HMMT-Feb 2013 team, Problem 4

Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers whose sum is 20 . Determine with proof the smallest possible value of \[ \sum_{1 \leq i<j \leq 5}\left\lfloor a_{i}+a_{j}\right\rfloor \]

Solution: We claim that the minimum is 72 . This can be achieved by taking $a_{1}=a_{2}=a_{3}=$ $a_{4}=0.4$ and $a_{5}=18.4$. To prove that this is optimal, note that \[ \sum_{1 \leq i<j \leq 5}\left\lfloor a_{i}+a_{j}\right\rfloor=\sum_{1 \leq i<j \leq 5}\left(a_{i}+a_{j}\right)-\left\{a_{i}+a_{j}\right\}=80-\sum_{1 \leq i<j \leq 5}\left\{a_{i}+a_{j}\right\} \] so it suffices to maximize \[ \sum_{1 \leq i<j \leq 5}\left\{a_{i}+a_{j}\right\}=\sum_{i=1}^{5}\left\{a_{i}+a_{i+2}\right\}+\sum_{i=1}^{5}\left\{a_{i}+a_{i+1}\right\} \] where $a_{6}=a_{1}$ and $a_{7}=a_{2}$, Taking each sum modulo 1 , it is clear that both are integers. Thus, the above sum is at most $2 \cdot 4=8$, and our original expression is at least $80-8=72$, completing the proof. $\fbox{72}$.

Answer: 72