Joe has exactly enough paint to paint the surface of a cube whose side length is 2.  It turns out that this is also exactly enough paint to paint the surface of a sphere.  If the volume of this sphere is $\frac{K \sqrt{6}}{\sqrt{\pi}}$, then what is $K$?
Since the cube has six sides, each of area $2^2 = 4$, the surface area of the cube is 24.  Since Joe's paint will exactly cover the cube and it will also exactly cover the sphere, the sphere also must have surface area 24.

If $r$ is the radius of the sphere, this tells us that \[ 4 \pi r^2 = 24 , \]or $r^2 = 6/\pi$, so \[ r = \sqrt{6/\pi} = \sqrt{6}/\sqrt{\pi}. \]Therefore the volume of the sphere is \[ \frac{4}{3} \pi r^3 =\frac{4}{3} \pi \Bigl( \frac{\sqrt{6}}{\sqrt{\pi}} \Bigr)^3 = \frac{4}{3} \pi \cdot \frac{6 \sqrt{6}}{\pi \sqrt{\pi}}
= \frac{8 \sqrt{6}}{\sqrt{\pi}} . \]Thus $\boxed{K=8}$.
8
Level 5
Geometry
Problem: Joe has exactly enough paint to paint the surface of a cube whose side length is 2.  It turns out that this is also exactly enough paint to paint the surface of a sphere.  If the volume of this sphere is $\frac{K \sqrt{6}}{\sqrt{\pi}}$, then what is $K$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of K, we first need to calculate the surface area of the cube. The formula for the surface area of a cube is given by \(6s^2\), where \(s\) is the side length. For a cube with a side length of 2, the surface area is:

\[
6(2^2) = 6 \times 4 = 24.
\]

This means Joe has enough paint to cover a surface area of 24 square units.

Next, we need to determine the radius of the sphere that has the same surface area. The formula for the surface area of a sphere is \(4\pi r^2\). Setting this equal to the surface area of the cube, we have:

\[
4\pi r^2 = 24.
\]

To solve for \(r^2\), we divide both sides by \(4\pi\):

\[
r^2 = \frac{24}{4\pi} = \frac{6}{\pi}.
\]

Taking the square root gives us the radius:

\[
r = \sqrt{\frac{6}{\pi}}.
\]

Now, we need to find the volume of the sphere, which is given by the formula \(\frac{4}{3}\pi r^3\). First, we calculate \(r^3\):

\[
r^3 = \left(\sqrt{\frac{6}{\pi}}\right)^3 = \frac{6\sqrt{6}}{\pi^{3/2}}.
\]

Now we can substitute \(r^3\) into the volume formula:

\[
V = \frac{4}{3}\pi \left(\frac{6\sqrt{6}}{\pi^{3/2}}\right).
\]

Simplifying this expression, we have:

\[
V = \frac{4 \times 6\sqrt{6}}{3\pi^{1/2}} = \frac{24\sqrt{6}}{3\sqrt{\pi}} = \frac{8\sqrt{6}}{\sqrt{\pi}}.
\]

According to the problem, the volume of the sphere is also given as \(\frac{K \sqrt{6}}{\sqrt{\pi}}\). By comparing the two expressions for volume, we can see that:

\[
\frac{K \sqrt{6}}{\sqrt{\pi}} = \frac{8\sqrt{6}}{\sqrt{\pi}}.
\]

This implies that \(K = 8\). Thus, the final answer is 8.<|END_OF_LATENT|> The final answer is $8$.