2017 AMC 10B Problems/Problem 6

Problem

What is the largest number of solid $2\text{-in} \times 2\text{-in} \times 1\text{-in}$ blocks that can fit in a $3\text{-in} \times 2\text{-in}\times3\text{-in}$ box?

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

Solution

We find that the volume of the larger block is $18$, and the volume of the smaller block is $4$. Dividing the two, we see that only a maximum of four $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is $\boxed{\textbf{(B) }4}$.

Video Solution

https://youtu.be/lgdWiCz6M3c

~savannahsolver

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions

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