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:
Answer (550):
A three-digit palindrome has the form aba, where a is one of the digits 1,2,3,…,9, and b is one of the digits 0,1,2,…,9. Thus there are 9⋅10=90 three-digit palindromes. Note that aba=101a+10b, so the sum of all the three-digit palindromes is
a=1∑9b=0∑9(101a+10b)=a=1∑9(1010a+10⋅45)=1010⋅45+10⋅45⋅9=1100⋅45=550⋅90.
The requested arithmetic mean is 90550⋅90=550.
OR
As above, there are 90 three-digit palindromes. Of these, 80 are less than 900 . For each three-digit palindrome n less than 900 , the number 999−n is a different three-digit palindrome. The sum of these palindromes is 999⋅280=999⋅40. The remaining three-digit palindromes are 909+k, for k=0,10,20,…,90, and their sum is 909⋅10+450. Thus the arithmetic mean of the three-digit palindromes is
90999⋅40+909⋅10+450=444+101+5=550.
OR
Of all the palindromes of the form aba, the mean value of a is 5 , and the mean value of b is 4.5 . Thus the arithmetic mean of all the three-digit palindromes is 100⋅5+10⋅4.5+5=550.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions