I'm starting to learn some elementary number theory and i came across a task i don't know how to solve.
$$x \equiv 5 (mod \ 6)$$ $$x \equiv A (mod \ 35)$$
and the second one
$$x \equiv A (mod \ 10)$$ $$x \equiv 14(mod \ 21)$$
If i'm thinking in good way then we have to find such A's that we get same solution x for both systems. It reminds me about chinese remainder theory because 6 and 35 are relatively prime and 10 and 21 are too.
But i have no idea how to use those facts to solve this problem.
I'd be really happy to know how to deal with such tasks. Cheers!
Hint $\ $ Using CRT to reduce each system to equivalent systems mod primes we obtain
$\ \ x \equiv\ 1\pmod{2},\ x\equiv 2\pmod 3,\ x\equiv A\pmod 5,\ x\equiv \color{#c00}A\pmod 7\ $ is equivalent to the $1$st. $\ \ x \equiv \color{#c00}A\pmod{2},\ x\equiv 2\pmod 3,\ x\equiv A\pmod 5,\ x\equiv\ 0\pmod 7\ $ is equivalent to the $2$nd.
Thus the two systems are equivalent $\iff \color{#c00}A \equiv\, \ldots$