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
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 (Generalization and Symmetry)
For any three-digit palindrome where and are digits with note that must be another palindrome by symmetry. This means 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
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.