2015 AMC 10B 第 25 题

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25.

一个长方体尺寸为 a×b×ca \times b \times c,其中 aabbcc 是整数,且 该长方体的体积和表面积在数值上相等。可能的有序三元组 (a,b,c)(a,b,c) 有多少个? 1abc.1\leq a \leq b \leq c.

A rectangular box measures a×b×c,a \times b \times c, where a,a, b,b, and cc are integers and 1abc.1\leq a \leq b \leq c. The volume and the surface area of the box are numerically equal. How many ordered triples (a,b,c)(a,b,c) are possible?

44

1010

1212

2121

2626

答案:B
知识点:丢番图方程西蒙最爱的因式分解技巧分类讨论
难度评级:2010
解答:

条件为 因为 abc6bcabc\le6bc,所以 a6a\le6。此外,a=1a=1a=2a=2 都没有正整数解,因此只需检查 a=3,4,5,6a=3,4,5,6abc=2(ab+ac+bc).abc=2(ab+ac+bc).

a=3a=3 时,(b6)(c6)=36(b-6)(c-6)=36,得到 (b,c)=(7,42)(b,c)=(7,42) (8,24)(8,24) (9,18)(9,18) (10,15)(10,15) (12,12)(12,12)。当 a=4a=4 时,(b4)(c4)=16(b-4)(c-4)=16,得到 (5,20),(6,12),(8,8)(5,20),(6,12),(8,8)

a=5a=5 时,(3b10)(3c10)=100(3b-10)(3c-10)=100,满足 abca\le b\le c 的唯一解是 (b,c)=(5,10)(b,c)=(5,10)。当 a=6a=6 时,(b3)(c3)=9(b-3)(c-3)=9,满足 abca\le b\le c 的唯一解是 (6,6)(6,6)

所以三元组总数为 5+3+1+1=105+3+1+1=10

所以正确答案是 B

The condition is abc=2(ab+ac+bc).abc=2(ab+ac+bc). Since abc6bcabc\le6bc, we have a6a\le6. Also a=1a=1 and a=2a=2 give no positive solutions, so test a=3,4,5,6a=3,4,5,6.

For a=3a=3, (b6)(c6)=36(b-6)(c-6)=36, giving (b,c)=(7,42),(b,c)=(7,42), (8,24),(8,24), (9,18),(9,18), (10,15),(10,15), (12,12)(12,12). For a=4a=4, (b4)(c4)=16(b-4)(c-4)=16, giving (5,20),(6,12),(8,8)(5,20),(6,12),(8,8).

For a=5a=5, (3b10)(3c10)=100(3b-10)(3c-10)=100, and the only solution with abca\le b\le c is (b,c)=(5,10)(b,c)=(5,10). For a=6a=6, (b3)(c3)=9(b-3)(c-3)=9, and the only solution with abca\le b\le c is (6,6)(6,6).

The total number of triples is 5+3+1+1=105+3+1+1=10.

Thus, the correct answer is B.

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