2002 AIME II Problems/Problem 1

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 $m+n=059$ .

2002 AIME II (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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2002 AIME II Problems/Problem 1

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