2009 AMC 12B 第 5 题

Question

Kiana 有两个年长的双胞胎哥哥。他们三个人年龄的乘积是

128

。他们三个人年龄的和是多少？

Kiana has two older twin brothers. The product of their three ages is

128.

What is the sum of their three ages?

10

12

16

18

24

128 = 2^7

，所以每个人的年龄都是

2

的幂。

128 = 2^7,

so every age is a power of

2

若每个双胞胎哥哥都是

t

岁，则

t^2

乘以 Kiana 的年龄等于

128

，且 Kiana 更小。

If each twin is

t

years old, then

t^2

times Kiana’s age is

128,

and Kiana is younger

因为

128 = 2^7

，每个年龄都是

2

的幂。双胞胎年龄相同，设为

t

，则 Kiana 的年龄为

\dfrac{128}{t^2}

。

取

t = 8

时，Kiana 的年龄是

\dfrac{128}{64} = 2

，比双胞胎小。（若双胞胎年龄更小，Kiana 就会更大，这是不允许的。）年龄和为

8 + 8 + 2 = 18

。

所以正确答案是 D。

Since

128 = 2^7,

each age is a power of

2.

The twins share an age

t,

so Kiana’s age is

\dfrac{128}{t^2}.

Taking

t = 8

gives Kiana

\dfrac{128}{64} = 2,

who is younger than the twins. (Smaller twins would make Kiana older, which is not allowed.) The sum is

8 + 8 + 2 = 18.

Thus, the correct answer is D.

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