Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Digits and Rationals.

## Digits and Rationals – AIME I, 1992

Let S be the set of all rational numbers r, 0<r<1, that have a repeating decimal expression in the form 0.abcabcabcabc…. where the digits a,b and c are not necessarily distinct. To write the elements of S as fractions in lowest terms find number of different numerators required.

- is 107
- is 660
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Digits

Prime

## Check the Answer

But try the problem first…

Answer: is 660.

AIME I, 1992, Question 5

Elementary Number Theory by David Burton

## Try with Hints

First hint

Let x=0.abcabcabcabc…..

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

\(\Rightarrow 999x=1000x-x=abc\)

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

Second Hint

numbers relatively prime to 999 gives us the numerators

\(\Rightarrow 999(1-\frac{1}{3})(1-\frac{1}{111})\)=660

Final Step

=660.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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