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

$\begin{matrix} & & \text{A} & \text{B} \ & & \times & 3 \ \hline & \text{C} & \text{A} & \text{B} \ \hline \end{matrix}$

$\begin{matrix} & & \text{A} & \text{B} \ & & \times & 3 \ \hline & \text{C} & \text{A} & \text{B} \ \hline \end{matrix}$

Firstly, we will find the value of B.

$B \times 3 = B\\[1em] 0 \times 3 = 0\\[1em] 5 \times 3 = 15\\[1em]$

So, 0 and 5 are possible values of B.

Now finding the value of A.

Lets take B = 0 then A x 3 = A

$\begin{matrix} & & \text{A} & \text{0} \ & & \times & 3 \ \hline & \text{C} & \text{A} & \text{0} \ \hline \end{matrix}$

0 x 3 = 0

5 x 3 = 15

As we need 2-digit number in answer. So A = 5 and C = 1.

Lets take B = 5

$\begin{matrix} & & \overset{1}{\text{A}} & \text{5} \ & & \times & 3 \ \hline & \text{C} & \text{A} & \text{5} \ \hline \end{matrix}$

A x 3 + 1 = A

⇒ A x 3 = A - 1

There is no value for A which satisfies the equation. Hence B ≠ 5

$\begin{matrix} & & \text{5} & \text{0} \ & & \times & 3 \ \hline & \text{1} & \text{5} & \text{0} \ \hline \end{matrix}$

Hence, A = 5, B = 0 and C = 1.