Difference between revisions of "2002 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
Many states use a sequence of three letters followed by a sequence of thee digits as their standard license-plate pattern. Given that teach three-letter three-digit arrangement is equally likely,  the probability that such a license plate will contain at least one palindromes (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>
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Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely,  the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>
  
 
== Solution ==
 
== Solution ==
{{solution}}
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We first have a slice of apple [[PIE]]:
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<math>\dfrac{1}{26}+\dfrac{1}{10}-\dfrac{1}{260}=\dfrac{35}{260}=\dfrac{7}{52}</math>
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7+52=59
  
 
== See also ==
 
== See also ==

Revision as of 14:53, 8 October 2007

Problem

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution

We first have a slice of apple PIE:

$\dfrac{1}{26}+\dfrac{1}{10}-\dfrac{1}{260}=\dfrac{35}{260}=\dfrac{7}{52}$

7+52=59

See also