Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (tagged)
(add soln)
Line 1: Line 1:
−
{{empty}}
 
 
== Problem ==
 
== 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 <math>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m+n</math>.
 
== Solution ==
 
== Solution ==
−
{{solution}}
+
We count the number of three-letter and three-digit palindromes, then subtract the number of license plates containing both types of palindrome. There are <math>10^3\cdot 26^2</math> letter palindromes, <math>10^2\cdot 26^3</math> digit palindromes, and <math>10^2\cdot26^2</math> both palindromes, while there are <math>10^326^3</math> possible plates, so the probability desired is <math>\frac{10^226^2(10+26-1)}{10^226^2\cdot 260}=\frac{35}{260}=\frac{7}{52}</math>. Thus <math>m+n=059</math>.
 
== See also ==
 
== See also ==
 
* [[2002 AIME II Problems/Problem 2 | Next problem]]
 
* [[2002 AIME II Problems/Problem 2 | Next problem]]
 
* [[2002 AIME II Problems]]
 
* [[2002 AIME II Problems]]

Revision as of 10:37, 12 March 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 $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

We count the number of three-letter and three-digit palindromes, then subtract the number of license plates containing both types of palindrome. There are $10^3\cdot 26^2$ letter palindromes, $10^2\cdot 26^3$ digit palindromes, and $10^2\cdot26^2$ both palindromes, while there are $10^326^3$ possible plates, so the probability desired is $\frac{10^226^2(10+26-1)}{10^226^2\cdot 260}=\frac{35}{260}=\frac{7}{52}$. Thus $m+n=059$.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_1&oldid=13956"