Numerical puzzle, №2

Writing the numbers out in full (ie expanding the powers of ten for each digit) and rearranging, we get the somewhat daunting

$3818393A + 181714B - 1961240C + 3780D = 0$

Firstly, let's look at the magnitudes of the coefficients. Only the coefficient of $C$ is comparable to the coefficient of $A$ ; if $C \neq 2A$ , there's no way to get close to zero using the other terms (remember that $1 \leq A,B,C,D \leq 9$ ). So $C=2A$ .

Now note that the coefficient of $A$ is the only odd one. This means that $A$ must be even (otherwise the total would be odd, and zero is not odd). So $A=2k$ for some $1\le k \le4$ , and $C=4k$ (so in fact, $k$ is either $1$ or $2$ ; we could just check both cases here, but the magnitude trick works again).

Substituting in and simplifying, we get

$90857 B + 1890 D - 104087 k=0$

By the same trick of looking at magnitudes, we see that $B=k$ . Substituting, simplifying:

$D-7k=0$

So we must have $k=1$ , $(A,B,C,D)=(2,1,4,7)$ (and checking, this does indeed work), leading to the digit sum of $\overline{CBABBDAA}$ being $\boxed{20}$ .

A=2, B=1, C=4, D=7, sum of digits of CBABBDAA is 20

$\text{Flatten}[\text{Table}[\text{If}[19 (\text{FromDigits}[\{a,b,c,d,c,b,a\}]+\text{FromDigits}[\{a,b,c,b,a\}]+\text{FromDigits}[\{a,b,a\}]+a)=\text{FromDigits}[\{c,b,a,b,b,d,a,a\}], \\ \text{Total}[\{c,b,a,b,b,d,a,a\}],\text{Nothing}],\{a,9\},\{b,9\},\{c,9\},\{d,9\}]][[1]] \Rightarrow 20$

Tried all 6561 possibilities, only 1 success, output the sum of digits as specified, 20. The Flatten function removes the failed cases and collapses the empty lists to nothing.