2022/23 AHSMC Part 1 Problem 14

Let $a_1, a_2, \ldots, a_{15}$ be the given positive integers and $M$ the least common multiple of these numbers. Since $$a_1 + a_2 + \cdots + a_{15} = 425 = 15 \cdot 28 + 5$$ and $M \ge a_1$, $\cdots$, $M \ge a_{15}$, we conclude that $15M \ge 15 \cdot 28 + 5$ and hence $M > 28$.

If $M = 29$ then $$a_1, a_2, \cdots, a_{15} \in \{1, 29\}.$$ If at least one of the numbers $a_1, a_2, \ldots, a_{15}$ is equal to $1$, then one obtains $$425 = a_1 + a_2 + \cdots + a_{15} \le 1 + 14 \cdot 29 = 407,$$ which is a contradiction. Hence, $a_1 = a_2 = \cdots = a_{15} = 29$, which is not possible since, otherwise, $$425 = a_1 + a_2 + \cdots + a_{15} = 15 \cdot 29 = 435.$$ If $M = 30$ we may take $a_1 = a_2 = \cdots = a_{14} = 30$, $a_{15} = 5$ and the conditions of the question are verified. The answer is (C).