In each of the following cases, find the least value/values of letters used in place of digits :

$\begin{matrix} & 6 & \text{A} & \text{B} & 5 \ + & \text{D} & 5 & 8 & \text{C} \ \hline & 9 & 3 & 5 & 1 \ \hline \end{matrix}$

$\begin{matrix} & 6 & \text{A} & \text{B} & 5 \ + & \text{D} & 5 & 8 & \text{C} \ \hline & 9 & 3 & 5 & 1 \ \hline \end{matrix}$

Firstly, we will find the value of C.

5 + C is a number whose ones digit is 1.

⇒ 5 + C = 1, 5 + C = 11 , 5 + C = 21; and so on.

⇒ C = 1 - 5, C = 11 - 5, C = 21 - 5; and so on.

⇒ C = -4, C = 6, C = 16; and so on.

Since, C is a digit. ∴ C = 6

$\begin{matrix} & 6 & \text{A} & \overset{1}{\text{B}} & 5 \ + & \text{D} & 5 & 8 & \text{6} \ \hline & 9 & 3 & 5 & 1 \ \hline \end{matrix}$

Secondly, we will find the value of B.

B + 8 + 1 is a number whose ones digit is 5.

⇒ B + 9 = 5 , B + 9 = 15 , B + 9 = 25; and so on.

⇒ B = 5 - 9 , B = 15 - 9, B = 25 - 9; and so on.

⇒ B = -4 , B = 6, B = 16; and so on.

Since, B is a digit. ∴ B = 6

$\begin{matrix} & 6 & \overset{1}{\text{A}} & \overset{1}{\text{6}} & 5 \ + & \text{D} & 5 & 8 & \text{6} \ \hline & 9 & 3 & 5 & 1 \ \hline \end{matrix}$

Thirdly, we will find the value of A.

A + 5 + 1 is a number whose ones digit is 3.

⇒ A + 6 = 3 , A + 6 = 13 , A + 6 = 23; and so on.

⇒ A = 3 - 6 , A = 13 - 6, A = 23 - 6; and so on.

⇒ A = -3 , A = 7, A = 17; and so on.

Since, A is a digit. ∴ A = 7

$\begin{matrix} & \overset{1}{6} & \overset{1}{\text{7}} & \overset{1}{\text{6}} & 5 \ + & \text{D} & 5 & 8 & \text{6} \ \hline & 9 & 3 & 5 & 1 \ \hline \end{matrix}$

Lastly, we will find the value of D.

6 + D + 1 is a number whose ones digit is 9.

⇒ D + 7 = 9 , D + 7 = 19 , D + 7 = 29; and so on.

⇒ D = 9 - 7 , D = 19 - 7, D = 29 - 7; and so on.

⇒ D = 2 , D = 12, D = 22; and so on.

Since, D is a digit. ∴ D = 2

$\begin{matrix} & \overset{1}{6} & \overset{1}{\text{7}} & \overset{1}{\text{6}} & 5 \ + & \text{2} & 5 & 8 & \text{6} \ \hline & 9 & 3 & 5 & 1 \ \hline \end{matrix}$

A = 7, B = 6, C = 6 and D = 2.