Question

设 $a$ 和 $b$ 是数字，不同时为九也不同时为零，循环小数 $0.\overline{ab}$ 被化为最简分数。可能出现多少个不同的分母？

Suppose that $a$ and $b$ are digits, not both nine and not both zero, and the repeating decimal $0.\overline{ab}$ is expressed as a fraction in lowest terms. How many different denominators are possible?

$3$

$4$

$5$

$8$

$9$

答案：C

知识点：循环小数因数

难度评级：1630

解答：

因为 $0.\overline{ab} = \dfrac{\overline{ab}}{99},$ 最简分母必须整除 $99 = 3^2\cdot 11.$ 它的因数为 $1, 3, 9, 11, 33, 99.$

分母为 $1$ 会要求 $\overline{ab} = 99,$ 即 $a = b = 9,$ 但这种情况被排除。其余各个分母都能达到：分子 $33, 11, 9, 3,$ 和 $1$ 分别约分得到分母 $3, 9, 11, 33,$ 和 $99,$ 所以共有 $5$ 种可能的分母。

所以正确答案是 C。

Since $0.\overline{ab} = \dfrac{\overline{ab}}{99},$ the reduced denominator divides $99 = 3^2\cdot 11.$ The divisors are $1, 3, 9, 11, 33, 99.$

The denominator $1$ would require $\overline{ab} = 99,$ i.e. $a = b = 9,$ which is excluded. Each is achievable: numerators $33, 11, 9, 3,$ and $1$ reduce to denominators $3, 9, 11, 33,$ and $99,$ respectively. Thus there are $5$ possible denominators.

Thus, the correct answer is C.

2002 AMC 12A 第 20 题

答案：C

难度评级：1630

解答：

其他年份的第 20 题