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 + 10B + A which is also 101A + 10B. | 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}. | + | 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 = <math>\boxed{550}</math>. |
+ | |||
+ | - ARCTICTURN | ||
==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 15:07, 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 = .
- ARCTICTURN
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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