6. In the table below, each of the three rows is a correct arithmetic operation (with $\div$ indicating division). Also, each of the three columns (read from top to bottom) is a correct arithmetic operation. However, the digits in the table have been replaced by letters. Different letters stand for different digits, and all digits are not equal to 0.

$$
\begin{array}{cccc}
\mathrm{ABC} & -\mathrm{ADF} & =\mathrm{F} \\
+ & - & - \\
\mathrm{ADD} & \div \mathrm{GC} & =\mathrm{C} \\
= & = & = \\
\mathrm{CEF} & \div \mathrm{GD} & =\mathrm{D}
\end{array}
$$

For which digit does the letter E stand?
A) 1
B) 3
C) 5
D) 7
E) 9
A6. E) 9

From the first row, it follows that

$$
\mathrm{ADF}+\mathrm{F}=\mathrm{ABC} .
$$

The digit $\mathrm{C}$ is therefore even and, moreover, not 0. The possibilities are thus: 2, 4, 6, and 8. From the middle row, it follows that

$$
\mathrm{C} \cdot \mathrm{GC}=\mathrm{ADD} .
$$

Since $2 \cdot 2=4, 4 \cdot 4=16, 6 \cdot 6=36$, and $8 \cdot 8=64$, it follows that the digit $\mathrm{D}$ is either 4 or 6. The last row then gives

$$
\mathrm{D} \cdot \mathrm{GD}=\mathrm{CEF} .
$$

Since $4 \cdot 4=16$ and $6 \cdot 6=36$, the digit $\mathrm{F}$ must be 6. Because the digits are different, $\mathrm{D}$ must be 4. The first row now gives

$$
\mathrm{A} 46+6=\mathrm{ABC} .
$$

We see that $\mathrm{B}$ is the digit 5 and $\mathrm{C}$ is the digit 2. The first column is now

$$
\mathrm{A} 52+\mathrm{A} 44=2 \mathrm{E} 6 .
$$

The digit $\mathrm{E}$ is thus 9. Additionally, $\mathrm{A}=1$ and $\mathrm{G}=7$.