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

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== 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?
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== Solution ==
 
== Solution ==
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<math>\Rightarrow x=\frac{abc}{999}</math>
 
<math>\Rightarrow x=\frac{abc}{999}</math>
  
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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 <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 arere multiples of <math>81</math> reduce to multiples of <math>3</math>. There aren't any numbers which are multiples of <math>372</math>, so we can't get numerators which are multiples of <math>37</math>. Therefore <math>648 + 12 = \boxed{660}</math>.
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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>.
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Note: You can use Euler's totient function to find the numbers relatively prime to 999 which gives 648.  
  
 
{{AIME box|year=1992|num-b=4|num-a=6}}
 
{{AIME box|year=1992|num-b=4|num-a=6}}
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 16:12, 2 January 2025

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

Note: You can use Euler's totient function to find the numbers relatively prime to 999 which gives 648.

1992 AIME (Problems • Answer Key • Resources)
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Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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