Difference between revisions of "2021 AIME II Problems/Problem 1"

Revision as of 12:10, 30 May 2022

Problem

Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)

Solution 1

Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=\boxed{550},$ which is the final answer.

~ math31415926535

Solution 2

For any palindrome $\underline{ABA},$ note that $\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ The average for $A$ is $5$ since $A$ can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ The average for $B$ is $4.5$ since $B$ is either $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ Therefore, the answer is $505 + 45 = \boxed{550}.$

- ARCTICTURN

Solution 3 (Symmetry and Generalization)

For every three-digit palindrome $\underline{ABA}$ with $A\in\{1,2,3,4,5,6,7,8,9\}$ and $B\in\{0,1,2,3,4,5,6,7,8,9\},$ note that $\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to $\begin{align*} \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\ &=1100. \end{align*}$ For instances: $\begin{align*} 171+929&=1100, \\ 262+838&=1100, \\ 303+797&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*}$ and so on.

From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1100}{2}=\boxed{550}.$

Remark

By the Multiplication Principle, there are $9\cdot10=90$ three-digit palindromes in total. Their sum is $1100\cdot45=49500,$ as we can match them into $45$ pairs such that each pair sums to $1100.$

~MRENTHUSIASM

Solution 4 (Similar to Solution 2: Very, Very Easy and Quick)

We notice that a three-digit palindrome looks like this: $\underline{aba}.$

And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\times{10}=90$ three-digit palindromes.

We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean.

How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$

Thus, all of these $90$ palindromes can be broken into this form.

Thus, the sum of these $90$ palindromes will be $101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$ ). The same logic explains why we multiply by $9$ when computing the total sum of $b.$

We get a sum of $45\times{1100},$ but don't compute this! Divide this by $90$ and you will get $\boxed{550}.$

~ $\alpha b \alpha$

Solution 5 (Extremely Fast Solution)

The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$

~MathIsFun286

~MathFun1000 (Rephrasing with more clarity)

Solution 6 (Two cases)

Case 1.

Consider palindromes of the form $5x5 = 505 + 10x.$ There are $10$ of them. The arithmetic mean of the first term is $505,$ and the second $\frac {10 \cdot(0 + 1 + ... + 9)}{10} = 45.$ The arithmetic mean of the sum is $505 + 45 = 550.$

Case 2.

Consider palindromes of the form $yxy,$ where $y= {1,2,3,4,6,7,8,9}.$ Let $u = 10 – y, v = 9 – x.$ Then $uvu$ is a symmetric palindrome that can be constructed for each $yxy.$ The arithmetic mean of each such pair is $550.$ For example, $\frac{737 + 363}{2} = 550.$ Thus, all palindromes are divided into groups of numbers with the arithmetic mean in each group equal to $550.$ The arithmetic mean of all numbers is also $550.$

~vvsss, www.deoma-cmd.ru

Remark

Visit the Discussion Page for questions and further generalizations.

~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=jDP2PErthkg

Video Solution by Interstigation

https://youtu.be/3_ik5N33CnQ

speedy 2 min video

@@ Line 63: / Line 63: @@
 ~MathFun1000 (Rephrasing with more clarity)
+==Solution 6 (Two cases)==
+<pre style="color: blue">
+Case 1.
+</pre>
+Consider palindromes of the form <math>5x5 = 505 + 10x.</math> There are <math>10</math> of them. The arithmetic mean of the first term is <math>505,</math> and the second <math>\frac {10 \cdot(0 + 1 + ... + 9)}{10} = 45.</math>
+The arithmetic mean of the sum is <math>505 + 45 = 550.</math>
+<pre style="color: blue">
+Case 2.
+</pre>
+Consider palindromes of the form <math>yxy,</math> where <math>y= {1,2,3,4,6,7,8,9}.</math>
+Let <math>u = 10 – y, v = 9 – x.</math>
+Then <math>uvu</math> is a symmetric palindrome that can be constructed for each <math>yxy.</math>
+The arithmetic mean of each such pair is <math>550.</math> For example, <math>\frac{737 + 363}{2} = 550.</math>
+Thus, all palindromes are divided into groups of numbers with the arithmetic mean in each group equal to <math>550.</math> The arithmetic mean of all numbers is also <math>550.</math>
+~vvsss, www.deoma-cmd.ru
 ==Remark==

2021 AIME II (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search