Multiplying Mystery Numbers

If $A > 1$ , then $\overline{AB5} \times 5 \geq 205 \times 5 > 1000$ , which is already a 4-digit number, thus $A \not > 1$ , or $A$ is forced to be equal to 1 only. And once again, we can simplify the cryptogram:

$\large{\begin{array}{cccccc} && & & 1 & B&5 \\ \times && & & & &5\\ \hline & & & & D & B&5\\ \hline \end{array}}$

By considering the last two digits, that is modulo 100, we can solve the cryptogram by modular arithmetic :

$\begin{aligned} &\phantom0&\overline{BC} \times C \equiv \overline{BC} \pmod{100} \\ && \Rightarrow \overline{B5} \times 5 \equiv \overline{B5} \pmod{100} \\ &&\Rightarrow 4 \times \overline{B5} \equiv0 \pmod{100} \\ &&\Rightarrow \overline{B5} \equiv 0 \pmod{25}\\ &&\Rightarrow B = 2 \text{ or } 7. \end{aligned}$

We are left ot check for two cases:
If $B = 2$ , then the product is $125 \times 5 = 625\Rightarrow D= 6\Rightarrow A+B+5+D = 1 + 2 + 5 + 6 = 14$ . If $B = 7$ , then the product is $175 \times 5 = 875\Rightarrow D= 6\Rightarrow A+B+5+D = 1 + 7+8+5= 21$ .

Since we want to minimize $A+B+5+D$ , then we take the smaller value of $14$ and $21$ , hence our answer is $\boxed{14}$ .