Problem 1299

Solution:

The answer is (B). Consider a sequence $b_{1}, b_{2}, \ldots, b_{2017}$ of positive integers, all distinct, whose sum is $2016 \cdot 2017 \cdot 2018$. Let $d$ be the greatest common divisor of the elements of this sequence. By definition, the numbers $a_{1}:=b_{1} / d, a_{2}:=b_{2} / d, \ldots, a_{2017}:=b_{2017} / d$ are also positive integers, all distinct, and their sum is therefore at least equal to the sum $1+2+\cdots+2017$. We then have
$$2016 \cdot 2017 \cdot 2018 = b_{1} + \cdots + b_{2017} = d \cdot (a_{1} + a_{2} + \cdots + a_{2017}) \geq d \cdot (1+2+\cdots+2017) = d \cdot \frac{2017 \cdot 2018}{2},$$
from which $d \leq 2 \cdot 2016$. On the other hand, letting $D = 2 \cdot 2016$, the sequence $D, 2D, 3D, \cdots, 2017D$ is one of the sequences that are written on the blackboard, and clearly the greatest common divisor of its elements is exactly $D$.