Find the number of real solutions to the equation
\[\frac{1}{x - 1} + \frac{2}{x - 2} + \frac{3}{x - 3} + \dots + \frac{100}{x - 100} = x.\]
If we multiply the left-hand side by $x-1$, we get
\[1 + \frac{2(x-1)}{x-2} + \frac{3(x-1)}{x-3} + \dots + \frac{100(x-1)}{x-100} = x(x-1).\] Now if we subtract $1$ from both sides, we get
\[\frac{2(x-1)}{x-2} + \frac{3(x-1)}{x-3} + \dots + \frac{100(x-1)}{x-100} = x(x-1) - 1.\] Right. So in each segment, there is exactly one intersection. So there is exactly one solution in that interval. So there is exactly one solution in that interval. Therefore, there are $99+1+1=101$ solutions total.

