2018 AMC 10B 第 23 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

23.

有多少个正整数有序对 (a,b)(a, b) 满足方程

ab+63=20lcm(a,b)+12gcd(a,b) \begin{aligned} a \cdot b + 63 &= 20 \cdot \operatorname{lcm}(a, b) \\ &\quad {}+ 12 \cdot \gcd(a, b) \end{aligned}\text{,}

其中 gcd(a,b)\gcd(a, b) 表示 aabb 的最大公因数,lcm(a,b)\operatorname{lcm}(a, b) 表示它们的最小公倍数。

How many ordered pairs (a,b)(a, b) of positive integers satisfy the equation

ab+63=20lcm(a,b)+12gcd(a,b), \begin{aligned} a \cdot b + 63 &= 20 \cdot \operatorname{lcm}(a, b) \\ &\quad {}+ 12 \cdot \gcd(a, b), \end{aligned}

where gcd(a,b)\gcd(a, b) denotes the greatest common divisor of aa and b,b, and lcm(a,b)\operatorname{lcm}(a, b) denotes their least common multiple?

00

22

44

66

88

答案:B
知识点:最小公倍数最大公约数西蒙最爱的因式分解技巧
难度评级:2120
小提示:

使用 ab=gcd(a,b)lcm(a,b)ab = \gcd(a,b) \cdot \operatorname{lcm}(a,b);令 x=lcm(a,b)x = \operatorname{lcm}(a,b)y=gcd(a,b)y = \gcd(a,b)

Use ab=gcd(a,b)lcm(a,b);ab = \gcd(a,b) \cdot \operatorname{lcm}(a,b); let x=lcm(a,b)x = \operatorname{lcm}(a,b) and y=gcd(a,b)y = \gcd(a,b)

大提示:

方程变为 (x12)(y20)=177=359(x - 12)(y - 20) = 177 = 3 \cdot 59;同时 yy 必须整除 xx

The equation becomes (x12)(y20)=177=359;(x - 12)(y - 20) = 177 = 3 \cdot 59; also yy must divide xx

解答:

回忆 ab=gcd(a,b)lcm(a,b)ab=\gcd(a,b)\operatorname{lcm}(a,b)。令 x=lcm(a,b)x=\operatorname{lcm}(a,b)y=gcd(a,b)y=\gcd(a,b),则原方程变为 (x12)(y20)=177=359(x-12)(y-20)=177=3\cdot59\text{。}

正因数分解给出 (x,y)=(13,197)(x,y)=(13,197)(189,21)(189,21)(15,79)(15,79)(71,23)(71,23)。负因数分解会使 xxyy 为负数,因此不可能。此外 yy 必须整除 xx,只有 (x,y)=(189,21)(x,y)=(189,21) 满足这一点。

写成 a=21ua=21ub=21vb=21v。于是 gcd(u,v)=1\gcd(u,v)=1uv=18921=9uv=\frac{189}{21}=9,所以 (u,v)=(1,9)(u,v)=(1,9)(9,1)(9,1)。因此符合条件的有序对是 (21,189)(21,189)(189,21)(189,21),正确答案是 B

Recall ab=gcd(a,b)lcm(a,b).ab=\gcd(a,b)\operatorname{lcm}(a,b). Let x=lcm(a,b)x=\operatorname{lcm}(a,b) and y=gcd(a,b).y=\gcd(a,b). The equation becomes (x12)(y20)=177=359.(x-12)(y-20)=177=3\cdot59.

The positive factor pairs give (x,y)=(13,197),(x,y)=(13,197), (189,21),(189,21), (15,79),(15,79), and (71,23).(71,23). The negative factor pairs make either xx or yy negative, so they are impossible. Also yy must divide x,x, and only (x,y)=(189,21)(x,y)=(189,21) passes.

Write a=21ua=21u and b=21v.b=21v. Then gcd(u,v)=1\gcd(u,v)=1 and uv=18921=9,uv=\frac{189}{21}=9, so (u,v)=(1,9)(u,v)=(1,9) or (9,1).(9,1). Hence the two ordered pairs are (21,189)(21,189) and (189,21),(189,21), and B is the correct answer.

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