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 I Problems/Problem 1"

m (→‎Solution 1)
 
Line 18: Line 18:
  
 
~Dhillonr25
 
~Dhillonr25
 +
 +
== Solution 4 ==
 +
 +
We can find the probability of getting a letter and number palindrome through Solution One, which gives us <math>\frac{1}{26},</math> and <math>\frac{1}{10},</math> respectively. Then, we can use casework to solve the question. We begin by creating the cases:
 +
 +
\begin{align*}
 +
\bullet\ \text{Case 1: The license plate includes only a letter palindrome, and no number palindrome} \\
 +
\bullet\ \text{Case 2: The license plate includes only a number palindrome, and no letter palindrome} \\
 +
\bullet\ \text{Case 3: The license plate includes both a number palindrome, and a letter palindrome}
 +
\end{align*}
 +
 +
We know that the complement of these probabilities gives us the probability that the numbers and letters are NOT palindromes, so we can use that in our cases to get:
 +
 +
\begin{align}
 +
\frac{1}{26} \cdot \frac{9}{10} &= \frac{9}{260} & \text{Case 1}\\
 +
\frac{25}{26} \cdot \frac{1}{10} &= \frac{25}{260} & \text{Case 2}\\
 +
\frac{1}{26} \cdot \frac{1}{10} &= \frac{1}{260} & \text{Case 3}
 +
\end{align}
 +
 +
Finally, we can add them all together to get: <math>\frac{9 + 25 + 1}{260} = \frac{35}{260} = \frac{7}{52} = \frac{m}{n}.</math> Thus, we have <math>m + n = \boxed{059}.</math>
 +
 +
~ Cheetahboy93
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 19:10, 16 September 2024

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 $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

Consider the three-digit arrangement, $\overline{aba}$. There are $10$ choices for $a$ and $10$ choices for $b$ (since it is possible for $a=b$), and so the probability of picking the palindrome is $\frac{10 \times 10}{10^3} = \frac 1{10}$. Similarly, there is a $\frac 1{26}$ probability of picking the three-letter palindrome.

By the Principle of Inclusion-Exclusion, the total probability is

$\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{59}$

Solution 2

Using complementary counting, we count all of the license plates that do not have the desired property. To not be a palindrome, the first and third characters of each string must be different. Therefore, there are $10\cdot 10\cdot 9$ three-digit non-palindromes, and there are $26\cdot 26\cdot 25$ three-letter non-palindromes. As there are $10^3\cdot 26^3$ total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is $\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}$. We subtract this from 1 to get $1-\frac{45}{52}=\frac{7}{52}$ as our probability. Therefore, our answer is $7+52=\boxed{59}$.

~minor edit by Yiyj1

Solution 3

Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is \[\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}\] thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{59}.$

~Dhillonr25

Solution 4

We can find the probability of getting a letter and number palindrome through Solution One, which gives us $\frac{1}{26},$ and $\frac{1}{10},$ respectively. Then, we can use casework to solve the question. We begin by creating the cases:

\begin{align*} \bullet\ \text{Case 1: The license plate includes only a letter palindrome, and no number palindrome} \\ \bullet\ \text{Case 2: The license plate includes only a number palindrome, and no letter palindrome} \\ \bullet\ \text{Case 3: The license plate includes both a number palindrome, and a letter palindrome} \end{align*}

We know that the complement of these probabilities gives us the probability that the numbers and letters are NOT palindromes, so we can use that in our cases to get:

\begin{align} \frac{1}{26} \cdot \frac{9}{10} &= \frac{9}{260} & \text{Case 1}\\ \frac{25}{26} \cdot \frac{1}{10} &= \frac{25}{260} & \text{Case 2}\\ \frac{1}{26} \cdot \frac{1}{10} &= \frac{1}{260} & \text{Case 3} \end{align}

Finally, we can add them all together to get: $\frac{9 + 25 + 1}{260} = \frac{35}{260} = \frac{7}{52} = \frac{m}{n}.$ Thus, we have $m + n = \boxed{059}.$

~ Cheetahboy93

Video Solution by OmegaLearn

https://youtu.be/jRZQUv4hY_k?t=98

~ pi_is_3.14

See also

2002 AIME I (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_I_Problems/Problem_1&oldid=227341"