AMC12 2009 B

AMC12 2009 B · Q5

AMC12 2009 B · Q5. It mainly tests Primes & prime factorization, Perfect squares & cubes.

Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?

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？

(A) 10 10

(B) 12 12

(D) 18 18

(E) 24 24

Answer

Correct choice: (D)

正确答案：（D）

Solution

The age of each person is a factor of $128 = 2^7$. So the twins could be $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ years of age and, consequently Kiana could be $128$, $32$, $8$, or $2$ years old, respectively. Because Kiana is younger than her brothers, she must be $2$ years old. So the sum of their ages is $2 + 8 + 8 = \boxed{18}$. The answer is $\mathrm{(D)}$.

每个人的年龄都是 $128 = 2^7$ 的一个因数。因此双胞胎可能分别为 $2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8$ 岁，于是 Kiana 的年龄分别可能为 $128$、$32$、$8$ 或 $2$ 岁。因为 Kiana 比她的哥哥们小，所以她一定是 $2$ 岁。因此他们年龄之和为 $2 + 8 + 8 = \boxed{18}$。答案是 $\mathrm{(D)}$。

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