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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.

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

But try the problem first…

Answer: is 660.

Source
Suggested Reading

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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