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

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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 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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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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== Solution 2==
 
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We can see that the denominator of the fraction will be 999b, where b is either 1, 1/3, 1/111, 1/9, or 1/333. This means that the numerator stays abc unless abc divides 999. Finding the numbers relatively prime to 999 gives us the numerators when the fraction cannot be simplified past 999.
 
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<math>\Rightarrow 999(1 - \frac{1}{3})(1- \frac{1}{111}) = 660</math>
 
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Now, look at a number that does divide 999:
 
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<math>\Rightarrow 555/999 = 5/9</math>
 
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We can see that if a number does divide 999, it will be simplified into a number which is relatively prime to 999 (since that is what simplifying is). There are 660 numbers less than or relatively prime to 999, so the answer is <math>\boxed{660}</math>.
 
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- krishkhushi09
 
  
 
{{AIME box|year=1992|num-b=4|num-a=6}}
 
{{AIME box|year=1992|num-b=4|num-a=6}}

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)
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Problem 4 		Followed by
Problem 6
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