Difference between revisions of "2017 AMC 10B Problems/Problem 6"
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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 <math>4</math> <math> | + | 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 <math>4</math> <math>2x2x1</math> blocks can fit inside the <math>3x3x2</math> box. <math>\boxed{\textbf{(B) }4}</math> |
{{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}} | ||
Revision as of 11:43, 16 February 2017
Problem
What is the largest number of solid 
by by 
blocks that can fit in a 
by by 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 
blocks can fit inside the 
box. 
| 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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