Difference between revisions of "2021 AIME II Problems/Problem 1"
Arcticturn (talk | contribs) (→Solution 2 (Symmetry)) |
MRENTHUSIASM (talk | contribs) (Solution 3 Finally done!) |
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| − | ==Solution 2 | + | ==Solution 2== |
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 = <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>. | ||
- ARCTICTURN | - ARCTICTURN | ||
| + | |||
| + | ==Solution 3 (Symmetry)== | ||
| + | For any three-digit palindrome <math>\overline{ABA},</math> where <math>A</math> and <math>B</math> are digits and <math>A\neq0,</math> note that <math>\overline{(10-A)(9-B)(10-A)}</math> 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 | ||
| + | <cmath>\begin{align*} | ||
| + | \overline{ABA}+\overline{(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] \\ | ||
| + | &=1000+90+10 \\ | ||
| + | &=1100. | ||
| + | \end{align*}</cmath> | ||
| + | For instances: | ||
| + | <cmath>\begin{align*} | ||
| + | 272+828&=1100, \\ | ||
| + | 414+686&=1100, \\ | ||
| + | 595+505&=1100, | ||
| + | \end{align*}</cmath> | ||
| + | and so on. | ||
| + | |||
| + | From this symmetry, the arithmetic mean of all the three-digit palindromes is <math>\frac{1110}{2}=\boxed{550}.</math> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
==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 15:23, 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
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)
For any three-digit palindrome 
where 
and 
are digits and 
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
![\begin{align*} \overline{ABA}+\overline{(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] \\ &=1000+90+10 \\ &=1100. \end{align*}](http://latex.artofproblemsolving.com/f/d/6/fd6c776254e3f02e57f957c964234ba0e2cd6722.png)
For instances:
and so on.
From this symmetry, the arithmetic mean of all the three-digit palindromes is 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
