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AIME 2021 II · 第 1 题

AIME 2021 II — Problem 1

专题
Contest Math
难度
L4
来源
AIME

题目详情

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 777777 or 383383.)

解析

Solution 1

Recall that the arithmetic mean of all the nn digit palindromes is just the average of the largest and smallest nn digit palindromes, and in this case the 22 palindromes are 101101 and 999999 and 101+9992=550,\frac{101+999}{2}=\boxed{550}, which is the final answer.

~ math31415926535

Solution 2

For any palindrome ABA,\underline{ABA}, note that ABA\underline{ABA} is 100A+10B+A=101A+10B.100A + 10B + A = 101A + 10B. The average for AA is 55 since AA can be any of 1,2,3,4,5,6,7,8,1, 2, 3, 4, 5, 6, 7, 8, or 9.9. The average for BB is 4.54.5 since BB is either 0,1,2,3,4,5,6,7,8,0, 1, 2, 3, 4, 5, 6, 7, 8, or 9.9. Therefore, the answer is 505+45=550.505 + 45 = \boxed{550}.

- ARCTICTURN

Solution 3 (Symmetry and Generalization)

For every three-digit palindrome ABA\underline{ABA} with A{1,2,3,4,5,6,7,8,9}A\in\{1,2,3,4,5,6,7,8,9\} and B{0,1,2,3,4,5,6,7,8,9},B\in\{0,1,2,3,4,5,6,7,8,9\}, note that (10A)(9B)(10A)\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

ABA+(10A)(9B)(10A)=[100A+10B+A]+[100(10A)+10(9B)+(10A)]=[100A+10B+A]+[1000100A+9010B+10A]=1000+90+10=1100.\begin{aligned} \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{aligned} For instances:

171+929=1100,262+838=1100,303+797=1100,414+686=1100,545+555=1100,\begin{aligned} 171+929&=1100, \\ 262+838&=1100, \\ 303+797&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{aligned} and so on.

From this symmetry, the arithmetic mean of all the three-digit palindromes is 11002=550.\frac{1100}{2}=\boxed{550}.

Remark

By the Multiplication Principle, there are 910=909\cdot10=90 three-digit palindromes in total. Their sum is 110045=49500,1100\cdot45=49500, as we can match them into 4545 pairs such that each pair sums to 1100.1100.

~MRENTHUSIASM

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

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

And we know aa can be any digit from 11 through 9,9, and bb can be any digit from 00 through 9,9, so there are 9×10=909\times{10}=90 three-digit palindromes.

We want to find the sum of these 9090 palindromes and divide it by 9090 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.101a+10b.

Thus, all of these 9090 palindromes can be broken into this form.

Thus, the sum of these 9090 palindromes will be 101×(1+2+...+9)×10+10×(0+1+2+...+9)×9,101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9}, because each aa will be in 1010 different palindromes (since for each a,a, there are 1010 choices for bb). The same logic explains why we multiply by 99 when computing the total sum of b.b.

We get a sum of 45×1100,45\times{1100}, but don't compute this! Divide this by 9090 and you will get 550.\boxed{550}.

~αbα\alpha b \alpha

Solution 5 (Extremely Fast Solution)

The possible values of the first and last digits each are 1,2,...,8,91, 2, ..., 8, 9 with a sum of 4545 so the average value is 5.5. The middle digit can be any digit from 00 to 99 with a sum of 45,45, so the average value is 4.5.4.5. The average of all three-digit palindromes is 5102+4.510+5=550.5\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.5x5 = 505 + 10x. There are 1010 of them. The arithmetic mean of the first term is 505,505, and the second 10(0+1+...+9)10=45.\frac {10 \cdot(0 + 1 + ... + 9)}{10} = 45. The arithmetic mean of the sum is 505+45=550.505 + 45 = 550.

Case 2

Consider palindromes of the form yxy,yxy, where y=1,2,3,4,6,7,8,9.y= {1,2,3,4,6,7,8,9}. Let u=10y,v=9x.u = 10 – y, v = 9 – x. Then uvuuvu is a symmetric palindrome that can be constructed for each yxy.yxy. The arithmetic mean of each such pair is 550.550. For example, 737+3632=550.\frac{737 + 363}{2} = 550.

Thus, all palindromes are divided into groups of numbers with the arithmetic mean in each group equal to 550.550.

The arithmetic mean of all numbers is also 550.550.

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