Recursive absolute value functions 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 $f_{100}(x)=0$?
I tried starting at $f_{100}(x) = 0$, then $f_{99}(x ) =1 $ or $-1$ and so on until i found the pattern, it so happens that $f_0(x) = 0,-2,+2,-4,+4...-100,+100$
I assumed that each of those had 3 solutions in each of them, this there are $3(101)=303$ solutions (because there are 3 intervals in the absolute value), but the answer is 301 solutions, what value of $f_0(x)$ from the list had only 1 value of x?
problem from:
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_24, they have solutions there but the 1st one is graphing and the 2nd one is quite like mine, but confusing.
 A: At first it has three roots at $-200,0,+200$. Now for example we investigate only around $x=0$ what happens. in each level the number of roots become one more and also change the even or odd like below:
$$f_0 \longrightarrow x=0$$
$$f_1 \longrightarrow x=-1,+1$$
$$f_2 \longrightarrow x=-2,0,+2$$
So for $f_n$ at all points we have:
$$f_n \longrightarrow x=\left\{\begin{array}{r} -200+n-2k+1 \\ 0+n-2k+1 \\ +200+n-2k+1 \\ \end{array}\right\} 1 \leqslant k \leqslant n$$
Therefore $f_{100}$ be:
$$\{ -300,...,-100 \} \ \cup \ \{ -100,...,+100 \} \ \cup \ \{ +100,...,+300 \}$$
So it has all even integers from $-300$ to $+300$ as roots. If you count number of roots around every points seperately, you get $303$ roots that is incorrect because of $\pm100$ were counted twice! So it has exactelt $301$ roots as shown. In fact for every $n\geqslant100$ it's same: it has all integer from $-(200+n)$ to $+(200+n)$ that are all even iff $n$ is even, (and all odd iff $n$ is odd). So it has $(n+201)$ roots exactly.
