2011/12 AHSMC Part 1 Problem 12

Let $x_1$, …, $x_{20}$ be the given numbers. If $d$ is the greatest common divisor of these numbers then $$x_1 + \cdots + x_{20} = d\left(\frac{x_1}{d} + \cdots + \frac{x_{20}}{d}\right) = 462 = 21 \cdot 22.$$ The value $d = 22$ is obtained if $x_1 = \cdots = x_{19} = d$ and $x_{20} = 2d$. For each $i$, $\frac{x_i}{d} \ge 1$. Hence $d \le \frac{462}{20} = 23.1$. Since $d$ divides $462$, the largest value for $d$ is indeed $22$. The answer is (B).