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

$\begin{matrix} & & \text{A} & \text{B} \ & & \times & 6 \ \hline & \text{B} & \text{B} & \text{B} \ \hline \end{matrix}$

$\begin{matrix} & & \text{A} & \text{B} \ & & \times & 6 \ \hline & \text{B} & \text{B} & \text{B} \ \hline \end{matrix}$

Firstly, we will find the value of B.

Hence, B x 6 = B

A number multiplied with 6 gives same number in its unit place.

$2 \times 6 = 12\\[1em] 4 \times 6 = 24\\[1em] 6 \times 6 = 36\\[1em] 8 \times 6 = 48\\[1em]$

Hence, B = 2 or 4 or 6 or 8

Now find the value of A

Lets take B = 2

Then

$\begin{matrix} & & \overset{1}{\text{A}} & \text{2} \ & & \times & 6 \ \hline & \text{2} & \text{2} & \text{2} \ \hline \end{matrix}$

1 carried on A means A x 6 + 1 = 22. Since no value of A satisfy the equation. Hence B ≠ 2.

Lets take B = 4

Then

$\begin{matrix} & & \overset{2}{\text{A}} & \text{4} \ & & \times & 6 \ \hline & \text{4} & \text{4} & \text{4} \ \hline \end{matrix}$

2 carried on A means A x 6 + 2 = 44 ⇒ A x 6 = 42 ⇒ A = 7.

Lets take B = 6

Then

$\begin{matrix} & & \overset{3}{\text{A}} & \text{6} \ & & \times & 6 \ \hline & \text{6} & \text{6} & \text{6} \ \hline \end{matrix}$

3 carried on A means A x 6 + 3 = 66. Since no value of A satisfy the equation. Hence B ≠ 6.

Lets take B = 8

Then

$\begin{matrix} & & \overset{4}{\text{A}} & \text{8} \ & & \times & 6 \ \hline & \text{8} & \text{8} & \text{8} \ \hline \end{matrix}$

4 carried on A means A x 6 + 4 = 88. Hence A = 14. But A is single digit number. So, B ≠ 8

$\begin{matrix} & & \overset{2}{\text{A}} & \text{4} \ & & \times & 6 \ \hline & \text{4} & \text{4} & \text{4} \ \hline \end{matrix}$

Hence, A = 7 and B = 4