6.
\(\;\;\; \begin{array} \hline \;\;\;A & B\\ \times& 5\\ \hline C \;\;A & B\\ \hline \end{array} \)

We can observe that we have equations:

(i)B\(\times\)5=B

Solving it we can have B= 0 or 5

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

\(∴B=0\)

(ii)A\(\times\)5=C A

we can satisfy this condition only if A=5 i.e., 5\(\times\)5=25.

So we have A=5, B=0 and C=2