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 <math>ABA,</math> is 100A + | + | For any palindrome <math>ABA,</math> note that <math>ABA,</math> 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 = \boxed{550}. | |
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==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2021|n=II|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:07, 22 March 2021
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.
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 = \boxed{550}.
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. 
