2014 AMC 10A 第 21 题

Question

正整数

a

和

b

使得直线

y=ax+5

与

y=3x+b

在

x

轴上交于同一点。所有可能交点的

x

坐标之和是多少？

Positive integers

a

and

b

are such that the graphs of

y=ax+5

and

y=3x+b

intersect the

x

-axis at the same point. What is the sum of all possible

x

-coordinates of these points of intersection?

-20

-18

-15

-12

-8

答案：E

知识点：一次方程因数系统列举

难度评级：1600

小提示：

令两条直线的横轴截距相等。

Set both x-intercepts equal

大提示：

15

的正整数因数对给出全部可能。

Positive integer factor pairs of

15

give all possibilities

解答：

两条直线与

x

轴相交时

y = 0

。因此

0 = ax + 5

且

0 = 3x + b\text{，}

解得

x = -\dfrac{5}{a}

和

x = -\dfrac{b}{3}\text{。}

令二者相等，得到

\dfrac{5}{a} = \dfrac{b}{3}

ab = 15\text{。}

因为

a

和

b

都是正整数，所以

(a, b)

只能是

(1, 15)\text{，}

(3, 5)\text{，}

(5, 3)\text{，}

(15, 1)\text{。}

对应的

x

坐标为

x = -5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}\text{。}

它们的和为

-8

。

所以正确答案是 E。

Note that the lines intersect the

x

-axis when

y = 0.

This gives us

0 = ax + 5

and

0 = 3x + b,

which when solved gives us

x = -\dfrac{5}{a}

and

x = -\dfrac{b}{3}.

Setting these equal to each other, we have

\dfrac{5}{a} = \dfrac{b}{3}

ab = 15.

We know that

a

and

b

are positive, which means that the only pairs of values

(a, b)

that satisfy the above equation are

(1, 15),

(3, 5),

(5, 3),

(15, 1).

Plugging these values back into the equations gives us

x

-values of

x = -5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}.

The sum of all these values is

-8.

Thus, E is the correct answer.

2014 AMC 10A 第 21 题

答案：E

难度评级：1600

小提示：

大提示：

解答：

其他年份的第 21 题