1993 AMC 12 第 19 题

Question

有多少个正整数有序对

(m,n)

满足

\frac4m+\frac2n=1\text{？}

How many ordered pairs

(m,n)

of positive integers are solutions to

\frac4m+\frac2n=1?

1

2

3

4

多于

4

个

more than

4

消去分母，并配成含

m-4

和

n-2

的乘积。

Clear denominators and complete a product involving

m-4

and

n-2

原方程等价于

(m-4)(n-2)=8

。

The equation is equivalent to

(m-4)(n-2)=8

消去分母并整理，得到

\begin{aligned} mn-4n-2m=0 &\quad\Longleftrightarrow\quad\\ (m-4)(n-2)=8 \end{aligned}\text{。}

两个因数都必须为正。

8

的四个有序正因数对给出

\begin{aligned} (m,n)={}&(5,10),(6,6),\\ &(8,4),(12,3) \end{aligned}\text{。}

因此正确答案是 D。

Clearing denominators and rearranging gives

\begin{aligned} mn-4n-2m=0 &\quad\Longleftrightarrow\quad\\ (m-4)(n-2)=8. \end{aligned}

Both factors must be positive. The four ordered positive factor pairs of

8

give

\begin{aligned} (m,n)={}&(5,10),(6,6),\\ &(8,4),(12,3). \end{aligned}

Thus the correct answer is D.

其他年份的第 19 题