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

$\begin{matrix} & 1 & 2 & \text{A} \ + & 6 & \text{A} & \text{B} \ \hline & \text{A} & 0 & 9 \ \hline \end{matrix}$

$\begin{matrix} & 1 & 2 & \text{A} \ + & 6 & \text{A} & \text{B} \ \hline & \text{A} & 0 & 9 \ \hline \end{matrix}$

Firstly, we will find the value of A.

Clearly, 2 + A is a number whose ones digit is 0.

⇒ 2 + A = 0, 2 + A = 10, 2 + A = 20; and so on.

⇒ A = 0 - 2, A = 10 - 2, A = 20 - 2; and so on.

⇒ A = -2, A = 8, A = 18; and so on.

Since, A is a digit. ∴ A = 8.

$\begin{matrix} & \overset{1}{1} & 2 & \text{8} \ + & 6 & \text{8} & \text{B} \ \hline & \text{8} & 0 & 9 \ \hline \end{matrix}$

Secondly, we will find the value of B.

Clearly, 8 + B is a number whose ones digit is 9.

⇒ 8 + B = 9 , 8 + B = 19 , 8 + B = 29; and so on.

⇒ B = 9 - 8 , B = 19 - 8 ,B = 29 - 8; and so on.

⇒ B = 1 , B = 11 , B = 21; and so on.

Since, B is a digit. ∴ B = 1.

$\begin{matrix} & \overset{1}{1} & 2 & \text{8} \ + & 6 & \text{8} & \text{1} \ \hline & \text{8} & 0 & 9 \ \hline \end{matrix}$

A = 8 and B = 1.