Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\]has $12$ distinct real solutions.
We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|=c\pm21.\]
Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values:
\[f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0  \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 020 \end{cases} \end{cases}.\]
We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ($f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$-axis) should facilitate the process of graphing.

Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$

For $f(x)=c+21$ to have $6$ distinct real solutions, we need $0\leq c<79.$
For $f(x)=c-21$ to have $6$ distinct real solutions, we need $21<c<121.$
Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=\boxed{057}$ such integers $c.$
~MRENTHUSIASM