Chinese Remainder Theorem with sets of residues?

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I've tried seeking this out in a few places and haven't come up with much that's useful.

(Edit: based on suggestion in the comments, I'm restricting the problem to coprime moduli. A restriction to only prime moduli might help more and would still be useful.)

The standard CRT gives one congruence for each modulus, and gives a unique answer modulo the product of the original moduli (for prime moduli). But what about a setup where you have a set of answers based on sets of congruences/residues? An example, where X is the set of results:

$$ x \in X \iff \begin{cases} x \equiv 0, 1, \text{or } 2 \pmod 5 \\ x \equiv 2, 4, \text{or } 6 \pmod 7 \\ x \equiv 3, 5, \text{or } 8 \pmod{11} \end{cases} $$

In this case, $$X = \{16, 25, 27, 30, 41, 60, 102, 107, 135, 137, 146, 151, 170, 181, 195,\\ 212, 247, 256, 261, 272, 291, 300, 305, 335, 366, 377, 382\}\pmod{385}$$

And of course the results set must have 27 members (3x3x3).

Are there any theorems or results for this sort of problem? Anything like:

  • There must be some subset of solutions between M and N
  • The [smallest/largest] solution must be [less than/greater than] M
  • Certain patterns in the congruences create certain patterns in the results

Or am I barking up an unsolved/uninteresting/proven unsolvable tree?