Circle $\omega_1$ and $\omega_2$ have centers $(0,6)$ and $(20,0)$, respectively.  Both circles have radius $30$, and intersect at two points $X$ and $Y$.  The line through $X$ and $Y$ can be written in the form $y = mx+b$.  Compute $100m+b$.

[i]Proposed by Evan Chen[/i]
1. We start with the equations of the two circles. The first circle $\omega_1$ has center $(0,6)$ and radius $30$, so its equation is:
   \[
   x^2 + (y - 6)^2 = 900
   \]
   The second circle $\omega_2$ has center $(20,0)$ and radius $30$, so its equation is:
   \[
   (x - 20)^2 + y^2 = 900
   \]

2. To find the line through the points of intersection $X$ and $Y$, we subtract the second equation from the first:
   \[
   x^2 + (y - 6)^2 - [(x - 20)^2 + y^2] = 900 - 900
   \]
   Simplifying the left-hand side:
   \[
   x^2 + y^2 - 12y + 36 - (x^2 - 40x + 400 + y^2) = 0
   \]
   \[
   x^2 + y^2 - 12y + 36 - x^2 + 40x - 400 - y^2 = 0
   \]
   \[
   -12y + 36 + 40x - 400 = 0
   \]
   \[
   40x - 12y - 364 = 0
   \]

3. Solving for $y$ in terms of $x$:
   \[
   40x - 12y = 364
   \]
   \[
   -12y = -40x + 364
   \]
   \[
   y = \frac{40}{12}x - \frac{364}{12}
   \]
   Simplifying the fractions:
   \[
   y = \frac{10}{3}x - \frac{91}{3}
   \]

4. The equation of the line through $X$ and $Y$ is:
   \[
   y = \frac{10}{3}x - \frac{91}{3}
   \]

5. To find $100m + b$, where $m = \frac{10}{3}$ and $b = -\frac{91}{3}$:
   \[
   100m + b = 100 \left(\frac{10}{3}\right) + \left(-\frac{91}{3}\right)
   \]
   \[
   = \frac{1000}{3} - \frac{91}{3}
   \]
   \[
   = \frac{1000 - 91}{3}
   \]
   \[
   = \frac{909}{3}
   \]
   \[
   = 303
   \]

The final answer is $\boxed{303}$