An even number of circles are nested, starting with a radius of $1$ and increasing by $1$ each time, all sharing a common point. The region between every other circle is shaded, starting with the region inside the circle of radius $2$ but outside the circle of radius $1.$ An example showing $8$ circles is displayed below. What is the least number of circles needed to make the total shaded area at least $2023\pi$?
[asy] filldraw(circle((0,0),8),gray); filldraw(circle((-1,0),7),white); filldraw(circle((-2,0),6),gray); filldraw(circle((-3,0),5),white); filldraw(circle((-4,0),4),gray); filldraw(circle((-5,0),3),white); filldraw(circle((-6,0),2),gray); filldraw(circle((-7,0),1),white); [/asy]
$\textbf{(A) } 46 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 64$
Notice that the area of the shaded region is $(2^2\pi-1^2\pi)+(4^2\pi-3^2\pi)+(6^2\pi-5^2\pi)+ \cdots + (n^2\pi-(n-1)^2 \pi)$ for any even number $n$.
Using the difference of squares, this simplifies to $(1+2+3+4+\cdots+n) \pi$. So, we are basically finding the smallest $n$ such that $\frac{n(n+1)}{2}>2023 \Rightarrow n(n+1) > 4046$. Since $60(61) > 60^2=3600$, the only option higher than $60$ is $\boxed{\textbf{(E) } 64}$.
~MrThinker