AMC12 2018 A

AMC12 2018 A · Q19

AMC12 2018 A · Q19. It mainly tests Fractions, Primes & prime factorization.

Let $A$ be the set of positive integers that have no prime factors other than 2, 3, or 5. The infinite sum $$ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \dots $$ of the reciprocals of all the elements of $A$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

令 $A$ 为仅由质因数 2、3 或 5 构成的正整数集合。无限和 $$ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \dots $$ 是 $A$ 中所有元素的倒数之和，可表示为 $\frac{m}{n}$，其中 $m$ 和 $n$ 互质正整数。求 $m+n$？

(A) 16 16

(B) 17 17

(D) 23 23

(E) 36 36

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Elements of set $A$ are of the form $2^i\cdot 3^j\cdot 5^k$ for nonnegative integers $i$, $j$, and $k$. Note that the product $\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+\cdots\right)$ will produce the desired sum. By the formula for infinite geometric series, this product evaluates to $\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}\cdot\frac{1}{1-\frac{1}{5}}=2\cdot\frac{3}{2}\cdot\frac{5}{4}=\frac{15}{4}.$ The requested sum is $15+4=19$.

答案（C）：集合 $A$ 的元素形如 $2^i\cdot 3^j\cdot 5^k$，其中 $i$、$j$、$k$ 为非负整数。注意乘积 $\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots\right)\left(1+\frac{1}{5}+\frac{1}{5^2}+\cdots\right)$ 会得到所需的和。由无穷等比级数公式，该乘积为 $\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}\cdot\frac{1}{1-\frac{1}{5}}=2\cdot\frac{3}{2}\cdot\frac{5}{4}=\frac{15}{4}.$ 所求的和为 $15+4=19$。

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