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Difference between revisions of "2017 AMC 10B Problems/Problem 6"

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==Problem==
 
==Problem==
  
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What is the largest number of solid <math>2</math>in x<math>2</math>in x<math>1</math>in blocks that can fit in a <math>3</math>-in by <math>2</math> in by <math>3</math>in box?
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What is the largest number of solid <math>2\text{-in} \times 2\text{-in} \times 1\text{-in}</math> blocks that can fit in a <math>3\text{-in} \times 2\text{-in}\times3\text{-in}</math> box?
  
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<math>\textbf{(A)}\ [3]\qquad\textbf{(B)}\ [4]\qquad\textbf{(C)}\ [5]\qquad\textbf{(D)}\ [6]\qquad\textbf{(E)}\ [7]</math>
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<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
  
 
==Solution==
 
==Solution==
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By simply finding the volume of the larger block, we see that its area is <math>18</math>. The volume of the smaller block is <math>4</math>. Dividing the two, we see that only a maximum of <math>4</math> <math>2</math>in x<math>2</math>in x<math>1</math>in blocks can fit inside a <math>3</math>-in by <math>2</math> in by <math>3</math>in box. <math>\qquad\textbf{(B)}\ [4]</math>
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We find that the volume of the larger block is <math>18</math>, and the volume of the smaller block is <math>4</math>. Dividing the two, we see that only a maximum of four <math>2</math> by <math>2</math> by <math>1</math> blocks can fit inside the <math>3</math> by <math>3</math> by <math>2</math> block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is <math>\boxed{\textbf{(B) }4}</math>.
  
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==Video Solution==
 +
https://youtu.be/lgdWiCz6M3c
  
−
{{AMC10 box|year=2017|ab=b|num-b=5|num-a=7}}
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~savannahsolver
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/XRfOULUmWbY
 +
 
 +
~IceMatrix
 +
 
 +
==See Also==
 +
 
 +
{{AMC10 box|year=2017|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:04, 1 May 2021

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

Video Solution by TheBeautyofMath

https://youtu.be/XRfOULUmWbY

~IceMatrix

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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