Question

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.

答案

Solution:

Each $1 \times b \times c$ layer has exactly $9$ red and $12$ green cubes, hence exactly $bc - 21$ yellow; each $a \times 1 \times c$ layer has exactly $20$ green and $25$ yellow, hence exactly $ac - 45$ red. Counting green cubes in the whole box both ways gives $12a = 20b,$ so $3a = 5b.$ Counting yellow both ways gives $a(bc - 21) = 25b = 15a,$ so $bc = 36.$ Counting red both ways gives $b(ac - 45) = 9a,$ and $\frac{9a}{b} = 15,$ so $ac = 60.$

Thus $a = \frac{60}{c}$ and $b = \frac{36}{c}$ are positive integers, so $c$ divides $\gcd(60, 36) = 12,$ and the volume is $abc = \frac{60 \cdot 36}{c} = \frac{2160}{c},$ smallest when $c = 12:$ volume $180$ with $(a, b, c) = (5, 3, 12).$

This is achievable: color every $1 \times 3 \times 12$ layer with three identical rows RRRGGGGYYYYY. Then each $1 \times 3 \times 12$ layer has $9$ red, $12$ green, and $15$ yellow cubes, and each $5 \times 1 \times 12$ layer has $15$ red, $20$ green, and $25$ yellow cubes. So the smallest possible volume is $180.$

2016 AIME II Problem 4

Solution:

Problem 4 in Other Years