Doppel 05.06.2023 13:40 Find the maximum value of an integer $B$ such that for every 9 distinct natural number with the sum of $2023$, there must exist a sum of 4 of the number that is greater than or equal to $B$
Seicchi28 05.06.2023 17:58 Solution The answer is $910$. To show that $B \le 910$, consider the following $9$ numbers, $\{229, 228, 227, 226, 225, 224, 223, 221, 220 \}$, which add up to $2023$, whose largest $4$ elements add up to $229+228+227+226 = 910$. Now, let $b_1 > b_2 > \dots > b_9$ be natural numbers that add up to $2023$. We'll prove that $b_1+b_2+b_3+b_4 \ge 910$. FTSOC, assume $b_1+b_2+b_3+b_4 \le 909$. Then, $909 \ge b_1 + b_2 + b_3 + b_4 \ge (b_4+3)+(b_4+2)+(b_4+1)+b_4 = 4b_4+6 \implies 225 \ge b_4$, which means $224 \ge b_5$, $223 \ge b_6$, etc., and, \[2023 = (b_1 + b_2 + b_3 + b_4) + b_5 + b_6 + b_7 + b_8 + b_9 \le 909 + 224 + 223 + 222 + 221 + 220 = 2019, \]contradiction.