Right $\vartriangle ABC$ has side lengths $6, 8$, and $10$. Find the positive integer $n$ such that the area of the region inside the circumcircle but outside the incircle of $\vartriangle ABC$ is $n\pi$.
[img]https://cdn.artofproblemsolving.com/attachments/d/1/cb112332069c09a3b370343ca8a2ef21102fe2.png[/img]
1. **Identify the side lengths and the type of triangle:**
   Given a right triangle $\vartriangle ABC$ with side lengths $6$, $8$, and $10$. Since $10^2 = 6^2 + 8^2$, it confirms that $\vartriangle ABC$ is a right triangle with the hypotenuse $10$.

2. **Calculate the circumradius $R$:**
   For a right triangle, the circumradius $R$ is half the length of the hypotenuse.
   \[
   R = \frac{10}{2} = 5
   \]

3. **Calculate the inradius $r$:**
   The area $A$ of $\vartriangle ABC$ can be calculated using the formula for the area of a right triangle:
   \[
   A = \frac{1}{2} \times 6 \times 8 = 24
   \]
   The semiperimeter $s$ of $\vartriangle ABC$ is:
   \[
   s = \frac{6 + 8 + 10}{2} = 12
   \]
   The inradius $r$ is given by:
   \[
   r = \frac{A}{s} = \frac{24}{12} = 2
   \]

4. **Calculate the area of the circumcircle:**
   The area of the circumcircle is:
   \[
   \text{Area of circumcircle} = \pi R^2 = \pi \times 5^2 = 25\pi
   \]

5. **Calculate the area of the incircle:**
   The area of the incircle is:
   \[
   \text{Area of incircle} = \pi r^2 = \pi \times 2^2 = 4\pi
   \]

6. **Calculate the area of the region inside the circumcircle but outside the incircle:**
   The desired area is:
   \[
   \text{Desired area} = \text{Area of circumcircle} - \text{Area of incircle} = 25\pi - 4\pi = 21\pi
   \]

The final answer is $\boxed{21}$.