Example of breaking RSA with CRT
We have the three cipher texts $m^3\equiv c_1\equiv 4\pmod{6}$, $m^3\equiv c_2\equiv 20\pmod{35}$ and $m^3\equiv c_3\equiv 142\pmod{143}$. An application of the Chinese remainder theorem tells us that $m^3\equiv 1000\pmod{30030}$, but because $m$ is less than $\root 3\of{30030}$ we know $m^3=1000$. A regular cube root now recovers $m=10$.