Difference between revisions of "2021 AIME II Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
− | Recall | + | 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 | ~ math31415926535 | ||
==Solution 2== | ==Solution 2== | ||
− | For any palindrome <math>\underline{ABA}</math> | + | 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 = | + | 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 | - ARCTICTURN | ||
==Solution 3 (Symmetry and Generalization)== | ==Solution 3 (Symmetry and Generalization)== | ||
− | For | + | For every three-digit palindrome <math>\underline{ABA}</math> with <math>A\in\{1,2,3,4,5,6,7,8,9\}</math> and <math>B\in\{0,1,2,3,4,5,6,7,8,9\},</math> note that <math>\underline{(10-A)(9-B)(10-A)}</math> 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 |
<cmath>\begin{align*} | <cmath>\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] \\ | \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ | ||
Line 31: | Line 31: | ||
and so on. | and so on. | ||
− | From this symmetry, the arithmetic mean of all the three-digit palindromes is <math>\frac{ | + | 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. 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 | ~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>\ | + | We notice that a three-digit palindrome looks like this: <math>\underline{aba}.</math> |
− | And we know a can be any | + | 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> | + | 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> |
− | |||
− | |||
~<math>\alpha b \alpha</math> | ~<math>\alpha b \alpha</math> | ||
Line 56: | Line 58: | ||
==Solution 5 (Extremely Fast Solution)== | ==Solution 5 (Extremely Fast Solution)== | ||
− | The | + | 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 | ~MathIsFun286 | ||
+ | |||
+ | ~MathFun1000 (Rephrasing with more clarity) | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | The arithmetic mean of all numbers is also <math>550.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Remark== | ||
+ | Visit the [https://artofproblemsolving.com/wiki/index.php/Talk:2021_AIME_II_Problems/Problem_1 Discussion Page] for questions and further generalizations. | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=jDP2PErthkg | https://www.youtube.com/watch?v=jDP2PErthkg | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/3_ik5N33CnQ | ||
+ | |||
+ | speedy 2 min video | ||
==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 14:36, 30 August 2022
Contents
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 or .)
Solution 1
Recall that the arithmetic mean of all the digit palindromes is just the average of the largest and smallest digit palindromes, and in this case the palindromes are and and which is the final answer.
~ math31415926535
Solution 2
For any palindrome note that is The average for is since can be any of or The average for is since is either or Therefore, the answer is
- ARCTICTURN
Solution 3 (Symmetry and Generalization)
For every three-digit palindrome with and note that 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 For instances: and so on.
From this symmetry, the arithmetic mean of all the three-digit palindromes is
Remark
By the Multiplication Principle, there are three-digit palindromes in total. Their sum is as we can match them into pairs such that each pair sums to
~MRENTHUSIASM
Solution 4 (Similar to Solution 2: Very, Very Easy and Quick)
We notice that a three-digit palindrome looks like this:
And we know can be any digit from through and can be any digit from through so there are three-digit palindromes.
We want to find the sum of these palindromes and divide it by to find the arithmetic mean.
How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts:
Thus, all of these palindromes can be broken into this form.
Thus, the sum of these palindromes will be because each will be in different palindromes (since for each there are choices for ). The same logic explains why we multiply by when computing the total sum of
We get a sum of but don't compute this! Divide this by and you will get
~
Solution 5 (Extremely Fast Solution)
The possible values of the first and last digits each are with a sum of so the average value is The middle digit can be any digit from to with a sum of so the average value is The average of all three-digit palindromes is
~MathIsFun286
~MathFun1000 (Rephrasing with more clarity)
Solution 6 (Two cases)
Case 1
Consider palindromes of the form There are of them. The arithmetic mean of the first term is and the second The arithmetic mean of the sum is
Case 2
Consider palindromes of the form where Let Then is a symmetric palindrome that can be constructed for each The arithmetic mean of each such pair is For example,
Thus, all palindromes are divided into groups of numbers with the arithmetic mean in each group equal to
The arithmetic mean of all numbers is also
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
speedy 2 min video
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.