Chinese remainder theorem for non-prime / non-coprime moduli

To use your example, trying to solve:

$$x\equiv 10\pmod{24}\\x\equiv 18\pmod{64}$$

You take each in terms of its prime factors:

$$x\equiv 10\pmod {3}\\ x\equiv 10\pmod{8}\\x\equiv 18\pmod {64} $$

Note that $x\equiv{18}\pmod {64}$ implies $x\equiv 10\pmod 8$, so we are left with:

$$x\equiv 10\pmod {3}\\x\equiv 18\pmod {64}$$

which is a pair of equations with coprime moduli.

You can always do this. Prime factorization is hard, however, so if you wanted an algorithm, you'd want to come up with a more general way than just prime-factorization. There are ways, but I'd have to call them up from memory. But the general gist is that you need to always reduce the problem to the co-prime case, and some way to check for contradictions that rule out solutions.

The general result is the following:

The system of congruences $ x\equiv a_i \mod{m_i} \quad(i=1,\dots,r) $ has a solution if and only if for each pair $(i,j)$, $\,\, a_i\equiv a_j \mod{m_i\wedge m_j} $. The solution is unique modulo $\mathrm{lcm}(m_1, \dots, m_r) $.