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

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== Solution ==
 
== Solution ==
 
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>.
 
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>.
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(Note: Any repeating sequence of <math>n</math> digits that looks like <math>0.a_1a_2a_3...a_{n-1}a_na_1a_2...</math> can be written as <math>\frac{a_1a_2...a_n}{10^n-1}</math>, where <math>a_1a_2...a_n</math> represents an <math>n</math> digit number.)
  
 
== See also ==
 
== See also ==

Revision as of 18:13, 22 April 2017

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 $n = 750$.


(Note: Any repeating sequence of $n$ digits that looks like $0.a_1a_2a_3...a_{n-1}a_na_1a_2...$ can be written as $\frac{a_1a_2...a_n}{10^n-1}$, where $a_1a_2...a_n$ represents an $n$ digit number.)

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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