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

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== Problem ==
 
== Problem ==
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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>m/n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers.  Find <math>m+n</math>.
 
== Solution ==
 
== Solution ==
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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 09:37, 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

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