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# Digits and Rationals | AIME I, 1992 | Question 5

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

Integers

Digits

Prime

## Check the Answer

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.

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