2021 AMC 10 A Fall

Complete problem set with solutions and individual problem pages

Problem 3 Easy

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?(2021 AMC Fall 10A, Question #3)

  • A.

    3

  • B.

    4

  • C.

    5

  • D.

    6

  • E.

    7

Answer:D

Solution 1:

The volume of the cube is V_{\text {cube }}=6^{3}=216, and the volume of a clay ball is V_{\text {ball }}=\frac{4}{3} \cdot \pi \cdot 2^{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 \approx 3.14, we have 12<4 \pi<13, or \left\lfloor\frac{81}{13}\right\rfloor \leq\left\lfloor\frac{81}{4 \pi}\right\rfloor \leq\left\lfloor\frac{81}{12}\right\rfloor. We simplify to get 6 \leq\left\lfloor\frac{81}{4 \pi}\right\rfloor \leq 6 from which \left\lfloor\frac{81}{4 \pi}\right\rfloor= (D) 6 .

Solution 2:

As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is \left\lfloor\frac{81}{4 \pi}\right\rfloor. By an underestimation \pi \approx 3, we have 4 \pi>12, or \frac{81}{4 \pi}<6 \frac{3}{4}. By an overestimation \pi \approx \frac{22}{7}, we have 4 \pi<\frac{88}{7}, or \frac{81}{4 \pi}>6 \frac{39}{88}. Together, we get 6<6 \frac{39}{88}<\frac{81}{4 \pi}<6 \frac{3}{4}<7 from which \left|\frac{81}{4 \pi}\right|=(\text{D}) 6.

Related Problems
Problem 1 2021 AMC 10 A Fall
Problem 2 2021 AMC 10 A Fall
Problem 4 2021 AMC 10 A Fall
Problem 5 2021 AMC 10 A Fall
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