14. Let $S=\{1,2,3,4, \ldots, 16\}$. In each of the following subsets of $S$,
$$
\{6\},\{1,2,3\},\{5,7,9,10,11,12\},\{1,2,3,4,5,6,7,8,9\}
$$
the sum of all the elements is a multiple of 3 . Find the total number of non-empty subsets $A$ of $S$ such that the sum of all elements in $A$ is a multiple of 3 .
14. Answer: 21855.
Let $S_{i}=\{x \in S: x=i(\bmod 3)\}$ for $i=0,1,2$. Note that $\left|S_{0}\right|=5,\left|S_{1}\right|=6$ and $\left|S_{2}\right|=5$. Let $\partial$ be the set of all subsets $A$ of $S$ such that $\sum_{x \in d} x$ is a multiple of 3 . Note that for any $A \subseteq S$,
$$
\sum_{x \in A} x=\sum_{i=0}^{2} \sum_{x \in A \cap s_{i}} x \equiv\left|A \cap S_{1}\right|+2\left|A \cap S_{2}\right|(\bmod 3) \text {. }
$$

Thus, $A \in \partial$ if and only if $\left|A \cap S_{1}\right|=\left|A \cap S_{2}\right|(\bmod 3)$. Thus, it is clear that $|\partial|=2^{\left|S_{0}\right|} m$, where
$$
\begin{array}{c}
m=\left\{\binom{6}{0}+\binom{6}{3}+\binom{6}{6}\right\}\left\{\binom{5}{0}+\binom{5}{3}\right\}+\left\{\binom{6}{1}+\binom{6}{4}\right\}\left\{\binom{5}{1}+\binom{5}{4}\right\} \\
+\left\{\binom{6}{2}+\binom{6}{5}\right\}\left\{\binom{5}{2}+\binom{5}{5}\right\}=683
\end{array}
$$

Hence $|\vec{\partial}|=2^{5} \times 683=21856$. Since we want only non-empty subsets, we have 21855 .