Sum of all $3-$ digit even palindrome

Question

Find the sum of all $3-$digit even palindromes.

Now number of even $3-$digit palindromes will be $10 \times 4=40$ but how to proceed to find sum?

Could someone give me some hint to proceed?

Answer 1

Such a number must end in $2$, $4$, $6$ or $8$ and begin with the same digit. It is not too difficult to add them up by using the formula for a finite arithmetic series. $$ (202+212+\dotsb+292)+(404+414+\dotsb+494)+\dotsb (808+818+\dotsb+898) \tag{1} $$ For example the term in the first bracket can be computed as $$ \frac{10(202+292)}{2}. $$ The general formula for a finite arithmetic series with $n$ terms, first term $a$ and last term $b$ is $$ \frac{n(a+b)}{2}. $$ You can compute the rest of (1) in this manner.

Answer 2

Each three digit even palindome will have the form $aba$ where $a=2,4,6,8$ and $b=0,1,\cdots,9$. So \begin{eqnarray*} S=\sum_{a=2,4,6,8} \sum_{b=0}^{9} (101a+10b)= \color{red}{2470}. \end{eqnarray*}