Let $S$ be a set of 9 integers. Does a subset $A$ $\subseteq$ $S$ and $A$ $\not =$ $\emptyset$ exist such that the sum of its elements is divisible by 9? What about another subset such that the sum of its elements is divisible by 10?
I’ve tried to construct $S$, and $A$ as a consequence so, for example, for $S=\{5, 4, 1, a, b, c, d, e, f\}$ and for $A=\{5, 4\}$ the first point of the problem is satisfied and for $A=\{5, 4, 1\}$ the second point is also satisfied.
On a general level, let $x$ and $y$ $\in$ $A$ such that $(x+y) \mid 9$ so it must exist a $z$ such that $9/(x+y) = z$ so $x+y$ should be $3, 1$ or $9$ but I don’t how to continue.
Say $S=\{s_1,s_2,\ldots,s_9\}$, and consider the following 10 values: \begin{align} 0&,\\ s_1&,\\ s_1+s_2&,\\ \vdots&\\ s_1+s_2+\cdots+s_9&.\\ \end{align}
By the pigeonhole principle, there must exist two values $s_1+\cdots+s_i$ and $s_1+\cdots+s_j$ ($i<j$) with the same remainder while divided by 9, then $s_{i+1}+\cdots+s_j$ must be divisible by 9, so we have found the set $A$: $A=\{s_{i+1},\ldots,s_j\}$.
For the case where the divisor is 10, such $A$ may not exist. For example, as David said in the comment, you can set $S=\{1,11,21,31,\ldots,81\}$. The last digit of the sum of any subset of $S$ cannot be 0, so is not divisible by 10.