AIME 1992 · 第 5 题

AIME 1992 — Problem 5

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$ , $0^{}_{}, that have a repeating decimal expansion in the form$ 0.abcabcabc\ldots=0.\overline{abc} $, where the digits$ a^{}_{} $,$ b^{}_{} $, and$ c^{}_{} $are not necessarily distinct. To write the elements of$ S^{}_{}$ as fractions in lowest terms, how many different numerators are required?

解析

Solution

We consider the method in which repeating decimals are normally converted to fractions with an example:

x=0.\overline{176}

\Rightarrow 1000x=176.\overline{176}

\Rightarrow 999x=1000x-x=176

\Rightarrow x=\frac{176}{999}

Thus, let $x=0.\overline{abc}$

\Rightarrow 1000x=abc.\overline{abc}

\Rightarrow 999x=1000x-x=abc

\Rightarrow x=\frac{abc}{999}

If $abc$ is not divisible by $3$ or $37$ , then this is in lowest terms. Let us consider the other multiples: $333$ multiples of $3$ , $27$ of $37$ , and $9$ of both $3$ and $37$ , so $999-333-27+9 = 648$ , which is the amount that are neither. The $12$ numbers that are multiples of $81$ reduce to multiples of $3$ . We have to count these since it will reduce to a multiple of $3$ which we have removed from $999$ , but, this cannot be removed since the numerator cannot cancel the $3$ .There aren't any numbers which are multiples of $37^2$ , so we can't get numerators which are multiples of $37$ . Therefore $648 + 12 = \boxed{660}$ .

Note: You can use Euler's totient function to find the numbers relatively prime to 999 which gives 648.