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

$\begin{matrix} & 2 & \text{A} & \text{B} \ + & \text{A} & \text{B} & 1 \ \hline & \text{B} & 1 & 8 \ \hline \end{matrix}$

$\begin{matrix} & 2 & \text{A} & \text{B} \ + & \text{A} & \text{B} & 1 \ \hline & \text{B} & 1 & 8 \ \hline \end{matrix}$

Firstly, we will find the value of B.

Clearly, B + 1 is a number whose ones digit is 8.

⇒ B + 1 = 8, B + 1 = 18, B + 1 = 28; and so on.

⇒ B = 8 - 1, B = 18 - 1, B = 28 - 1; and so on.

⇒ B = 7, B = 17, B = 27; and so on.

Since, B is a digit. ∴ B = 7.

$\begin{matrix} & 2 & \text{A} & \text{7} \ + & \text{A} & \text{7} & 1 \ \hline & \text{7} & 1 & 8 \ \hline \end{matrix}$

Secondly, we will find the value of A.

Clearly, A + 7 is a number whose ones digit is 1.

⇒ A + 7 = 1, A + 7 = 11, A + 7 = 21; and so on.

⇒ A = 1 - 7, A = 11 - 7, A = 21 - 7; and so on.

⇒ A = -6 , A = 4 , A = 14; and so on.

Since, A is a digit. ∴ A = 4.

$\begin{matrix} & 2 & \text{4} & \text{7} \ + & \text{4} & \text{7} & 1 \ \hline & \text{7} & 1 & 8 \ \hline \end{matrix}$

A = 4 and B = 7.