Difference between revisions of "2021 AIME II Problems/Problem 1"
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==Solution 2 (Symmetry)== | ==Solution 2 (Symmetry)== | ||
− | For any palindrome <math>ABA,</math> note that | + | For any palindrome <math>ABA,</math> note that <math>ABA,</math> is 100A + 10b + A which is also 101A + 10B. |
<b>Will finish up soon. No edit please. Thanks. | <b>Will finish up soon. No edit please. Thanks. |
Revision as of 16:03, 22 March 2021
Contents
[hide]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 (Symmetry)
For any palindrome note that
is 100A + 10b + A which is also 101A + 10B.
Will finish up soon. No edit please. Thanks.
~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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.