Difference between revisions of "2016 AIME II Problems/Problem 4"
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==Solution== | ==Solution== | ||
By counting the number of green cubes <math>2</math> different ways, we have <math>12a=20b</math>, or <math>a=\dfrac{5}{3} b</math>. Notice that there are only <math>3</math> possible colors for unit cubes, so for each of the <math>1 \times b \times c</math> layers, there are <math>bc-21</math> yellow cubes, and similarly there are <math>ac-45</math> red cubes in each of the <math>1 \times a \times c</math> layers. Therefore, we have <math>a(bc-21)=25b</math> and <math>b(ac-45)=9a</math>. We check a few small values of <math>a,b</math> and solve for <math>c</math>, checking <math>(a,b)=(5,3)</math> gives <math>c=12</math> with a volume of <math>180</math>, <math>(a,b)=(10,6)</math> gives <math>c=6</math> with a volume of <math>360</math>, and <math>(a,b)=(15,9)</math> gives <math>c=4</math>, with a volume of <math>540</math>. Any higher <math>(a,b)</math> will <math>ab>180</math>, so therefore, the minimum volume is <math>\boxed{180}</math>. | By counting the number of green cubes <math>2</math> different ways, we have <math>12a=20b</math>, or <math>a=\dfrac{5}{3} b</math>. Notice that there are only <math>3</math> possible colors for unit cubes, so for each of the <math>1 \times b \times c</math> layers, there are <math>bc-21</math> yellow cubes, and similarly there are <math>ac-45</math> red cubes in each of the <math>1 \times a \times c</math> layers. Therefore, we have <math>a(bc-21)=25b</math> and <math>b(ac-45)=9a</math>. We check a few small values of <math>a,b</math> and solve for <math>c</math>, checking <math>(a,b)=(5,3)</math> gives <math>c=12</math> with a volume of <math>180</math>, <math>(a,b)=(10,6)</math> gives <math>c=6</math> with a volume of <math>360</math>, and <math>(a,b)=(15,9)</math> gives <math>c=4</math>, with a volume of <math>540</math>. Any higher <math>(a,b)</math> will <math>ab>180</math>, so therefore, the minimum volume is <math>\boxed{180}</math>. | ||
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== Solution 2 == | == Solution 2 == |
Latest revision as of 18:55, 18 October 2020
Contents
Problem
An rectangular box is built from unit cubes. Each unit cube is colored red, green, or yellow. Each of the layers of size parallel to the faces of the box contains exactly red cubes, exactly green cubes, and some yellow cubes. Each of the layers of size parallel to the faces of the box contains exactly green cubes, exactly yellow cubes, and some red cubes. Find the smallest possible volume of the box.
Solution
By counting the number of green cubes different ways, we have , or . Notice that there are only possible colors for unit cubes, so for each of the layers, there are yellow cubes, and similarly there are red cubes in each of the layers. Therefore, we have and . We check a few small values of and solve for , checking gives with a volume of , gives with a volume of , and gives , with a volume of . Any higher will , so therefore, the minimum volume is .
Solution 2
The total number of green cubes is given by .
Let be the number of red cubes on each one of the layers then the total number of red cubes is . Substitute gives .
Repeating the procedure on the number of yellow cubes on each of the layers gives .
Therefore and . Multiplying yields .
Since is fixed, is minimized when is maximized, which occurs when , are minimized (since each of , is fixed). Thus
~ Nafer
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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