## anonymous one year ago solve the following system. $2x\equiv 3 (mod 5)$ $4x\equiv 2 (mod 6)$ $3x\equiv 2 (mod 7)$

1. ganeshie8

familiar with chinese remainder theorem ?

2. IrishBoy123

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3. anonymous

I have done an assignment but it only had something like $x\equiv 3 (mod 5)$ $x\equiv 2 (mod 6)$ $x\equiv 2 (mod 7)$. But am not sure how to use it to solve the problem. @ganeshie8

4. ganeshie8

solve each of the linear congruence to get a system with simpler congruences, for example : $$2x\equiv 3 \pmod 5$$ simplifies to $$x\equiv 4\pmod{5}$$

5. anonymous

@ganeshie8 Please show me a step on how you got the 4

6. ganeshie8

I have just guessed it. Plugin x=4 and observe that it satisfies the congruence $2*4\equiv 8\equiv 3\pmod{5}$

7. anonymous

@ganeshie8 Thanks. I will try doing that and then apply the Chinese remainder theorem

8. ganeshie8

For the second congruence, $$4x\equiv 2 \pmod 6$$, dividing $$2$$ through out gives $2x\equiv 1\pmod{3}$ then it is easy to eyeball the solution : $x\equiv 2\pmod{3}$

9. anonymous

$3x\equiv 2 (mod 7)$ simplifues to $x\equiv -4 (mod 7)$ Does that look correct? @ganeshie8

10. anonymous

Probably $x\equiv 10 (mod 7)$

11. ganeshie8

$$x\equiv -4\pmod{7}$$ is same as $$x\equiv 3 \pmod{7}$$

12. ganeshie8

so the given system, after transforming into simple linear congruences is $x\equiv 4\pmod{5}\\x\equiv 2\pmod{3}\\x\equiv 3\pmod{7}$

13. anonymous

I got it. Thanks alot @ganeshie8