Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1992 AIME Problems/Problem 5"

m
(→‎Solution 2)
 
(12 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
Let <math>S^{}_{}</math> be the set of all rational numbers <math>r^{}_{}</math>, <math>0^{}_{}<r<1</math>, that have a repeating decimal expansion in the form <math>0.abcabcabc\ldots=0.\overline{abc}</math>, where the digits <math>a^{}_{}</math>, <math>b^{}_{}</math>, and <math>c^{}_{}</math> are not necessarily distinct. To write the elements of <math>S^{}_{}</math> as fractions in lowest terms, how many different numerators are required?
 
Let <math>S^{}_{}</math> be the set of all rational numbers <math>r^{}_{}</math>, <math>0^{}_{}<r<1</math>, that have a repeating decimal expansion in the form <math>0.abcabcabc\ldots=0.\overline{abc}</math>, where the digits <math>a^{}_{}</math>, <math>b^{}_{}</math>, and <math>c^{}_{}</math> are not necessarily distinct. To write the elements of <math>S^{}_{}</math> as fractions in lowest terms, how many different numerators are required?
 +
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
We consider the method in which repeating decimals are normally converted to fractions with an example:
 +
 
 +
<math>x=0.\overline{176}</math>
 +
 
 +
<math>\Rightarrow 1000x=176.\overline{176}</math>
 +
 
 +
<math>\Rightarrow 999x=1000x-x=176</math>
 +
 
 +
<math>\Rightarrow x=\frac{176}{999}</math>
 +
 
 +
Thus, let <math>x=0.\overline{abc}</math>
 +
 
 +
<math>\Rightarrow 1000x=abc.\overline{abc}</math>
 +
 
 +
<math>\Rightarrow 999x=1000x-x=abc</math>
 +
 
 +
<math>\Rightarrow x=\frac{abc}{999}</math>
 +
 
 +
If <math>abc</math> is not divisible by <math>3</math> or <math>37</math>, then this is in lowest terms. Let us consider the other multiples: <math>333</math> multiples of <math>3</math>, <math>27</math> of <math>37</math>, and <math>9</math> of both <math>3</math> and <math>37</math>, so <math>999-333-27+9 = 648</math>, which is the amount that are neither. The <math>12</math> numbers that are multiples of <math>81</math> reduce to multiples of <math>3</math>. We have to count these since it will reduce to a multiple of <math>3</math> which we have removed from <math>999</math>, but, this cannot be removed since the numerator cannot cancel the <math>3</math>.There aren't any numbers which are multiples of <math>37^2</math>, so we can't get numerators which are multiples of <math>37</math>. Therefore <math>648 + 12 = \boxed{660}</math>.
  
−
== See also ==
 
−
* [[1992 AIME Problems/Problem 4 | Previous Problem]]
 
  
−
* [[1992 AIME Problems/Problem 6 | Next Problem]]
+
{{AIME box|year=1992|num-b=4|num-a=6}}
  
−
* [[1992 AIME Problems]]
+
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:45, 4 March 2024

Problem

Let $S^{}_{}$ be the set of all rational numbers $r^{}_{}$, $0^{}_{}<r<1$, 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?


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 $648 + 12 = \boxed{660}$.


1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_5&oldid=216463"