Difference between revisions of "2017 AMC 10B Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | 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 | + | 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? |
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math> | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math> | ||
==Solution== | ==Solution== | ||
− | 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 | + | 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>. |
+ | ==Video Solution== | ||
+ | https://youtu.be/lgdWiCz6M3c | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/XRfOULUmWbY | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==See Also== | ||
{{AMC10 box|year=2017|ab=B|num-b=5|num-a=7}} | {{AMC10 box|year=2017|ab=B|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:04, 1 May 2021
Problem
What is the largest number of solid blocks that can fit in a box?
Solution
We find that the volume of the larger block is , and the volume of the smaller block is . Dividing the two, we see that only a maximum of four by by blocks can fit inside the by by block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is .
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
~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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