An algebra problem by Kim Ian Macay

Consider

$\begin{aligned} \overline{ab} + \overline{ba} & = \overline{dbc} \\ 10a + b + 10b + a & = 100d + 10b + c \\ 11a + b & = 100d + c & \small \color{#3D99F6} \text{Since } d \ne 0 \implies d = 1 \text{ and } a =9 \\ \implies 99 + b & = 100 + c \\ \color{#3D99F6} b & \color{#3D99F6} = c + 1 & \end{aligned}$

From $d + e = 1+e=5$ , $\implies e = 4$ .

From $d+b+c = \overline{de}$ , $\implies 1+c+1+c = 10 + 4$ $\implies c = 6$ and $b = 7$ .

Therefore, $a=9$ , $b=7$ , $c=6$ , $d=1$ , and $e=4$ ; and $a+b+c+d+e = 9+7+6+1+4 = \boxed{27}$ .

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Note that the clues: $d + c = b$ and $a + b = \overline{dc}$ are unnecessary.

easy, the maximum possible sum of 2 digits is 18 and its minimum is 2, so a and b are single digits, having a sum of dc. d could only be equal to 1 since a and b are having a two digit sum not more the 18 so d is 1 not 2 not 0. the rest will follow. COMMON SENSE