2019 AMC 10A 第 3 题

Question

Ana 和 Bonita 的生日是同月同日，但出生年份相差

n

年。去年 Ana 的年龄是 Bonita 的

5

倍。今年 Ana 的年龄是 Bonita 年龄的平方。求

n

。

Ana and Bonita were born on the same date in different years,

n

years apart. Last year Ana was

5

times as old as Bonita. This year Ana’s age is the square of Bonita’s age. What is

n?

3

5

9

12

15

设 Bonita 今年

b

岁。

Let Bonita’s current age be

b

同时使用去年的倍数关系和今年的平方关系。

Use last year’s relation and this year’s square relation together

设 Ana 今年

a

岁，Bonita 今年

b

岁。于是

\begin{aligned} a-1&=5(b-1),\\ a&=b^2 \end{aligned}\text{。}

代入可得

\begin{aligned} b^2-1&=5b-5,\\ b^2-5b+4&=0,\\ (b-4)(b-1)&=0 \end{aligned}\text{。}

可以看出

b \neq 1

，因为那会使 Ana 和 Bonita 同龄，所以

b = 4

。

于是

a = 4^2 = 16

，而年龄差

n = 16 - 4 = 12

。

所以正确答案是 D。

Let

a

be Ana’s current age and

b

be Bonita’s current age. Then

\begin{aligned} a-1&=5(b-1),\\ a&=b^2. \end{aligned}

Substitution gives

\begin{aligned} b^2-1&=5b-5,\\ b^2-5b+4&=0,\\ (b-4)(b-1)&=0. \end{aligned}

We can see that

b \neq 1

since that would make Ana and Bonita the same age, so we know that

b = 4.

This gives us that

a = 4^2 = 16

and

n = 16 - 4 = 12.

Thus, D is the correct answer.

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