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Difference between revisions of "2021 AIME II Problems/Problem 1"

(→‎Solution 3 (Symmetry and Generalization))
m (Undo revision 150261 by Dearpasserby (talk) I think there is no need to make things more complicated. So, I will undo the channges.)
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==Solution 3 (Symmetry and Generalization)==
 
==Solution 3 (Symmetry and Generalization)==
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For any three-digit palindrome <math>\overline{ABA},</math> where <math>A</math> and <math>B</math> are digits with <math>A\neq0,</math> note that <math>\overline{(10-A)(9-B)(10-A)}</math> must be another palindrome by symmetry. The mapping from 3-digit palindromes to 3-digit palindromes, <math>f: \overline{ABA} \rightarrow \overline{(10-A)(9-B)(10-A)}</math>, is a bijection. Different palindromes are mapped to different palindromes, and each palindrome has a preimage. In particular, because <math>f^2=id</math>, <math>f^{-1}(x)=f(x)</math>. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to  
+
For any three-digit palindrome <math>\overline{ABA},</math> where <math>A</math> and <math>B</math> are digits with <math>A\neq0,</math> note that <math>\overline{(10-A)(9-B)(10-A)}</math> must be another palindrome by symmetry. The mapping from 3-digit palindromes to 3-dit palindromes, <math>f: \overline{ABA} \rightarrow \overline{(10-A)(9-B)(10-A)}</math>, is a bijection. Different palindromes are mapped to different palindromes, and each palindrome has a preimage. In particular, because <math>f^2=id</math>, <math>f^{-1}(x)=f(x)</math>. 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*}
 
\overline{ABA}+\overline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\
 
\overline{ABA}+\overline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\

Revision as of 12:26, 25 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 $777$ or $383$.)

Solution 1

Recall* the the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=550$ and $\boxed{550}$ is the final answer.

~ math31415926535

  • This relies on the fact that whenever x is a palindrome, 1100-x is also a palindrome. Once you realize this bijective relationship you will immediately obtain the mean, although it honestly may not be easy to see it on spot.
  • Refer to Solution 3 (Symmetry and Generalization) for the note above.

-Note by Ross Gao and MRENTHUSIASM

Solution 2

For any palindrome $\overline{ABA}$, note that $\overline{ABA}$, 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 = $\boxed{550}$.

- ARCTICTURN

Solution 3 (Symmetry and Generalization)

For any three-digit palindrome $\overline{ABA},$ where $A$ and $B$ are digits with $A\neq0,$ note that $\overline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. The mapping from 3-digit palindromes to 3-dit palindromes, $f: \overline{ABA} \rightarrow \overline{(10-A)(9-B)(10-A)}$, is a bijection. Different palindromes are mapped to different palindromes, and each palindrome has a preimage. In particular, because $f^2=id$, $f^{-1}(x)=f(x)$. Therefore, 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*} For instances: \begin{align*} 101+999&=1100, \\ 262+838&=1100, \\ 373+727&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*} and so on.

From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1100}{2}=\boxed{550}.$

~MRENTHUSIASM

Solution 4

\begin{align*} \sum_{A = 1}^9 \sum_{B = 0}^9 \overline{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 \\ &= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\ &= 1010 \cdot 45 + 90 \cdot 45 \\ &= \end{align*}

- 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)
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First Problem 		Followed by
Problem 2
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All AIME Problems and Solutions

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