2016 AIME II 第 4 题

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4.

一个 a×b×ca \times b \times c 的长方体由 abca \cdot b \cdot c 个单位立方体组成。每个单位立方体被涂成红色、绿色或黄色。平行于长方体 (b×c)(b \times c) 面的 1×b×c1 \times b \times c 层共有 aa 层,每层恰有 99 个红色立方体、1212 个绿色立方体,以及若干黄色立方体。平行于 (a×c)(a \times c) 面的 a×1×ca \times 1 \times c 层共有 bb 层,每层恰有 2020 个绿色立方体、2525 个黄色立方体,以及若干红色立方体。求这个长方体可能的最小体积。

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.

答案:180
知识点:双重计数整除性
难度评级:2450
解答:

每个 1×b×c1 \times b \times c 层恰有 99 个红色和 1212 个绿色立方体,因此恰有 bc21bc - 21 个黄色立方体;每个 a×1×ca \times 1 \times c 层恰有 2020 个绿色和 2525 个黄色立方体,因此恰有 ac45ac - 45 个红色立方体。用两种方式统计整个长方体中的绿色立方体,得 12a=20b12a = 20b,所以 3a=5b3a = 5b。统计黄色立方体得 a(bc21)=25b=15aa(bc - 21) = 25b = 15a,所以 bc=36bc = 36。统计红色立方体得 b(ac45)=9ab(ac - 45) = 9a,且 9ab=15\frac{9a}{b} = 15,所以 ac=60ac = 60

因此 a=60ca = \frac{60}{c}b=36cb = \frac{36}{c} 是正整数,所以 cc 整除 gcd(60,36)=12\gcd(60, 36) = 12。体积为 abc=6036c=2160cabc = \frac{60 \cdot 36}{c} = \frac{2160}{c},当 c=12c = 12 时最小。此时体积为 180180,且 (a,b,c)=(5,3,12)(a, b, c) = (5, 3, 12)

这可以实现:把每个 1×3×121 \times 3 \times 12 层涂成三行相同的 RRRGGGGYYYYY。于是每个 1×3×121 \times 3 \times 12 层有 99 个红色、1212 个绿色和 1515 个黄色立方体;每个 5×1×125 \times 1 \times 12 层有 1515 个红色、2020 个绿色和 2525 个黄色立方体。所以最小可能体积是 180180

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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