Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$.  For how many values of $x$ is $f_{100}(x)=0$?
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$
1. Draw the graph of $f_0(x)$ by dividing the domain into three parts.

2. Apply the recursive rule a few times to find the pattern.
Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same.

3. Extrapolate to $f_{100}$. Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$.

$f_{100}(x)$ reaches $0$ at $x = -300$, then zigzags between $0$ and $-1$, hitting $0$ at every even $x$, before leaving $0$ at $x = 300$.
This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$. This is a $601$-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{601+1}{2} = \boxed{\textbf{(C) }301}$.
(Revised by Flamedragon & Jason,C & [emerald_block](https://artofproblemsolving.com/wiki/index.php/User:Emerald_block))