Difference between revisions of "2002 AIME II Problems/Problem 1"
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== 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>. | 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>. | ||
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== 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 | + | We count the number of three-letter and three-digit palindromes, then subtract the number of license plates containing both types of palindrome. |
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+ | 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> palindromes that contain both letters and digits. | ||
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+ | Since there are <math>10^3\cdot26^3</math> possible plates, the probability desired is <math>\frac{10^2\cdot26^2(10+26-1)}{10^2\cdot26^2\cdot 260}=\frac{35}{260}=\frac{7}{52}</math>. Thus <math>m+n=059</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2002|n=II|before=First Question|num-a=2}} | {{AIME box|year=2002|n=II|before=First Question|num-a=2}} | ||
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Revision as of 12:39, 19 April 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 count the number of three-letter and three-digit palindromes, then subtract the number of license plates containing both types of palindrome.
There are letter palindromes, digit palindromes, and palindromes that contain both letters and digits.
Since there are possible plates, the probability desired is . Thus .
See also
2002 AIME II (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 |