2018 AMC 12A 第 19 题

Question

设

A

为质因数只有

2

、

3

或

5

的正整数集合。所有

A

中元素倒数的无穷和

\begin{aligned} &\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ &\quad {}+ \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} \\ &\quad {}+ \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} \\ &\quad {}+ \frac{1}{18} + \frac{1}{20} + \cdots \end{aligned}

可表示为

\tfrac{m}{n}

，其中

m

和

n

是互质的正整数。

m + n

是多少？

Let

A

be the set of positive integers that have no prime factors other than

2,

3,

or

5.

The infinite sum

\begin{aligned} &\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ &\quad {}+ \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} \\ &\quad {}+ \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} \\ &\quad {}+ \frac{1}{18} + \frac{1}{20} + \cdots \end{aligned}

of the reciprocals of all the elements of

A

can be expressed as

\tfrac{m}{n},

where

m

and

n

are relatively prime positive integers. What is

m + n?

16

17

19

23

36

答案：C

知识点：等比数列求和质因数分解

难度评级：1930

小提示：

每个元素都是

2^i 3^j 5^k

，所以这个和可分解为三个独立等比级数的乘积

Every element is

2^i 3^j 5^k,

so the sum factors as a product of three separate geometric series

大提示：

计算

\dfrac{1}{1 - \frac12} \cdot \dfrac{1}{1 - \frac13} \cdot \dfrac{1}{1 - \frac15}

并化简

Evaluate

\dfrac{1}{1 - \frac12} \cdot \dfrac{1}{1 - \frac13} \cdot \dfrac{1}{1 - \frac15}

and reduce the result

解答：

A

中每个元素都可唯一写成

2^i 3^j 5^k

，其中

i, j, k \ge 0

，所以所有倒数的和可分解为

\begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}\text{。} \end{aligned}

这等于

2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}

。因为

\gcd(15, 4) = 1

，所以

m + n = 15 + 4 = 19

。

所以正确答案是 C。

Each element of

A

is uniquely

2^i 3^j 5^k

with

i, j, k \ge 0,

so summing all reciprocals factors as

\begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}. \end{aligned}

This equals

2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.

With

\gcd(15, 4) = 1,

m + n = 15 + 4 = 19.

Thus, the correct answer is C.

2018 AMC 12A 第 19 题

答案：C

难度评级：1930

小提示：

大提示：

解答：

其他年份的第 19 题