What have you looked at so far
abcba
any digit can go in the middle, How many is that?
a can't be 0
So there are 9 choices for a and 10 choices for b.
So how many altogether?
Wait! The question is not asking for how many 5-digit palindromes there are. It wants the sum of all those palindromes instead. For example, were I to change the 5 to a 2, the situation would be like this. There are nine 2-digit palindromes: 11, 22, 33, 44, 55, 66, 77, 88, and 99, but the sum of all the 2-digit palindromes is $11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 = 495$.
Let's see
abcba
any digit can go in the middle, How many is that? 10
a can't be 0
So there are 9 choices for a and 10 choices for b.
So how many altogether?
10*9*10=900 exist.
Oh dear, Bginner is right, they are asking for the sum of them
ok
sum of 100c is 100(0+1+2+...9)=100(45)=4500
sum of 10001+20002+ ..... +90009 = 10000(1+2+...9) + (1+2+...9) = 10000*45+45 = 10001*45=450045
sum of 0000+1010+2020 .... 9090 = 45(1000+10) = 45*1010 = 45450
4500+450045+45450 = 499995 that is what I get but I havn't checked it. I could easily have miscounted.