2021 AMC Fall 10A Problem 3

Because the clay can be reshaped, the only constraint is volume: the total clay volume cannot exceed the cube's volume.

The cube has volume $6^3 = 216$. One ball has volume
$$\frac{4}{3}\pi \cdot 2^3 = \frac{32\pi}{3} \approx 33.5.$$

The number of balls that fit is
$$\left\lfloor \frac{216}{32\pi/3} \right\rfloor = \left\lfloor \frac{648}{32\pi} \right\rfloor = \left\lfloor \frac{20.25}{\pi} \right\rfloor.$$
Since $6\pi \approx 18.85 < 20.25 < 21.99 \approx 7\pi$, the quotient is between $6$ and $7$, so $6$ balls fit and a seventh does not.

The answer is $\boxed{\textbf{(D)}\ 6}$.

"Reshaped but not compressed" is code for "volume is conserved and shape is irrelevant." Once you read it that way, the problem is a single division. Bracketing $20.25$ between $6\pi$ and $7\pi$ avoids any decimal division.

• Treating the balls as rigid spheres of diameter 4 that must be packed geometrically inside the side-6 cube, which would allow only one.
• Using diameter $4$ in place of radius $2$ in the sphere formula, or rounding $6.45$ up to $7$.