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?

$10$

$12$

$16$

$18$

$24$

答案：D

知识点：质因数分解 2的幂

难度评级：1080

解答：

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 $\dfrac{128}{64} = 2$ ，确实比双胞胎小；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.

2009 AMC 12B 第 5 题

答案：D

难度评级：1080

解答：

其他年份的第 5 题