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

(Proof 2 (Generalization of Solution 3))
m (Remark)
 
(11 intermediate revisions by the same user not shown)
Line 3: Line 3:
  
 
==Solution 1==
 
==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}=550</math> and <math>\boxed{550}</math> is the final answer.
  
 
~ math31415926535
 
~ math31415926535
Line 35: Line 35:
 
<u><b>Remark</b></u>
 
<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 the sum of each pair is <math>1100.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 64: Line 64:
 
~MathIsFun286
 
~MathIsFun286
  
==Video Solution==
+
==Remark==
https://www.youtube.com/watch?v=jDP2PErthkg
+
Visit the [https://artofproblemsolving.com/wiki/index.php/Talk:2021_AIME_II_Problems/Problem_1 Discussion Page] for questions and further generalizations.
 
 
==Remarks (Further Generalizations)==
 
<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> 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:
 
 
 
===Proof 1 (Generalization of Solution 2)===
 
<b>Proof in progress. Please do not edit--a million thanks.</b>
 
  
 
~MRENTHUSIASM
 
~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*}
 
10\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) \\
 
&=10\left(10^{2k-2}+1\right)+9\left(\frac{10^{2k-2}-10}{9}\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
 
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2021|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2021|n=II|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:16, 3 May 2021

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 $\frac{101+999}{2}=550$ and $\boxed{550}$ is the final answer.

~ math31415926535

Solution 2

For any palindrome $\underline{ABA}$, note that $\underline{ABA}$ is 100A + 10B + A, which is also 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 $\frac{1110}{2}=\boxed{550}.$

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 the sum of each pair is $1100.$

~MRENTHUSIASM

Solution 4 (Very, Very Easy and Quick)

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

And we know a can be any number from 1-9, and b can be any number from 0-9, 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 there is a times 9 when computing the total sum of b.

We get a sum of $45\times{1100}$

But don't compute this! There's no need. Divide this by 90 and you will get $\boxed{550}$

~$\alpha b \alpha$

Solution 5 (Extremely Fast Solution)

The average values of the first and last digits are each $5,$ and the average value of the middle digit is $4.5,$ so the average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$

~MathIsFun286

Remark

Visit the Discussion Page for questions and further generalizations.

~MRENTHUSIASM

Video Solution

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

See Also

2021 AIME II (ProblemsAnswer KeyResources)
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

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

Invalid username
Login to AoPS