Difference between revisions of "1989 AIME Problems/Problem 3"

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== Problem ==
 
== Problem ==
Suppose <math>n_{}^{}</math> is a positive integer and <math>d_{}^{}</math> is a single digit in base 10. Find <math>n_{}^{}</math> if
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Suppose <math>n</math> is a [[positive integer]] and <math>d</math> is a single [[digit]] in [[base 10]]. Find <math>n</math> if
 
<center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center>
 
<center><math>\frac{n}{810}=0.d25d25d25\ldots</math></center>
  
 
== Solution ==
 
== Solution ==
{{solution}}
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Repeating decimals represent [[rational number]]s.  To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math>0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}</math>.  Thus <math>\frac{n}{810} = \frac{100d + 25}{999}</math> so <math>n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}</math>.  Since 750 and 37 are [[relatively prime]], <math>4d + 1</math> must be [[divisible]] by 37, and the only digit for which this is possible is <math>d = 9</math>.  Thus <math>4d + 1 = 37</math> and <math>n = 750</math>.
  
 
== See also ==
 
== See also ==
* [[1989 AIME Problems/Problem 4|Next Problem]]
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{{AIME box|year=1989|num-b=2|num-a=4}}
* [[1989 AIME Problems/Problem 2|Previous Problem]]
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* [[1989 AIME Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 10:16, 25 February 2007

Problem

Suppose $n$ is a positive integer and $d$ is a single digit in base 10. Find $n$ if

$\frac{n}{810}=0.d25d25d25\ldots$

Solution

Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, $0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$. Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$. Since 750 and 37 are relatively prime, $4d + 1$ must be divisible by 37, and the only digit for which this is possible is $d = 9$. Thus $4d + 1 = 37$ and $n = 750$.

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AIME Problems and Solutions
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