Difference between revisions of "2021 AIME II Problems/Problem 1"
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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{1110}{2}=\boxed{550}.</math> | ||
| − | + | (note: this method doesn't work because 1110/2 is actually 555) | |
<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. 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> | 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 | ||
Revision as of 15:58, 5 February 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
![\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*}](http://latex.artofproblemsolving.com/1/0/3/10347cd562e5b7658dee454874cdae46e341ce4a.png)
For instances:
and so on.
From this symmetry, the arithmetic mean of all the three-digit palindromes is 
(note: this method doesn't work because 1110/2 is actually 555)
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)
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. 
