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

(Solution 2)
(Solution 2)
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Repeating the procedure on the number of yellow cubes <math>y</math> on each of the <math>a</math> layers gives <math>y=15</math>.
 
Repeating the procedure on the number of yellow cubes <math>y</math> on each of the <math>a</math> layers gives <math>y=15</math>.
  
Therefore <math>bc=9+12+15=36</math> and <math>ac=15+20+25=60</math>. Multiplying yields <math>abc^2=36\cdot60</math>.
+
Therefore <math>bc=9+12+15=36</math> and <math>ac=15+20+25=60</math>. Multiplying yields <math>abc^2=2160</math>.
  
To
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Since <math>abc^2</math> is fixed, <math>abc</math> is minimized when <math>c</math> maximized, which occurs when <math>a</math>, <math>b</math> are minimized (since each of <math>ac</math>, <math>bc</math> is fixed). Thus <math>(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2016|n=II|num-b=3|num-a=5}}

Revision as of 16:51, 14 March 2020

Problem

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 $1 \times a \times c$ 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

Solution 2

The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$.

Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$. Substitute $a=\frac{5}{3}b$ gives $r=15$.

Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$.

Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$. Multiplying yields $abc^2=2160$.

Since $abc^2$ is fixed, $abc$ is minimized when $c$ maximized, which occurs when $a$, $b$ are minimized (since each of $ac$, $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$

See also

2016 AIME II (ProblemsAnswer KeyResources)
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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