2016 AMC 10A 第 25 题

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25.

有多少个正整数有序三元组 (x,y,z)(x,y,z) 同时满足 以及 ? lcm(x,y)=72,\text{lcm}(x,y) = 72, lcm(x,z)=600, \text{lcm}(x,z) = 600, lcm(y,z)=900?\text{lcm}(y,z)=900?

How many ordered triples (x,y,z)(x,y,z) of positive integers satisfy lcm(x,y)=72,\text{lcm}(x,y) = 72,lcm(x,z)=600, \text{lcm}(x,z) = 600, and lcm(y,z)=900?\text{lcm}(y,z)=900?

1515

1616

2424

2727

6464

答案:A
知识点:最小公倍数质因数分解分类讨论
难度评级:2390
解答:

先分解给定的最小公倍数: 因为 lcm(x,y)\operatorname{lcm}(x,y) 没有因子 55,所以 xxyy 都没有因子 55,而 zz 必须含 525^272=2332,600=23352,900=223252. \begin{gathered} 72=2^3\cdot3^2, \\ \quad 600=2^3\cdot3\cdot5^2, \\ \quad 900=2^2\cdot3^2\cdot5^2. \end{gathered}

x=2a3bx=2^a3^by=2c3dy=2^c3^dz=2e3f52z=2^e3^f5^2lcm(y,z)\operatorname{lcm}(y,z)33 的指数迫使 d=2d=2lcm(x,z)\operatorname{lcm}(x,z)22 的指数迫使 a=3a=3

剩余独立条件为 max(b,f)=1\max(b,f)=1max(c,e)=2\max(c,e)=2。前者有 33 个有序选择,后者有 55 个有序选择,所以共有 35=153\cdot5=15 个三元组。

所以正确答案是 A

Factor the given least common multiples: 72=2332,600=23352,900=223252. \begin{gathered} 72=2^3\cdot3^2, \\ \quad 600=2^3\cdot3\cdot5^2, \\ \quad 900=2^2\cdot3^2\cdot5^2. \end{gathered} Since lcm(x,y)\operatorname{lcm}(x,y) has no factor of 55, neither xx nor yy has a factor of 55, and zz must contain 525^2.

Write x=2a3bx=2^a3^b, y=2c3dy=2^c3^d, and z=2e3f52z=2^e3^f5^2. The power of 33 in lcm(y,z)\operatorname{lcm}(y,z) forces d=2d=2, and the power of 22 in lcm(x,z)\operatorname{lcm}(x,z) forces a=3a=3.

The remaining independent conditions are max(b,f)=1\max(b,f)=1 and max(c,e)=2\max(c,e)=2. These have 33 and 55 ordered choices, respectively, so there are 35=153\cdot5=15 triples.

Thus, the correct answer is A.

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