Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(add soln)
m (→‎Problem: added spacing)
 
(14 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
== 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>.
+
Given that<br>
 +
<cmath>\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*}</cmath>
 +
 
 +
How many distinct values of <math>z</math> are possible?
 +
 
 
== 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 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>.
+
We express the numbers as <math>x=100a+10b+c</math> and <math>y=100c+10b+a</math>.  From this, we have
 +
<cmath>\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\
 +
\end{eqnarray*}</cmath>
 +
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).
 +
 
 
== See also ==
 
== See also ==
−
* [[2002 AIME II Problems/Problem 2 | Next problem]]
+
{{AIME box|year=2002|n=II|before=First Question|num-a=2}}
−
* [[2002 AIME II Problems]]
+
 
 +
[[Category: Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 21:56, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_1&oldid=117676"