2021 AMC Fall 10A Problem 3

1. "Reshaped but not compressed" means you can mold the clay into any shape, but you cannot squeeze it into less space. So the only thing that matters is volume: all the clay together has to fit into the cube's volume.
2. Volume of the cube: $6 \times 6 \times 6 = 216$.
3. Volume of one ball, using $V = \frac{4}{3}\pi r^3$ with $r = 2$:
$$\frac{4}{3}\pi \times 8 = \frac{32\pi}{3}.$$
With $\pi \approx 3.14$, this is about $\frac{32 \times 3.14}{3} \approx 33.5$.
4. Check how many balls' worth of clay fits:
- $6$ balls: $6 \times 33.5 = 201$, which is less than $216$. They fit.
- $7$ balls: $7 \times 33.5 \approx 234.5$, which is more than $216$. They do not fit.
5. So the most is $6$ balls.

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

The phrase "can be reshaped" removes all the geometry of packing round balls into a box. What is left is one comparison of volumes. To be sure the rounding of $\pi$ does not matter, check the two nearest whole numbers: $6$ balls use about $201$, safely under $216$, and $7$ use about $234$, safely over.

Avoid decimals. The number of balls is at most $216 \div \frac{32\pi}{3} = \frac{648}{32\pi} = \frac{20.25}{\pi}$. Since $6\pi \approx 18.8$ is less than $20.25$, and $7\pi \approx 22.0$ is more than $20.25$, the answer is between $6$ and $7$, so $6$ balls fit.

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 $\boxed{\textbf{(D)}\ 6}$.