jmc

We can represent the sum as $$\sum _{n=1}^{1987}n(1988-n).$$This is equal to $$\begin{array}{rl}\sum _{n=1}^{1987}(1988n-n^2)& =1988\sum _{n=1}^{1987}n-\sum _{n=1}^{1987}{n}^2\\ & =1988\cdot \frac{1987\cdot 1988}{2}-\frac{1987\cdot 1988\cdot 3975}{6}\\ & =\frac{1987\cdot 1988}{6}(3\cdot 1988-3975)\\ & =\frac{1987\cdot 2\cdot 994}{6}\cdot 1989\\ & =\frac{1987\cdot 994}{3}\cdot 1989\\ & =1987\cdot 994\cdot 663.\end{array}$$Thus, $x=\boxed{663}.$