Question

一名学生把数 $66$ 乘以循环小数其中 $a$ 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 $0.5$ 。 $1.\overline{ab} = 1.ababab\ldots,$

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

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

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

解答：

其他年份的第 5 题