Difference between revisions of "2021 AIME II Problems/Problem 1"
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MRENTHUSIASM (talk | contribs) m (We use underline to denote successive digits, as overline is "overloading".) |
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==Solution 2== | ==Solution 2== | ||
− | For any palindrome <math>\ | + | For any palindrome <math>\underline{ABA}</math>, note that <math>\underline{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 = <math>\boxed{550}</math>. | 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>. | ||
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==Solution 3 (Symmetry and Generalization)== | ==Solution 3 (Symmetry and Generalization)== | ||
− | For any three-digit palindrome <math>\ | + | For any three-digit palindrome <math>\underline{ABA},</math> where <math>A</math> and <math>B</math> are digits with <math>A\neq0,</math> note that <math>\underline{(10-A)(9-B)(10-A)}</math> must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \ | + | \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ |
&=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ | &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ | ||
&=1000+90+10 \\ | &=1000+90+10 \\ | ||
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==Solution 4== | ==Solution 4== | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | \sum_{A = 1}^9 \sum_{B = 0}^9 \ | + | \sum_{A = 1}^9 \sum_{B = 0}^9 \underline{ABA} &= \sum_{A = 1}^9 \sum_{B = 0}^9 \left( 101 A + 10 B \right) \\ |
&= \sum_{A = 1}^9 \sum_{B = 0}^9 101 A + \sum_{A = 1}^9 \sum_{B = 0}^9 10 B \\ | &= \sum_{A = 1}^9 \sum_{B = 0}^9 101 A + \sum_{A = 1}^9 \sum_{B = 0}^9 10 B \\ | ||
&= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\ | &= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\ |
Revision as of 21:59, 25 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 (Symmetry and Generalization)
For any three-digit palindrome where
and
are digits with
note that
must be another palindrome by symmetry. Therefore, 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
Video Solution
https://www.youtube.com/watch?v=jDP2PErthkg
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.