2016 AIME II Problem 4

Attempt Problem 4 of the 2016 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AIME II solutions, or check the answer key.

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4.

An a×b×ca \times b \times c rectangular box is built from abca \cdot b \cdot c unit cubes. Each unit cube is colored red, green, or yellow. Each of the aa layers of size 1×b×c1 \times b \times c parallel to the (b×c)(b \times c)-faces of the box contains exactly 99 red cubes, exactly 1212 green cubes, and some yellow cubes. Each of the bb layers of size a×1×ca \times 1 \times c parallel to the (a×c)(a \times c)-faces of the box contains exactly 2020 green cubes, exactly 2525 yellow cubes, and some red cubes. Find the smallest possible volume of the box.

Answer: 180
Concepts:double countingdivisibility
Difficulty rating: 2450
Solution:

Each 1×b×c1 \times b \times c layer has exactly 99 red and 1212 green cubes, hence exactly bc21bc - 21 yellow; each a×1×ca \times 1 \times c layer has exactly 2020 green and 2525 yellow, hence exactly ac45ac - 45 red. Counting green cubes in the whole box both ways gives 12a=20b,12a = 20b, so 3a=5b.3a = 5b. Counting yellow both ways gives a(bc21)=25b=15a,a(bc - 21) = 25b = 15a, so bc=36.bc = 36. Counting red both ways gives b(ac45)=9a,b(ac - 45) = 9a, and 9ab=15,\frac{9a}{b} = 15, so ac=60.ac = 60.

Thus a=60ca = \frac{60}{c} and b=36cb = \frac{36}{c} are positive integers, so cc divides gcd(60,36)=12,\gcd(60, 36) = 12, and the volume is abc=6036c=2160c,abc = \frac{60 \cdot 36}{c} = \frac{2160}{c}, smallest when c=12:c = 12: volume 180180 with (a,b,c)=(5,3,12).(a, b, c) = (5, 3, 12).

This is achievable: color every 1×3×121 \times 3 \times 12 layer with three identical rows RRRGGGGYYYYY. Then each 1×3×121 \times 3 \times 12 layer has 99 red, 1212 green, and 1515 yellow cubes, and each 5×1×125 \times 1 \times 12 layer has 1515 red, 2020 green, and 2525 yellow cubes. So the smallest possible volume is 180.180.

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