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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!

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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 $\iff \color{#c00}A \equiv\, \ldots$

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  • $\begingroup$ If i'm thinking in right direction, only possible A's are that $2|A-1$ and $7|(x-A)$ and those two makes: A must be odd and A must be a multiply of 7, am i right? $\endgroup$ – Krzysztof Lewko May 13 '14 at 19:00
  • $\begingroup$ @Chris The $2$nd should be $\, 7\mid 0 -A,\, $ so we reduce to solving $\, \color{#c00}A\equiv 1\pmod 2,\ \color{#c00}A\equiv 0\pmod 7 $ $\endgroup$ – Bill Dubuque May 13 '14 at 19:03

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