2016 AMC 12A 第 22 题

先试着解答 2016 AMC 12A 第 22 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2016 AMC 12A 解答,或核对答案

所有题目均经美国数学协会(MAA)官方合法授权使用。

22.

有多少个正整数有序三元组 (x,y,z)(x,y,z) 满足 lcm(x,y)=72\text{lcm}(x,y)=72lcm(x,z)=600\text{lcm}(x,z)=600,且 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:2160
解答:

因为 lcm(x,y)=2332\text{lcm}(x,y)=2^3\cdot3^2,且 lcm(x,z)=23352\text{lcm}(x,z)=2^3\cdot3\cdot5^2,所以 525^2 整除 zz,而 xxyy 都不能被 55 整除。同理,323^2 整除 yy,而 xxzz 都不能被 323^2 整除;此外,xx 必须含有因子 232^3

x=233jx=2^3\cdot3^{\,j}y=2k32y=2^{\,k}\cdot3^2z=2m3n52z=2^{\,m}\cdot3^{\,n}\cdot5^2,最小公倍数条件要求 max(j,n)=1\max(j,n)=1,且 max(k,m)=2\max(k,m)=2(1,0),(0,1),(1,1)(1,0),(0,1),(1,1)(2,0),(2,1),(2,2),(0,2),(1,2)(2,0),(2,1),(2,2),(0,2),(1,2) 种选择, 有 种选择,因此共有 35=153\cdot5=15 个有序三元组。

所以正确答案是 A

Because lcm(x,y)=2332\text{lcm}(x,y)=2^3\cdot3^2 and lcm(x,z)=23352,\text{lcm}(x,z)=2^3\cdot3\cdot5^2, the factor 525^2 divides zz while neither xx nor yy is divisible by 5.5. Also 323^2 divides y,y, while neither xx nor zz is divisible by 32,3^2, and xx must have the factor 23.2^3.

Writing x=233j,x=2^3\cdot3^{\,j}, y=2k32,y=2^{\,k}\cdot3^2, and z=2m3n52,z=2^{\,m}\cdot3^{\,n}\cdot5^2, the lcm conditions require max(j,n)=1\max(j,n)=1 and max(k,m)=2.\max(k,m)=2. The first pair can be (1,0),(0,1),(1,1),(1,0),(0,1),(1,1), and the second can be (2,0),(2,1),(2,2),(0,2),(1,2).(2,0),(2,1),(2,2),(0,2),(1,2). Thus there are 35=153\cdot5=15 ordered triples.

Thus, the correct answer is A.

← 第 21 题#21
完整试卷

其他年份的第 22 题