Do you know your ABCD's?

• $F \ne 1$ . If $F =1$ then $\overline{ABCDEF} \times F > \overline{ABCDEF} \ne \overline{GGGGGG}$ .
• $F \ne 2$ . If $F=2$ then $F \times F = 4 = G$ . $E$ must be $7$ so that $\overline{EF} \times F = 144$ . We find that we cannot find a $D \times 2$ that ends with $3$ so that $\overline{D72} \times 2 = \overline{^0_1 44}$ .
• $F \ne 3$ . If $F=3$ then $F \times F = 9 = G$ . We cannot find an $E \times 3$ (other than $3 = F$ ) that ends with $9$ so that $\overline{E3} \times 3 = \overline{^0_1 99}$ .
• $F \ne 4$ . If $F = 4$ , then $F \times F = 16$ and $G = 6$ , but we cannot find an $E \times 4$ that ends with $5$ so that $\overline{E4} \times 4 = \overline{^0_1 66}$ .
• Therefore, $F \ge 5$ and $A$ can only be $1$ so that $\overline{ABCDEF} \times F \le 999999$ .
• $F \ne 5$ because we cannot find $\overline{E5} \times 5 = \overline{^0_1 55}$ .
• $F \ne 6$ because we cannot find $\overline{E6} \times 6 = \overline{^0_1 66}$ .
• $F \ne 8$ . If $F = 8$ , then $G = 4$ , but $A \times F + ^0_1 = 1 \times 8 + ^0_1 = 8$ or $9$ and not $4$ .
• $F \ne 9$ or else $\overline{ABCDEF} \times F > 999999$ .
• Therefore $F$ must be $\boxed{7}$ , and $G = 9$ .

$\Rightarrow 999999 \div 7 = 142857 \quad \Rightarrow A+B+C+D+E+F+G= 1+4+2+8+5+7+9 = \boxed{36}$ .

Hence, $A= 1,B=4,C=2,D=8,E=5,F=7,G=9$ . The sum is $\boxed{36}$ .

GGGGGG=111111*G=3 * 7 * 13 * 37 * G, which means that F can only be 3 or 7 and G=9. Since when F=3, A=B=C=D=E=F, F=7. 999999/7=142857