1970 AMC 12 第 10 题

Question

设

F=0.4818181\ldots

是数字

8

和

1

循环出现的无限循环小数。将

F

写成最简分数后，分母比分子大：

Let

F=0.4818181\ldots

be an infinite repeating decimal with the digits

8

and

1

repeating. When

F

is written as a fraction in lowest terms, the denominator exceeds the numerator by:

13

14

29

57

126

将开头的

0.4

与后面的循环部分分开

Separate the initial

0.4

from the repeating tail

循环部分为

0.081(1+0.01+0.01^2+\cdots)

The tail is

0.081(1+0.01+0.01^2+\cdots)

我们有

\begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110} \end{aligned}\text{。}

分母比分子大

110-53=57

。

因此，正确答案是 D。

We have

\begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110}. \end{aligned}

The denominator exceeds the numerator by

110-53=57.

Therefore, the correct answer is D.

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