Dec 7, 2014 at 13:32 | comment | added | Tehmas | thanks for replying. Your answer is a lot simpler. Will this subtracting/adding always work out the problem? | |
| Dec 7, 2014 at 13:21 | history | answered | Fmonkey2001 | CC BY-SA 3.0 |