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

(Solution 2 (Symmetry))
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==Solution 2 (Symmetry)==
 
==Solution 2 (Symmetry)==
For any palindrome <math>ABA,</math> note that ... is also...
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For any palindrome <math>ABA,</math> note that <math>ABA,</math> is 100A + 10b + A which is also 101A + 10B.  
  
 
<b>Will finish up soon. No edit please. Thanks.
 
<b>Will finish up soon. No edit please. Thanks.

Revision as of 16:03, 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 $777$ 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

Solution 2 (Symmetry)

For any palindrome $ABA,$ note that $ABA,$ is 100A + 10b + A which is also 101A + 10B.

Will finish up soon. No edit please. Thanks.

~MRENTHUSIASM

See also

2021 AIME II (ProblemsAnswer KeyResources)
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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