Difference between revisions of "2021 AIME II Problems/Problem 1"
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Symmetry and Generalization)) |
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Symmetry and Generalization)) |
||
| Line 14: | Line 14: | ||
==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] \\ | ||
Revision as of 06:39, 8 April 2021
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 the 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 
and 
is the final answer.
~ math31415926535
Solution 2
For any palindrome 
, note that , 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 = .
- 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 
Remark
By the Multiplication Principle, there are 
three-digit palindromes in total. We can match them into 
pairs such that the sum of each pair is 
Therefore, the arithmetic mean of all the three-digit palindromes is 
~MRENTHUSIASM
Solution 4 (Very, Very Easy and Quick)
We notice that a three-digit palindrome looks like this 
And we know a can be any number from 1-9, and b can be any number from 0-9, so there are 
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 
, 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 
But don't compute this! There's no need. Divide this by 90 and you will get
~
Solution 5 (Extremely Fast Solution)
The average values of the first and last digits are each 
and the average value of the middle digit is 
so the average of all three-digit palindromes is 
~MathIsFun286
Video Solution
https://www.youtube.com/watch?v=jDP2PErthkg
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. 
