2002 AIME I Problems/Problem 1: Difference between revisions
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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= | <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 
, where 
and 
are relatively prime positive integers. Find 
Solution
We first have a slice of apple PIE:
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
