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# 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