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

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== Solution ==
 
== Solution ==
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{{solution}}
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We consider the method in which repeating decimals are normally converted to fractions with an example:
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<math>x=0.\overline{176}</math>
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<math>\Rightarrow 1000x=176.\overline{176}</math>
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<math>\Rightarrow 999x=1000x-x=176</math>
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<math>\Rightarrow x=\frac{176}{999}</math>
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Thus, let <math>x=0.\overline{abc}</math>
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<math>\Rightarrow 1000x=abc.\overline{abc}</math>
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<math>\Rightarrow 999x=1000x-x=abc</math>
 +
 
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<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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=4|num-a=6}}
 
{{AIME box|year=1992|num-b=4|num-a=6}}

Revision as of 22:15, 11 November 2007

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 $3$ and $37$, so $999-333-27+9 = 648$, which is the amount that are neither. The $12$ numbers that arere multiples of $81$ reduce to multiples of $3$. There aren't any numbers which are multiples of $372$, so we can't get numerators which are multiples of $37$. Therefore $648 + 12 = \boxed{660}$.

See also

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
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