Difference between revisions of "2021 AIME II Problems/Problem 1"
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- A bit too complicated of a solution - somebody please fix. - ARCTICTURN | - 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== | ==See Also== |
Revision as of 17:52, 22 March 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 (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 |
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