Four-Digit Subtraction - Revised

Rewrite the equation:

$CADB + ABCD = DCBA$

For the unit digits, since $A < B < D$ , it must follow that $B+D = A+10$ .

For the tens digits, since $B < C < D$ , it must follow that $D+C+1 = B+10$ .

For the hundreds digits, since $A < B < C$ , it must follow that $A+B+1 = C$ .

For the thousands digits, we have $C+A = D$ .

Adding all the four equations yields $2(A+B+C+D)+2 = (A+B+C+D)+20$ .

Therefore, $A+B+C+D = 20-2 = \boxed{18}$ .

Note: the solution to the system of equations is $(A,B,C,D) = (1,4,6,7)$ .

The sum of the digits of a number is congruent to its remainder modulo 9. That is,

$A + B + C + D \equiv \overline{ABCD} \equiv \overline{DCBA} \equiv \overline{CADB} \mod 9$

Since $\overline{ABCD} \equiv \overline{DCBA} \bmod 9$ , $\overline{ABCD} - \overline{DCBA} \equiv 0\bmod 9$ . Thus $\overline{CADB} \equiv 0\bmod 9$ and, indeed, all our numbers are divisible by 9. This means that the sum $A+B+C+D$ must be divisible by 9.

The only digit that could be zero is $A$ , but that would give us a leading zero in $\overline{ABCD}$ . Thus, the smallest possible value of $A+B+C+D$ is $1+2+3+4=10>9$ . Similarly, the greatest possible value is $6+7+8+9 = 30<36$ . Thus, the only possible sums are $18$ and $27$ .

$A+B+C+D=27$ is only possible if $D=9$ and $A\geq 3$ . But then the leftmost digit in the subtraction would imply that $C\leq 6$ . Since $4+5+6+9=24$ , this is impossible.

Thus the only possibility is $A+B+C+D=\boxed{18}$ .

The difference can be represented by

$\begin{aligned} 1000D+100C+10B+A - 1000A-100B-10C-D & = 1000C+100A+10D+B \\ \implies 1099A+91B+910C & = 989D \\ {\color{#3D99F6}7}(157A+13B+130C) & = 23 \times 43\times \color{#3D99F6}D \\ \implies D & = 7 \end{aligned}$

Since the LHS is a multiple of 7, the RHS must also be a multiple of 7, therefore, $D=7$ . Since $A<B<C<D$ , we note $1 \le A \le 4$ , $2 \le B \le 5$ , and $3 \le C \le 6$ . From the above, we have:

$\begin{aligned} 157A+13B+130C & = 989 & \small \color{#3D99F6} \text{Consider the last digit} \\ \implies 7A + 3B & \equiv 9 \text{ (mod 10)} \end{aligned}$

$\implies \begin{cases} A=1 & \implies \color{#3D99F6} B = 4 & \small \color{#3D99F6} \text{Acceptable} \\ A=2 & \implies \color{#3D99F6} B = 5 & \small \color{#3D99F6} \text{Acceptable} \\ A=3 & \implies \color{#D61F06} B = 6 & \small \color{#D61F06} \text{Unacceptable} \\ A=4 & \implies \color{#D61F06} B = 7 & \small \color{#D61F06} \text{Unacceptable} \end{cases}$

From $C = \dfrac {989 - 157A-13B}{130}$ $\implies \begin{cases} A=1, & B = 4 & \color{#3D99F6} \implies C = 6 \\ A=2, & B = 5 & \color{#D61F06} \implies C = 4\frac 9{13} \quad \small \text{No integer solution} \end{cases}$

Therefore, $A+B+C+D=1+4+6+7 = \boxed{18}$ .