Difference between revisions of "2016 AIME II Problems/Problem 4"

Revision as of 12:06, 15 January 2017

An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.

Solution

By counting the number of green cubes $2$ different ways, we have $12a=20b$ , or $a=\dfrac{5}{3} b$ . Notice that there are only $3$ possible colors for unit cubes, so for each of the $1 \times b \times c$ layers, there are $bc-21$ yellow cubes, and similarly there are $ac-45$ red cubes in each of the $1 \times a \times b$ layers. Therefore, we have $a(bc-21)=25b$ and $b(ac-45)=9a$ . We check a few small values of $a,b$ and solve for $c$ , checking $(a,b)=(5,3)$ gives $c=12$ with a volume of $180$ , $(a,b)=(10,6)$ gives $c=6$ with a volume of $360$ , and $(a,b)=(15,9)$ gives $c=4$ , with a volume of $540$ . Any higher $(a,b)$ will $ab>180$ , so therefore, the minimum volume is $\boxed{180}$ .

Solution by Shaddoll

@@ Line 2: / Line 2: @@
 ==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 b</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>.
 Solution by Shaddoll

2016 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2016 AIME II Problems/Problem 4"

Revision as of 12:06, 15 January 2017

Solution

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