20. We can find sets of 13 distinct positive integers that add up to 2142. Find the largest possible greatest common divisor of these 13 distinct positive integers.
20. Answer: 21
Solution. Let $d$ be the greatest common divisor (gcd) of these 13 distinct positive integers. Then these 13 integers can be represented as $d a_{1}, d a_{2}, \cdots, d a_{13}$, where $\operatorname{god}\left(a_{1}, a_{2}, \cdots, a_{13}\right)=$ 1. Let $S$ denote $a_{1}+a_{2}+\cdots+a_{13}$. Then $S d=2142$. In order for $d$ to be the largest possible, $S$ must be the smallest. Since $S \geq 1+2+3+\cdots+13=91$ and that $S$ divides 2142 , and that $2142=2 \times 3 \times 7 \times 51$, the smallest possible value of $S$ can be $2 \times 51=102$, and the largest value of $d$ is thus 21 . By choosing $\left(a_{1}, a_{2}, a_{3}, \cdots, a_{12}, a_{13}\right)=(1,2,3, \cdots, 12,24)$, we conclude that $d=21$ is possible.