solve the following system.
$$2x\equiv 3 (mod 5)$$
$$4x\equiv 2 (mod 6)$$
$$3x\equiv 2 (mod 7)$$

Question

solve the following system.
$$2x\equiv 3 (mod 5)$$
$$4x\equiv 2 (mod 6)$$
$$3x\equiv 2 (mod 7)$$

ganeshie8 · Answer

familiar with chinese remainder theorem ?

irishboy123 · Answer

.

anonymous · Answer

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

ganeshie8 · Answer

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

anonymous · Answer

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

ganeshie8 · Answer

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

anonymous · Answer

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

ganeshie8 · Answer

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 :
$$x\equiv 2\pmod{3}$$

anonymous · Answer

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

anonymous · Answer

Probably $$x\equiv 10 (mod 7)$$

ganeshie8 · Answer

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

ganeshie8 · Answer

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

anonymous · Answer

I got it. Thanks alot @ganeshie8