5. \( \begin{array} \hline \;\;\;A & B\\ \times& 3\\ \hline C \;\;A & B\\ \hline \end{array} \)

We can observe that we have equations:

(i)B\(\times\)3=B

Solving it we can have B=0 or 5

5 is not taken because, for satisfying the condition A\(\times\)3=CA, we should not take carry. If carry is taken the ten’s value becomes A+1 not A.

\(∴B=0\)

(ii)A\(\times\)3=C A

we can satisfy this condition only if A=5 i.e., 5\(\times\)3=15.

So we have A=5, B=0 and C=1