Double Cryptarithm!

Consider the second product above. From the thousands digit, we see that $G \cdot B \leq G$ and, hence, $B=1$ . Then since $G \cdot G$ ends in $B=1$ , we have $G=9$ . Now the product is $1FD9 \cdot 9 = 9FD1$ . We cannot have $F>1$ , because the product $1FD9 \cdot 9$ would have then five digits instead of 4, and we cannot have $F=1$ since $B=1$ . Thus $F=0$ . From $10D9 \cdot 9 = 9D01$ , we find that the only choice for D is $D=8$ .

The first product is now $A1C8 \cdot E = 8C1A$ , with $A,C$ and E having values in $\{2,3,4,5,6,7\}$ . Since $A1C8 \cdot E$ is even, we see that $8C1A$ comes from adding a carried digit to the product $A \cdot E$ . This carried digit is atmost 1, since it comes from $1C8 \cdot E$ ; therefore, $A \cdot E$ is either $7$ or $8$ . Since $A$ is even, we must have $A \cdot E = 8$ . Now either $A=2$ and $E=4$ , or $A=4$ and $E=2$ . If $A=4$ and $E=2$ , we have $41C8 \cdot 2 = 8C14$ , which is impossible. We conclude that $A=2, E=4$ , which gives us $21C8 \cdot 4 = 8C12$ .

Finally we verify that the only choice for $C$ is $C=7$ . Thus $A=2, B=1, C=7, D=8, E=4, F=0, G=9$ .

And the sum $A+B+C+D+E+F+G = \boxed{31}$ .