Find the number of real solutions to the equation
\[\frac{1}{x - 1} + \frac{2}{x - 2} + \frac{3}{x - 3} + \dots + \frac{100}{x - 100} = x.\]
Let
\[f(x) = \frac{1}{x - 1} + \frac{2}{x - 2} + \frac{3}{x - 3} + \dots + \frac{100}{x - 100}.\]Consider the graph of $y = f(x).$

[asy]
unitsize(1 cm);

real func(real x) {
  return((1/(x - 1) + 2/(x - 2) + 3/(x - 3) + 4/(x - 4) + 5/(x - 5) + 6/(x - 6))/15);
}

draw((-2,0)--(8,0));
draw((0,-2)--(0,2));
draw((1,-2)--(1,2),dashed);
draw((2,-2)--(2,2),dashed);
draw((3,-2)--(3,2),dashed);
draw((5,-2)--(5,2),dashed);
draw((6,-2)--(6,2),dashed);
draw((-2,-2/4)--(8,8/4));
draw(graph(func,-2,0.99),red);
draw(graph(func,1.01,1.99),red);
draw(graph(func,2.01,2.99),red);
draw(graph(func,5.01,5.99),red);
draw(graph(func,6.01,8),red);

limits((-2,-2),(8,2),Crop);

label("$1$", (1,0), SW);
label("$2$", (2,0), SW);
label("$3$", (3,0), SE);
label("$99$", (5,0), SW);
label("$100$", (6,0), SE);
label("$y = x$", (8,2), E);
label("$y = f(x)$", (8,func(8)), E, red);
[/asy]

The graph of $y = f(x)$ has vertical asymptotes at $x = 1,$ $x = 2,$ $\dots,$ $x = 100.$  In particular, $f(x)$ approaches $-\infty$ as $x$ approaches $n$ from the left, and $f(x)$ approaches $\infty$ as $x$ approaches $n$ from the right, for $1 \le n \le 100.$  Furthermore, $y = 0$ is a vertical asymptote.  In particular, $f(x)$ approaches 0 as $x$ approaches both $\infty$ and $-\infty.$

Thus, the graph of $y = f(x)$ intersects the graph of $y = x$ exactly once on each of the intervals $(-\infty,1),$ $(1,2),$ $(2,3),$ $\dots,$ $(99,100),$ $(100,\infty).$  Therefore, there are a total of $\boxed{101}$ real solutions.
101
Level 5
Intermediate Algebra
Problem: Find the number of real solutions to the equation
\[\frac{1}{x - 1} + \frac{2}{x - 2} + \frac{3}{x - 3} + \dots + \frac{100}{x - 100} = x.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of real solutions to the given equation, we first analyze the left-hand side, which is a sum of rational functions. Each term \(\frac{k}{x - k}\) for \(k = 1, 2, \ldots, 100\) has a vertical asymptote at \(x = k\). Therefore, the function is undefined at these points, leading to potential discontinuities in the overall function.

Next, we consider the behavior of the left-hand side as \(x\) approaches each integer \(k\). As \(x\) approaches \(k\) from the left, \(\frac{k}{x - k} \to -\infty\), and as \(x\) approaches \(k\) from the right, \(\frac{k}{x - k} \to +\infty\). This indicates that the function will cross the line \(y = x\) at least once between each pair of consecutive integers \(k\) and \(k+1\).

The left-hand side is a continuous function on the intervals \((-\infty, 1)\), \((1, 2)\), \((2, 3)\), ..., \((99, 100)\), and \((100, \infty)\). In each of these intervals, the function transitions from \(-\infty\) to \(+\infty\), guaranteeing at least one solution in each interval.

Counting the intervals, we have:
1. One solution in \((-\infty, 1)\)
2. One solution in each of the intervals \((1, 2)\), \((2, 3)\), ..., \((99, 100)\) — there are 99 such intervals.
3. One solution in \((100, \infty)\)

Thus, the total number of intervals is \(1 + 99 + 1 = 101\). Therefore, the final answer is \(101\).<|END_OF_LATENT|> The final answer is $101$.