<s>[INST] simple congruence system problem. This is excercise 1 in George E. Andrews number theory. page 51. How am i supposed to solve this? Thanks very much in advance. I used the cancellation law to get isolate the x. But I don’t know what to do next.
$5x\equiv 4 \pmod 3$
$7x\equiv 6 \pmod 5$
$9x\equiv 8 \pmod 7$
 [/INST] $$5 x \equiv 1 \pmod 3 \implies x \equiv 2 \pmod 3$$
$$7 x \equiv 1 \pmod 5 \implies x \equiv 3 \pmod 5$$
$$9 x \equiv 1 \pmod 7 \implies x \equiv 4 \pmod 7$$
$$x \equiv 2 \pmod 3 \text{ and }x \equiv 3 \pmod 5 \implies x \equiv a \pmod{15}$$
Since $x \equiv 2 \pmod 3$, we have $a \in \{2,5,8,11,14\}$. Since $x$ is also $3 \pmod 5$, we get that $$x \equiv 8 \pmod {15}$$
$$x \equiv 8 \pmod {15} \text{ and }x \equiv 4 \pmod 7 \implies x \equiv b \pmod{105}$$
Since $x \equiv 8 \pmod 15$, we have $b \in \{8,23,38,53,68,83,98\}$. Since $x$ is also $4 \pmod 7$, we get that $$x \equiv 53 \pmod {105}$$
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