Junior Balkan Mathematical Olympiad 2021 - Problem 2 | Solution - 2 July 2021



Problem 2. For any set $A = \{x_1, x_2, x_3, x_4, x_5\}$ of five distinct positive integers denote by $S_A$ the sum of its elements, and denote by $T_A$ the number of triples $(i, j, k)$ with $1 \le i < j < k \le 5$ for which $x_i + x_j + x_k$ divides $S_A$ . Find the largest possible value of $T_A$ .

Solution For distinct indices $i,j,k,m,n$ we have $x_i + x_j + x_k \mid x_i + x_j + x_k + x_m + x_n \Leftrightarrow x_i + x_j + x_k \mid x_m + x_n$ . Without loss of generality let $x_1 < x_2 < x_3 < x_4 < x_5$ . Then $x_3 + x_4 + x_5 > x_1 + x_2$ , $x_2 + x_4 + x_5 > x_1 + x_3$ , $x_1 + x_4 + x_5 > x_2 + x_3$ , $x_1 + x_3 + x_5 > x_2 + x_4$ and $x_2 + x_3 + x_5 > x_1 + x_4$ . Also, $x_1 + x_2 + x_5 \leq x_3 + x_4$ and $x_2 + x_3 + x_4 \leq x_1 + x_5$ cannot simultaenously hold since otherwise adding them and cancelling gives $x_2 \leq 0$ , which is false. Therefore we can have at most $4$ sums. An example with $4$ sums is with $x_1 = 1$ , $x_2 = 2$ , $x_3 = 3$ , $x_4 = 4$ and $x_5$ be such that $6$ , $7$ , $8$ and $9$ divide $x_5 + 10$ , so e.g. $x_5 = 6 \cdot 7 \cdot 8 \cdot 9 - 10 = 3014$ . (An alternative is $x_5 = \mbox{lcm}(6,7,8,9) - 10 = 494.$ )