2018 AMC 12A Problem #12

Mathematica code

Select[Subsets[Range@12,{6}],!Or@@(IntegerQ[#2/#]&@@#&/@Subsets[#,{2}])&]

returns all the $S$ sets

{4, 5, 6, 7, 9, 11}, {4, 6, 7, 9, 10, 11}, {5, 6, 7, 8, 9, 11}, {5, 7, 8, 9, 11, 12}, {6, 7, 8, 9, 10, 11}, {7, 8, 9, 10, 11, 12}

So the answer is $4$

Trial and error is the fastest way to solve this. When choosing numbers to add to the set, we must go from largest to smallest when possible. For example, realize that choosing $4$ would eliminate $2$ other integers ( $8$ and $12$ ), but choosing $12$ only eliminates $4$ , making it a better option in general.

If $3$ was in the set, we find that $11, 10, 8,$ and $7$ can be added into the set, but no more. Therefore $3$ does not work. We can find a valid set $S$ if $\boxed{4}$ is the lowest integer: $\{4, 6, 7, 9, 10, 11\}$ .