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