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Find all solutions: $$\begin{cases} x\equiv 39 \mod(189) \\ x\equiv 25 \mod(539) \\ x\equiv 399 \mod(1089) \end{cases}$$

But $189=3^3\cdot7$, $539=11\cdot7^2$ and $1089=3^2\cdot 11^2$, so I can't use here Chinese remainder theorem.

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up vote 2 down vote accepted

You can find a system of congruences equivalent to the original system, but with pairwise relatively prime moduli.

For example, the first congruence is equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 39\pmod{7}$.

The second original congruence is equivalent to the system $x\equiv 25\pmod{11}$ $x\equiv 25\pmod{7^2}$.

Note that from the first congruence we had $x\equiv 4\pmod{7}$. If $ \equiv 25\pmod{7^2}$ then $x\equiv 4\pmod{7}$, our first two original congruences are equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 25\pmod{7^2}$, $x\equiv 25\pmod{11}$.

The third original congruence can be similarly dealt with. When we are through we end up with a system with moduli $3^3$, $7^2$, and $11$.

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