2021 Fall AMC 10A Problems/Problem 3

Problem

What is the maximum number of balls of clay of radius $2$ that can completely fit inside a cube of side length $6$ assuming the balls can be reshaped but not compressed before they are packed in the cube?

$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution 1 (Inequality)

The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac43\cdot\pi\cdot2^3=\frac{32}{3}\pi.$

Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is $\left\lfloor\frac{V_{\text{cube}}}{V_{\text{ball}}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.$ Approximating with $\pi\approx3.14,$ we have $\left\lfloor\frac{81}{13}\right\rfloor \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq \left\lfloor\frac{81}{12}\right\rfloor,$ or $6 \leq \left\lfloor\frac{81}{4\pi}\right\rfloor \leq 6.$ Clearly, we get $\left\lfloor\frac{81}{4\pi}\right\rfloor=\boxed{\textbf{(D) }6}.$

~NH14 ~MRENTHUSIASM

Solution 2 (Arithmetic)

A sphere with radius $2$ has volume $\frac {32\pi}{3}$ . A cube with side length $6$ has volume $216$ . If $\pi$ was $3$ , it would fit 6.75 times inside. Since $\pi$ is approximately $5$ % larger than $3$ , it is safe to assume that the $3$ balls of clay can fit $6$ times inside. Therefore, our answer is $\boxed {(D)6}$ .

~Arcticturn

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2021 Fall AMC 10A Problems/Problem 3

Contents

Problem

Solution 1 (Inequality)

Solution 2 (Arithmetic)

See Also