Which option can be a correct set-builder form of C = {0, 3, 6, 9}?
The governing concept is expressing a roster-form set through a defining property together with suitable bounds. Every element of C is a multiple of 3. Among numbers satisfying 0 ≤ x < 10, the multiples of 3 are exactly 0, 3, 6, and 9. Thus option A produces precisely the given set: the lower bound includes 0, while the strict upper bound excludes 10 and larger values. Option B gives the factors of 9, namely {1, 3, 9}, so it includes 1 and omits 0 and 6. Option C gives odd digits such as 1, 3, 5, 7, and 9. Option D gives prime digits such as 2, 3, 5, and 7; neither matches C.
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