Difference between revisions of "Talk:2021 AIME II Problems/Problem 1"
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− | ==Further Generalizations== | + | ==Remarks (Further Generalizations)== |
<i><b>More generally, for every positive integer <math>\boldsymbol{k,}</math> the arithmetic mean of all the <math>\boldsymbol{(2k-1)}</math>-digit palindromes is <cmath>\boldsymbol{\frac{10^{2k-1}+10^{2k-2}}{2}.}</cmath></b></i> In this problem we have <math>k=2,</math> from which the answer is <math>\frac{10^3+10^2}{2}=550.</math> | <i><b>More generally, for every positive integer <math>\boldsymbol{k,}</math> the arithmetic mean of all the <math>\boldsymbol{(2k-1)}</math>-digit palindromes is <cmath>\boldsymbol{\frac{10^{2k-1}+10^{2k-2}}{2}.}</cmath></b></i> In this problem we have <math>k=2,</math> from which the answer is <math>\frac{10^3+10^2}{2}=550.</math> | ||
Revision as of 19:12, 3 May 2021
Remarks (Further Generalizations)
More generally, for every positive integer the arithmetic mean of all the
-digit palindromes is
In this problem we have
from which the answer is
Note that all -digit palindromes are of the form
where
and
Using this notation, we will prove the bolded claim in two different ways:
Proof 1 (Generalization of Solution 2)
The arithmetic mean of all values for is
and the arithmetic mean of all values for each of
is
Together, the arithmetic mean of all the
-digit palindromes is
~MRENTHUSIASM
Proof 2 (Generalization of Solution 3)
Note that must be another palindrome by symmetry. Therefore, we can pair each
-digit palindrome
uniquely with another
-digit palindrome
so that they sum to
From this symmetry, the arithmetic mean of all the
-digit palindromes is
As a side note, the total number of -digit palindromes is
by the Multiplication Principle. Their sum is
as we can match them into
pairs such that the sum of each pair is
~MRENTHUSIASM