2017 AIME I Problem 12

$1$ can never appear (since $1 \cdot 1 = 1$), and if the least element is at least $4,$ every product of two elements exceeds $10$

With least element $2$ or $3,$ only a few conflicts matter: $2 \cdot 2 = 4,$ $2 \cdot 3 = 6,$ $2 \cdot 5 = 10,$ and $3 \cdot 3 = 9.$ Handle $\{3, 6, 9\}$ and $\{5, 10\}$ separately.

Since $1 \cdot 1 = 1,$ no product-free set contains $1.$ Split by the least element $t.$ If $t \ge 4,$ any product of two elements is at least $16 \gt 10,$ so every subset of $\{4, 5, \ldots, 10\}$ works: $2^7 = 128$ subsets, including the empty set.

If $t = 2:$ then $4 \notin S$ (as $2 \cdot 2 = 4$), while $7$ and $8$ are unrestricted ($2$ choices each). Among $\{3, 6, 9\},$ the constraints $2 \cdot 3 = 6$ and $3 \cdot 3 = 9$ leave exactly $\varnothing,$ $\{3\},$ $\{6\},$ $\{9\},$ $\{6, 9\}$ — $5$ choices. Among $\{5, 10\},$ the constraint $2 \cdot 5 = 10$ leaves $3$ choices. That gives $2 \cdot 2 \cdot 5 \cdot 3 = 60$ sets. If $t = 3:$ then $9 \notin S$ (as $3 \cdot 3 = 9$), and any subset of $\{4, 5, 6, 7, 8, 10\}$ may be added since all other products exceed $10:$ $2^6 = 64$ sets.

In total there are $128 + 60 + 64 = 252$ product-free subsets.