2021 AMC 12A Spring 第 5 题

Question

一名学生把数

66

乘以循环小数

1.\overline{ab} = 1.ababab\ldots\text{，}

其中

a

和

b

是数字。他没有注意到循环记号，只是计算了

66

乘以有限小数

1.ab

。后来他发现自己的答案比正确答案小

0.5

。

两位整数

\overline{ab}

是多少？

When a student multiplied the number

66

by the repeating decimal

1.\overline{ab} = 1.ababab\ldots,

where

a

and

b

are digits, he did not notice the notation and just multiplied

66

by the terminating decimal

1.ab.

Later he found that his answer was

0.5

less than the correct answer.

What is the two-digit integer

\overline{ab}?

15

30

45

60

75

答案：E

知识点：循环小数一次方程

难度评级：1370

小提示：

设

n = \overline{ab}

，则

1.\overline{ab} = 1 + \dfrac{n}{99}

，而

1.ab = 1 + \dfrac{n}{100}

Write

n = \overline{ab}.

Then

1.\overline{ab} = 1 + \dfrac{n}{99}

and

1.ab = 1 + \dfrac{n}{100}

大提示：

两个乘积的差是

66\left(\dfrac{n}{99} - \dfrac{n}{100}\right) = 0.5

The difference of the two products is

66\left(\dfrac{n}{99} - \dfrac{n}{100}\right) = 0.5

解答：

设

n = \overline{ab}

为这个两位整数。则

1.\overline{ab} = 1 + \dfrac{n}{99}

，而有限小数为

1.ab = 1 + \dfrac{n}{100}

。正确乘积减去学生的乘积为

\begin{aligned} &66\left(\frac{n}{99} - \frac{n}{100}\right) \\ &= 66 \cdot \frac{n}{9900} = \frac{n}{150} \end{aligned}\text{。}

令

\dfrac{n}{150} = 0.5

得到

n = 75

。

因此，正确答案是 E。

Let

n = \overline{ab}

be the two-digit integer. Then

1.\overline{ab} = 1 + \dfrac{n}{99}

while the terminating value is

1.ab = 1 + \dfrac{n}{100}.

The correct product minus the student’s product is

\begin{aligned} &66\left(\frac{n}{99} - \frac{n}{100}\right) \\ &= 66 \cdot \frac{n}{9900} = \frac{n}{150}. \end{aligned}

Setting

\dfrac{n}{150} = 0.5

gives

n = 75.

Thus, the correct answer is E.

2021 AMC 12A Spring 第 5 题

答案：E

难度评级：1370

小提示：

大提示：

解答：

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