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Difference between revisions of "1992 AIME Problems/Problem 5"
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== Problem ==
 
== Problem ==
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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 ==
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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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== See also ==
+
<math>x=0.\overline{176}</math>
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* [[1992 AIME Problems/Problem 4 | Previous Problem]]
 
  
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* [[1992 AIME Problems/Problem 6 | Next Problem]]
+
<math>\Rightarrow 1000x=176.\overline{176}</math>
  
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* [[1992 AIME Problems]]
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<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 <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==
 +
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.
 +
 
 +
<math>\Rightarrow 999(1 - \frac{1}{3})(1- \frac{1}{111}) = 660</math>
 +
 
 +
Now, look at a number that does divide 999:
 +
 
 +
<math>\Rightarrow 555/999 = 5/9</math>
 +
 
 +
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>.
 +
- krishkhushi09
 +
 
 +
{{AIME box|year=1992|num-b=4|num-a=6}}
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[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Revision as of 19:27, 22 April 2020

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

Solution 2

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.

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

Now, look at a number that does divide 999:

$\Rightarrow 555/999 = 5/9$

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 $\boxed{660}$. - krishkhushi09

1992 AIME (Problems • Answer Key • Resources)
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