, 2013

Solution:
Answer: $34$

We do casework based on the largest element of $S$. Call a set $n$-free if none of its subsets have elements summing to $n$.

Case 1: The largest element of $S$ is $6$. Then $4 \notin S$. If $5 \notin S$, then we wish to find all $4$-free subsets of $\{1,2,3\}$ (note that $1+2+3=6<10$). We just cannot include both $1,3$, so we have $2(2^{2}-1)=6$ choices here.

If $5 \in S$, then we want $4,5$-free subsets of $\{1,2,3\}$. The only $4$-but-not-$5$-free subset is $\{2,3\}$, so we have $6-1$ choices here, for a case total of $6+5=11$.

Case 2: The largest element of $S$ is $5$. We seek $5,10$-free subsets of $\{1,2,3,4\}$. We just cannot have both $1,4$ or both $2,3$ (note that getting $10$ requires the whole set), so we have $(2^{2}-1)^{2}=9$ subsets in this case.

Case 3: The largest element of $S$ is at most $4$. (So we want a $4$-free subset of $\{1,2,3,4\}$.) The only way to sum to $10$ with $1,2,3,4$ is by using all the terms, so we simply discount the empty set and $\{1,2,3,4\}$, for a total of $2^{4}-2=14$ subsets.

In conclusion, the total number of subsets is $11+9+14=34$.