Digits and Rationals | AIME I, 1992 | Question 5

Join Trial or Access Free Resources

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


Answer: is 660.

AIME I, 1992, Question 5

Elementary Number Theory by David Burton

Try with Hints


Let x=0.abcabcabcabc.....

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

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

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

numbers relatively prime to 999 gives us the numerators

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

=660.

Subscribe to Cheenta at Youtube


More Posts

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

linkedin facebook pinterest youtube rss twitter instagram facebook-blank rss-blank linkedin-blank pinterest youtube twitter instagram