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

m (See also)
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<math>\dfrac{1}{26}+\dfrac{1}{10}-\dfrac{1}{260}=\dfrac{35}{260}=\dfrac{7}{52}</math>
 
<math>\dfrac{1}{26}+\dfrac{1}{10}-\dfrac{1}{260}=\dfrac{35}{260}=\dfrac{7}{52}</math>
  
7+52=59
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<math>7+52=059</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|before=First Question|num-a=2}}
 
{{AIME box|year=2002|n=I|before=First Question|num-a=2}}

Revision as of 21:26, 28 February 2008

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:

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

$7+52=059$

See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions