## anonymous 4 years ago Find all solutions of the following linear congruence. $$3x\equiv2\mod7$$ First of all, we notice that $$(3,7)=1$$. Therefore, we will only have $$1$$ solution. We now need to obtain a solution of the linear, diophantine equation $$3x-7y=2$$. The Euclidean algorithm gives: $$7=3\cdot2+1$$, $$3=1\cdot3+0$$. Hence, $$7\cdot1-3\cdot2=1$$ and $$7\cdot2-3\cdot4=2$$. Therefore, a particular solution to the diophantine equation is $$x_0=-4$$, $$y_0=-2$$ and all solutions of the linear congruences are given by $$x\equiv-4\equiv3\mod7$$. Am I right?

1. myininaya

I have a way but you might now like it

2. myininaya

not*

3. anonymous

i am sure this is right, but i would write the following $3x\equiv2\mod7$ $3x\equiv9\mod7$ $x\equiv3\mod7$

4. myininaya

$3x \equiv 2 \mod 7 =>3x=2+7k$ k=0, there is no integer x k=1, x=3 k=2, there is no integer x k=3, there is no integer x k=4, x=10 k=5, there is no integer x k=6, there is no integer x k=7, x=17 so on... so we have x=...,3,10,17,17+7,(17+7)+7, and so on...

5. anonymous

that is the usual gimmick. keep adding until you can divide to solve the congruence

6. myininaya

x=3+7i where i is an integer

7. myininaya

which is what satellite said

8. anonymous

in two lines lol

9. myininaya

shhh... don't make fun of my inefficient way ;)

10. anonymous

speed kills