2021 AIME II Problems/Problem 1

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* the 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 any three-digit palindrome $\underline{ABA},$ where $A$ and $B$ are digits with $A\neq0,$ 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: $\begin{align*} 101+999&=1100, \\ 262+838&=1100, \\ 373+727&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*}$ and so on.

From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1110}{2}=\boxed{550}.$

~MRENTHUSIASM

Solution 4

$\begin{align*} \sum_{A = 1}^9 \sum_{B = 0}^9 \underline{ABA} &= \sum_{A = 1}^9 \sum_{B = 0}^9 \left( 101 A + 10 B \right) \\ &= \sum_{A = 1}^9 \sum_{B = 0}^9 101 A + \sum_{A = 1}^9 \sum_{B = 0}^9 10 B \\ &= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\ &= 1010 \cdot 45 + 90 \cdot 45 \\ &= \end{align*}$

- A bit too complicated of a solution - somebody please fix. - ARCTICTURN

Doriding is the original author. I will wait for him to come back. ~MRENTHUSIASM

Solution 5 (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 $\boxed{550}$

~ $\alpha b \alpha$

Video Solution

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

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