Let's Make Digits Distinct!

Suppose that there are some digits that work for the given cryptarithm. Let $A, B, C, D, E,F, G$ be distinct nonnegative digits, such that $\begin{array}{rl} \overline{A} \times \overline{B} &= \overline{CD}\\ \overline{CD} + \overline{G} &= \overline{EF} \end{array}$ where $\overline{EF}$ is the number being divided by $\overline{A}$ , and $\overline{G}$ is the remainder. Since $\overline{EF}$ and $\overline{CD}$ both have different digits, then $E = C + 1$ , so $\begin{array}{rl} \overline{CD} + \overline{G} &= \overline{EF}\\ 10C + D + G &= 10E + F\\ 10C + D + G &= 10C + 10 + F\\ D + G &= 10 + F\\ D &= 10 + F - G \end{array}$ where $F < G$ , so $D < 10$ . However, $\begin{array}{rl} \overline{A} \times \overline{B} &= \overline{CD}\\ &= 10C + D\\ &= 10C + 10 + F - G\\ &= 10\left(C + 1\right) + \left(F - G\right)\\ &= 10E + (F - G) \end{array}$ Contradiction: This can't happen since $F - G$ is the negative ones digit, and since $C = E$ violates the assumption that all digits are different.

Thus, it is $\boxed{\text{impossible}}$ to fill all boxes with distinct digits.