Circles $\omega$ and $\gamma$ are drawn such that $\omega$ is internally tangent to $\gamma$, the distance between their centers are $5$, and the area inside of $\gamma$ but outside of $\omega$ is $100\pi$. What is the sum of the radii of the circles?

[asy]
size(3cm);
real lw=0.4, dr=0.3;
real r1=14, r2=9;
pair A=(0,0), B=(r1-r2,0);
draw(A--B,dashed);
draw(circle(A,r1),linewidth(lw)); draw(circle(B,r2),linewidth(lw));
filldraw(circle(A,dr)); filldraw(circle(B,dr));
label("$5$",(A+B)/2,dir(-90));
label("$\gamma$",A+r1*dir(135),dir(135)); label("$\omega$",B+r2*dir(135),dir(135));
[/asy]

[i]2020 CCA Math Bonanza Individual Round #2[/i]
1. Let the radius of circle $\omega$ be \( r \) and the radius of circle $\gamma$ be \( R \).
2. Since circle $\omega$ is internally tangent to circle $\gamma$, the distance between their centers is \( R - r = 5 \).
3. The area inside circle $\gamma$ but outside circle $\omega$ is given as \( 100\pi \).
4. The area of circle $\gamma$ is \( \pi R^2 \) and the area of circle $\omega$ is \( \pi r^2 \).
5. The area inside circle $\gamma$ but outside circle $\omega$ is the difference of these areas:
   \[
   \pi R^2 - \pi r^2 = 100\pi
   \]
6. Dividing both sides by \( \pi \), we get:
   \[
   R^2 - r^2 = 100
   \]
7. We also know from the distance between the centers:
   \[
   R - r = 5
   \]
8. We now have a system of equations:
   \[
   \begin{cases}
   R^2 - r^2 = 100 \\
   R - r = 5
   \end{cases}
   \]
9. We can use the identity \( R^2 - r^2 = (R - r)(R + r) \) to rewrite the first equation:
   \[
   (R - r)(R + r) = 100
   \]
10. Substituting \( R - r = 5 \) into the equation, we get:
    \[
    5(R + r) = 100
    \]
11. Solving for \( R + r \), we divide both sides by 5:
    \[
    R + r = 20
    \]
12. Therefore, the sum of the radii of the circles is:
    \[
    \boxed{20}
    \]