[Combinatorics]
Problem: A set $S$ contains thirteen distinct positive integers whose sum is 120 . Compute the largest possible value for the median of $S$.

Solution: Let $S_{L}$ be the set of the least six integers in $S$, let $m$ be the median of $S$, and let $S_{G}$ be the set of the greatest six integers in $S$. In order to maximize the median, the elements of $S_{L}$ should be as small as possible, so start with $S_{L}=\{1,2,3,4,5,6\}$. Then the sum of $S_{L}$ 's elements is 21, leaving 99 as the sum of $m$ and the six elements of $S_{G}$. If $m=11$ and $S_{G}=\{12,13,14,15,16,17\}$, then the sum of all thirteen elements of $S$ is 119 . It is impossible to increase $m$ any further, because then the smallest set of numbers for $S_{G}$ would be $\{13,14,15,16,17,18\}$, and the sum would be at least 126 . To get the sum to be exactly 120, simply increase either 6 to 7 or 17 to 18 . The answer is $\mathbf{1 1 .}$

Final answer: 11