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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>.
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Given that<br>
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<cmath>\begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\
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    &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\
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    &(3)& z=|x-y|.
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\end{eqnarray*}</cmath>
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How many distinct values of <math>z</math> are possible?
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== 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>.
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We express the numbers as <math>x=100a+10b+c</math> and <math>y=100c+10b+a</math>.  From this, we have
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<cmath>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\
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\end{eqnarray*}</cmath>
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Because <math>a</math> and <math>c</math> are digits, and <math>a</math> and <math>c</math> are both between 1 and 9 (from condition 1), there are <math>\boxed{009}</math> possible values (since all digits except <math>9</math> can be expressed this way).
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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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* [[2002 AIME II Problems]]
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[[Category: Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 20:56, 11 February 2020

Problem

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

How many distinct values of $z$ are possible?

Solution

We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$. From this, we have \begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ \end{eqnarray*} Because $a$ and $c$ are digits, and $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\boxed{009}$ possible values (since all digits except $9$ can be expressed this way).

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

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