Cryptogram with multiple solutions

There are two solutions. $\Large \begin{array}{ccc} \begin{array}{rrrr} & 7 & 6 & 5 \\ \times & & & 7 \\ \hline 5 & 3 & 5 & 5 \end{array} & & \begin{array}{rrrr} & 6 & 7 & 4 \\ \times & & & 6 \\ \hline 4 & 0 & 4 & 4 \end{array} \end{array}$

---

Consider the last digit. There are only a few cases in which $C \times A$ ends in the digit $C$ :

• $C = 0$ , but zero is not allowed at the beginning of a number.

• $A = 1$ , but $ABC\times 1 =ABC$ , so this cannot work.

• $C = 5$ and $A$ is odd

• $A = 6$ and $C$ is even

(Systematic way of making this list: $A \times C$ is $C$ plus a multiple of 10, so that $(A-1)\times C$ is a multiple of 10. This means that between $A-1$ and $C$ there is a factor 2 and a factor 5. Working out the four possible placements of these factors gives you the four items above.)

Case C = 5 : the product $AB5 \times A$ lies between $5055$ and $5955$ . This requires A = 7. We find the solution $7B5\times 7 = 5D55\ \ \ \therefore\ \ \ 4935 + 70\cdot B = 5055 + 100\cdot D\ \ \ \therefore\ \ \ 70B - 100D = 120\ \ \ \therefore\ \ \ B = 6,\ D = 3.$

Case A = 6 : the product is $6BC \times 6$ lies between 3600 and 4200; thus C = 4. We find the solution $6B4\times 6 = 4D44\ \ \ \therefore\ \ \ 3624 + 60\cdot B = 4044 + 100\cdot D\ \ \ \therefore\ \ \ 60B - 100D = 420\ \ \ \therefore\ \ \ B = 7,\ D = 0.$