Given:
$$Ax=b$$
$$A = \begin{pmatrix} 1&1&1&1&1&1&1&1&1&1&1&1\\ 1&1&0&1&1&1&0&0&0&0&0&0\\ 0&0&1&1&0&1&1&1&0&0&0&0\\ 0&0&0&0&1&1&0&1&1&1&0&0\\ 0&0&0&0&0&0&1&1&0&1&1&1\\ 1&1&0&0&0&0&0&0&1&1&0&1\\ 0&1&1&1&0&0&0&0&0&0&1&1 \end{pmatrix}$$ $$ b = \begin{pmatrix} 78\\ 33\\ 33\\ 33\\ 33\\ 33\\ 33 \end{pmatrix} $$
$$ \{x_0, x_2, x_4, x_6, x_8, x_{10}\} ∩ \{6, 12\} = ∅, \\ ∀_i\ x_i ∈ ℕ \\ ∀_i\ x_i ∈ [1, 12] \\ ∀_{i ≠ j}\ x_i ≠ x_j $$
How can $x$ be solved for? For now, I've set up a matrix equation and am brute forcing permutations of:
$$ x_0 ∈ [1, 5]∪[7, 11] \\ x_2 ∈ [1, 5]∪[7, 11] \\ x_4 ∈ [1, 5]∪[7, 11] \\ 6 ∈ [x_1, x_3, x_5, x_7, x_9, x_{11}] \\ 12 ∈ [x_1, x_3, x_5, x_7, x_9, x_{11}] $$
Is there a "more mathematical" way to find a solution?
Here's how you can solve it with the constraint programming solver in SAS:
There are exactly 12 feasible solutions, one of which is: $$(x_0, x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}) = (1, 9, 2, 12, 8, 3, 11, 5, 10, 7, 4, 6)$$