AMC10 2009 B

AMC10 2009 B · Q6

AMC10 2009 B · Q6. It mainly tests Primes & prime factorization, Counting divisors.

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

Kiana 有两个年龄相同的双胞胎哥哥。他们三人的年龄乘积是 128。他们三人的年龄和是多少？

(A) 10 10

(B) 12 12

(D) 18 18

(E) 24 24

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (C): The three operations can be performed in any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. Thus at most four distinct values can be obtained. It is easily checked that the values of the four expressions $$(2\times 3) + (4\times 5) = 26,$$ $$((2\times 3 + 4)\times 5) = 50,$$ $$2\times (3 + (4\times 5)) = 46,$$ $$2\times (3 + 4)\times 5 = 70$$ are in fact all distinct.

答案（C）：这三个运算可以按任意 $3! = 6$ 种顺序进行。然而，如果加法在最先或最后进行，那么按任意顺序做乘法都会得到相同结果。因此最多能得到四个不同的值。很容易检验，下面四个表达式的值分别为 $$(2\times 3) + (4\times 5) = 26,$$ $$((2\times 3 + 4)\times 5) = 50,$$ $$2\times (3 + (4\times 5)) = 46,$$ $$2\times (3 + 4)\times 5 = 70$$ 并且它们确实互不相同。

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