2021 AIME II Problems/Problem 1
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
- This relies on the fact that whenever x is a palindrome, 1100-x is also a palindrome. Once you realize this bijective relationship you will immediately obtain the mean, although it honestly may not be easy to see it on spot.
- Refer to Solution 3 (Symmetry and Generalization) for the note above.
-Note by Ross Gao and MRENTHUSIASM
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 any three-digit palindrome where and are digits with 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
~MRENTHUSIASM
Solution 4
- 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
Video Solution
https://www.youtube.com/watch?v=jDP2PErthkg
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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