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

Latest revision as of 14:36, 30 August 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 $550.$

vladimir.shelomovskii@gmail.com, vvsss

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 3: / Line 3: @@
 ==Solution 1==
-Recall the the arithmetic mean of all the <math>n</math> digit palindromes is just the average of the largest and smallest <math>n</math> digit palindromes, and in this case the <math>2</math> palindromes are <math>101</math> and <math>999</math> and <math>\frac{101+999}{2}=550</math> and <math>\boxed{550}</math> is the final answer.
+Recall that the arithmetic mean of all the <math>n</math> digit palindromes is just the average of the largest and smallest <math>n</math> digit palindromes, and in this case the <math>2</math> palindromes are <math>101</math> and <math>999</math> and <math>\frac{101+999}{2}=\boxed{550},</math> which is the final answer.
 ~ math31415926535
 ==Solution 2==
-For any palindrome <math>\underline{ABA}</math>, note that <math>\underline{ABA}</math> is 100A + 10B + A, which is also 101A + 10B.
+For any palindrome <math>\underline{ABA},</math> note that <math>\underline{ABA}</math> is <math>100A + 10B + A = 101A + 10B.</math>
-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 = <math>\boxed{550}</math>.
+The average for <math>A</math> is <math>5</math> since <math>A</math> can be any of <math>1, 2, 3, 4, 5, 6, 7, 8,</math> or <math>9.</math> The average for <math>B</math> is <math>4.5</math> since <math>B</math> is either <math>0, 1, 2, 3, 4, 5, 6, 7, 8,</math> or <math>9.</math> Therefore, the answer is <math>505 + 45 = \boxed{550}.</math>
 - ARCTICTURN
@@ Line 31: / Line 31: @@
 and so on.
-From this symmetry, the arithmetic mean of all the three-digit palindromes is <math>\frac{1110}{2}=\boxed{550}.</math>
+From this symmetry, the arithmetic mean of all the three-digit palindromes is <math>\frac{1100}{2}=\boxed{550}.</math>
 <u><b>Remark</b></u>
-By the Multiplication Principle, there are <math>9\cdot10=90</math> three-digit palindromes in total. We can match them into <math>45</math> pairs such that the sum of each pair is <math>1100.</math> Therefore, the arithmetic mean of all the three-digit palindromes is <math>\frac{1100\cdot45}{90}=550.</math>
+By the Multiplication Principle, there are <math>9\cdot10=90</math> three-digit palindromes in total. Their sum is <math>1100\cdot45=49500,</math> as we can match them into <math>45</math> pairs such that each pair sums to <math>1100.</math>
 ~MRENTHUSIASM
-==Solution 4 (Very, Very Easy and Quick)==
+==Solution 4 (Similar to Solution 2: Very, Very Easy and Quick)==
-We notice that a three-digit palindrome looks like this <math>\overline{aba}</math>
+We notice that a three-digit palindrome looks like this: <math>\underline{aba}.</math>
-And we know a can be any number from 1-9, and b can be any number from 0-9, so there are <math>9\times{10}=90</math> three-digit palindromes
+And we know <math>a</math> can be any digit from <math>1</math> through <math>9,</math> and <math>b</math> can be any digit from <math>0</math> through <math>9,</math> so there are <math>9\times{10}=90</math> three-digit palindromes.
-We want to find the sum of these 90 palindromes and divide it by 90 to find the arithmetic mean
+We want to find the sum of these <math>90</math> palindromes and divide it by <math>90</math> 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
+How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: <math>101a+10b.</math>
-Thus, all of these 90 palindromes can be broken into this form
+Thus, all of these <math>90</math> palindromes can be broken into this form.
-Thus, the sum of these 90 palindromes will be <math>101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9}</math>, because each a will be in 10 different palindromes (since for each a, there are 10 choices for b). The same logic explains why there is a times 9 when computing the total sum of b.
+Thus, the sum of these <math>90</math> palindromes will be <math>101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},</math> because each <math>a</math> will be in <math>10</math> different palindromes (since for each <math>a,</math> there are <math>10</math> choices for <math>b</math>). The same logic explains why we multiply by <math>9</math> when computing the total sum of <math>b.</math>
-We get a sum of <math>45\times{1100}</math>
+We get a sum of <math>45\times{1100},</math> but don't compute this! Divide this by <math>90</math> and you will get <math>\boxed{550}.</math>
-But don't compute this! There's no need. Divide this by 90 and you will get <math>\boxed{550}</math>
 ~<math>\alpha b \alpha</math>
@@ Line 60: / Line 58: @@
 ==Solution 5 (Extremely Fast Solution)==
-The average values of the first and last digits are each <math>5,</math> and the average value of the middle digit is <math>4.5,</math> so the average of all three-digit palindromes is <math>5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.</math>
+The possible values of the first and last digits each are <math>1, 2, ..., 8, 9</math> with a sum of <math>45</math> so the average value is <math>5.</math> The middle digit can be any digit from <math>0</math> to <math>9</math> with a sum of <math>45,</math> so the average value is <math>4.5.</math> The average of all three-digit palindromes is <math>5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.</math>
 ~MathIsFun286
-==Video Solution==
+~MathFun1000 (Rephrasing with more clarity)
-https://www.youtube.com/watch?v=jDP2PErthkg
+==Solution 6 (Two cases)==
+<i><b>Case 1</b></i>
+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>
+<i><b>Case 2</b></i>
+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>
-==Remarks (Further Generalizations)==
+The arithmetic mean of all numbers is also <math>550.</math>
-<i><b>More generally, for every positive integer <math>\boldsymbol{k,}</math> the arithmetic mean of all the <math>\boldsymbol{(2k-1)}</math>-digit palindromes is <cmath>\boldsymbol{\frac{10^{2k-1}+10^{2k-2}}{2}.}</cmath></b></i> In this problem we have <math>k=2,</math> from which the answer is <math>\frac{10^3+10^2}{2}=550.</math>
-Note that all <math>(2k-1)</math>-digit palindromes are of the form <cmath>\underline{D_1D_2D_3\cdots D_k\cdots D_3D_2D_1}=D_1\left(10^{2k-2}+1\right)+D_2\left(10^{2k-3}+10\right)+D_3\left(10^{2k-4}+10^2\right)+\cdots+D_k\left(10^{k-1}\right),</cmath> where <math>D_1\in\{1,2,3,4,5,6,7,8,9\}</math> and <math>D_2,D_3,\cdots,D_k\in\{0,1,2,3,4,5,6,7,8,9\}.</math> Using this notation, we will prove the bolded claim in two different ways:
+'''vladimir.shelomovskii@gmail.com, vvsss'''
-===Proof 1 (Generalization of Solution 2)===
+==Remark==
-The arithmetic mean of all values of <math>D_1</math> is <math>5.</math> For each of <math>D_2,D_3,\cdots,D_k,</math> the arithmetic mean of all values is <math>4.5.</math> Together, the arithmetic mean of all the <math>(2k-1)</math>-digit palindromes is
+Visit the [https://artofproblemsolving.com/wiki/index.php/Talk:2021_AIME_II_Problems/Problem_1 Discussion Page] for questions and further generalizations.
-<cmath>\begin{align*}
-\left(10^{2k-2}+1\right)+4.5\left(10^{2k-3}+10\right)+4.5\left(10^{2k-4}+10^2\right)+\cdots+4.5\left(10^{k-1}\right)&=5\left(10^{2k-2}+1\right)+4.5\left(10+10^2+\cdots+10^{k-1}+\cdots+10^{2k-4}+10^{2k-3}\right) \\
-&=5\left(10^{2k-2}+1\right)+4.5\left(\frac{10^{2k-2}-10}{9}\right) \\
-&=5\left(10^{2k-2}+1\right)+\frac{10^{2k-2}-10}{2} \\
-&=\frac{10\left(10^{2k-2}+1\right)}{2}+\frac{10^{2k-2}-10}{2} \\
-&=\frac{10^{2k-1}+10}{2}+\frac{10^{2k-2}-10}{2} \\
-&=\frac{10^{2k-1}+10^{2k-2}}{2}.
-\end{align*}</cmath>
 ~MRENTHUSIASM
-===Proof 2 (Generalization of Solution 3)===
+==Video Solution==
-Note that <cmath>\underline{\left(10-D_1\right)\left(9-D_2\right)\left(9-D_3\right)\cdots \left(9-D_k\right)\cdots \left(9-D_3\right)\left(9-D_2\right)\left(10-D_1\right)}=\left(10-D_1\right)\left(10^{2k-2}+1\right)+\left(9-D_2\right)\left(10^{2k-3}+10\right)+\left(9-D_3\right)\left(10^{2k-4}+10^2\right)+\cdots+\left(9-D_k\right)\left(10^{k-1}\right),</cmath> must be another palindrome by symmetry. Therefore, we can pair each <math>(2k-1)</math>-digit palindrome <math>\underline{D_1D_2D_3\cdots D_k\cdots D_3D_2D_1}</math> uniquely with another <math>(2k-1)</math>-digit palindrome <math>\underline{\left(10-D_1\right)\left(9-D_2\right)\left(9-D_3\right)\cdots \left(9-D_k\right)\cdots \left(9-D_3\right)\left(9-D_2\right)\left(10-D_1\right)}</math> so that they sum to
+https://www.youtube.com/watch?v=jDP2PErthkg
-<cmath>\begin{align*}
-\left(10^{2k-2}+1\right)+9\left(10^{2k-3}+10\right)+9\left(10^{2k-4}+10^2\right)+\cdots+9\left(10^{k-1}\right)&=10\left(10^{2k-2}+1\right)+9\left(10+10^2+\cdots+10^{k-1}+\cdots+10^{2k-4}+10^{2k-3}\right) \\
+==Video Solution by Interstigation==
-&=10\left(10^{2k-2}+1\right)+9\left(\frac{10^{2k-2}-10}{9}\right) \\
+https://youtu.be/3_ik5N33CnQ
-&=10\left(10^{2k-2}+1\right)+\left(10^{2k-2}-10\right) \\
-&=10^{2k-1}+10+10^{2k-2}-10 \\
-&=10^{2k-1}+10^{2k-2}.
-\end{align*}</cmath>
-From this symmetry, the arithmetic mean of all the <math>(2k-1)</math>-digit palindromes is <math>\frac{10^{2k-1}+10^{2k-2}}{2}.</math>
-~MRENTHUSIASM
+speedy 2 min video
 ==See Also==
 {{AIME box|year=2021|n=II|before=First Problem|num-a=2}}
 {{MAA Notice}}

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