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>.
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Given that<br>
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<center><math>\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\
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    &(2)& \text{y is the number formed by reversing the digits of x; and}\\
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    &(3)& z=|x-y|. \end{eqnarray*}</math></center>  
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How many distinct values of <math>z</math> are possible?
  
 
== Solution ==
 
== Solution ==

Revision as of 14:37, 9 August 2008

Problem

Given that

$\begin{eqnarray*}&(1)& \text{x and y are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\
   &(2)& \text{y is the number formed by reversing the digits of x; and}\\
&(3)& z=|x-y|. \end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

How many distinct values of $z$ are possible?

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 $10^2\cdot26^2$ palindromes that contain both letters and digits.

Since there are $10^3\cdot26^3$ possible plates, the probability desired is $\frac{10^2\cdot26^2(10+26-1)}{10^2\cdot26^2\cdot 260}=\frac{35}{260}=\frac{7}{52}$. Thus $m+n=059$.

See also

2002 AIME II (ProblemsAnswer KeyResources)
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