For

$\begin{matrix} & 1 & \text{A} \ + & \text{A} & 3 \ \hline 1 & 0 & 1 \ \hline \end{matrix}$

the value of digit A is :

1. 18

2. 6

3. 28

4. none of these

$\begin{matrix} & 1 & \text{A} \ + & \text{A} & 3 \ \hline 1 & 0 & 1 \ \hline \end{matrix}$

A + 3 must give a number whose ones digit is 1.

⇒ A + 3 = 1, A + 3 = 11, A + 3 = 21, ……………

⇒ A = 1 - 3, A = 11 - 3, A = 21 - 3, ……………

⇒ A = -2, A = 8, A = 18, ……………

⇒ A = 8, as A is a digit.

$\begin{matrix} & \overset{1}{1} & \text{8} \ + & \text{8} & 3 \ \hline 1 & 0 & 1 \ \hline \end{matrix}$

Hence, option 4 is the correct option.