Average Digits

From here we list out all the possibilities.

Answer=1+2+3+4+5+6+7+8+9=45

We assume that $\overline { ABC } =100A+10B+C$

Consider $\overline { ABC }$ with $A\neq 0$ ; we know that $B=\frac { A+C }{ 2 }$ . This implies that A+C is an even number: it can be possible if A and Care both even or both odd. If A and C are both even A could be 2, 4, 6, 8 (4 digits) and C could be 0, 2, 4, 6, 8 (5 digits) so we have 4x5 possibilities. If A and C are both odd A could be 1, 3, 5, 7, 9 (5 digits) just like C which it could be 1, 3, 5, 7, 9 (5 digits) so we have 5x5 possibilities. We have 5x4+5x5= 45 3-digit integers that have the property that their central digit is the average of the other two digits.