2013 AMC 10B 第 9 题

Question

三个正整数都大于

1

，乘积为

27000

，且两两互质。它们的和是多少？

Three positive integers are each greater than

1,

have a product of

27000 ,

and are pairwise relatively prime. What is their sum?

100

137

156

160

165

每个素数幂因子必须完整放入三个数中的某一个。

Put each prime-power factor wholly into one of the three numbers

两两互质意味着同一个素数不能出现在两个不同的数中。

Pairwise relatively prime means no prime can appear in two different numbers

分解后可知质因数只有

2

、

3

、

5

。因为三个数两两互质，每个素数幂只能完整属于其中一个数。

又因为三个数都大于

1

，所以每个数都必须含有其中一种质因数。

又因为

27000=2^3\cdot 3^3\cdot 5^3\text{，}

所以三个数必为

2^3,3^3,5^3

，和为

160

。

所以正确答案是 D。

Only one number is a multiple of

2,

only one is a multiple of

3,

and only one is a multiple of

5.

Since each positive integer is greater than

1,

each must be a multiple of one of these primes.

Now

27000=2^3\cdot 3^3\cdot 5^3.

Therefore, the numbers must be

2^3,3^3,5^3,

and their sum is

160.

Thus, the correct answer is D.

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