China Western Mathematical Olympiad

Define $S(\varnothing )=0$. Let $T_0=\{3,6\}$, $T_1=\{1,4,7\}$, $T_2=\{2,5,8\}$. For $A\subseteq T$, let $A_0=A\cap T_0$, $A_1=A\cap T_1$, $A_2=A\cap T_2$, then

$$S(A)=S(A_0)+S(A_1)+S(A_2)\equiv |A_1|-|A_2|(\mathrm{mod}3),$$

So $3|S(A)$ if and only if $|A_1|\equiv |A_2|(\mathrm{mod}3)$. It follows that $$(|A_1|,|A_2|)=(0,0),(0,3),(3,0),(3,3),(1,1),(2,2).$$

The number of nonempty subsets $A$ so that $3|S(A)$ is $$\begin{gathered} 2^2\left(\genfrac{}{}{0ex}{}{3}{0}\right)\left(\genfrac{}{}{0ex}{}{3}{0}\right)+\left(\genfrac{}{}{0ex}{}{3}{0}\right)\left(\genfrac{}{}{0ex}{}{3}{3}\right)+\left(\genfrac{}{}{0ex}{}{3}{3}\right)\left(\genfrac{}{}{0ex}{}{3}{0}\right)+\left(\genfrac{}{}{0ex}{}{3}{3}\right)\left(\genfrac{}{}{0ex}{}{3}{3}\right) \\ +\left(\genfrac{}{}{0ex}{}{3}{1}\right)\left(\genfrac{}{}{0ex}{}{3}{1}\right)+\left(\genfrac{}{}{0ex}{}{3}{2}\right)\left(\genfrac{}{}{0ex}{}{3}{2}\right)-1=87. \end{gathered}$$

If $3|S(A)$ and $5|S(A)$, then $15|S(A)$. Since $S(T)=36$, so the value $S(A)$ is 15 or 30 (if $3|S(A)$ and $5|S(A)$).

Furthermore, $$\begin{array}{rl}15& =8+7=8+6+1=8+5+2=8+4+3\\ & =8+4+2+1=7+6+2=7+5+3\\ & =7+5+2+1=7+4+3+1=6+5+4\\ & =6+5+3+1=6+4+3+2\\ & =5+4+3+2+1,\end{array}$$ $$36-30=6=5+1=4+2=3+2+1.$$ So the number of $A$ such that $3|S(A)$, $5|S(A)$, and $A\ne \varnothing$ is 17.

The answer is $87-17=70$.