Problem: For any of the $16$ subsets with cardinality $3$ of $\mathbb Z_7$, show that there always exists two sets, so that if you sum the elements of one set, it is the same as the other. For instance, $\sum ${$1,2,3$}=$\sum ${$5,1,0$}.
My solution: I first set bounds on $\sum X$, where $X$ is a subset of $\mathbb Z_7$. Well, we know that the minimal sum is $0+1+2=3$, so $3\leq \sum X$. And we also know that the maximum value of this sum is $4+5+6=15$, so we have the bounds $3\leq \sum X \leq 15$. Thus $\sum X$ can take at most $12$ different values. Let $A$ be the set of all possible values of the $12$ digit sums of $X$. Similarly, denote $B$ as the set of all possible values of the $12$ digit sums of $Y$. We then have that the problem reduces to the following statement: $A \cap B \neq 0$.
Assume that $A \cap B =0$. Then if we denote $C$ as the set of all possible sums of all possible subsets of $\mathbb Z_7$, we have that $|C|=12$. Also note that by the inclusion-exclusion principle, $|A \cup B|$=$|A|+|B| -$$|A \cap B|$=$24$. But then notice that $|A \cup B|$=$|C|$=$12$. A contradiction.
After this, one can make the more general statement that this could work for any two sets, for any value of the sums of those sets, and be far more general.
I would like to know whether my solution is correct and that if one could try and apply the pigeonhole principle to this problem.
For any $3$-element subset of the set $\{0,1,2,3,4,5,6\}$, you correctly note that its sum is at least $0+1+2=3$ and at most $4+5+6=15$. This leaves at most $13$ possible values for the sum of such a $3$-element subset. Hence $16$ such subsets cannot all have distinct sums, meaning that there are at least $2$ subsets with the same sum.