Difference between revisions of "2002 AIME I Problems/Problem 1"
(→Problem) |
|||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | Many states use a sequence of three letters followed by a sequence of | + | 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 == | ||
| − | {{ | + | We first have a slice of apple [[PIE]]: |
| + | |||
| + | <math>\dfrac{1}{26}+\dfrac{1}{10}-\dfrac{1}{260}=\dfrac{35}{260}=\dfrac{7}{52}</math> | ||
| + | |||
| + | 7+52=59 | ||
== See also == | == See also == | ||
Revision as of 15:53, 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 
, where 
and 
are relatively prime positive integers. Find 
Solution
We first have a slice of apple PIE:
7+52=59
