2016 AIME II Problems/Problem 4

Revision as of 16:38, 14 March 2020 by Nafer (talk | contribs) (Solution 2)

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 $y=15$.

See also

2016 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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All AIME Problems and Solutions
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