$(B \times 3) = B$ Then, $B$ can either be $0$ or $5.$ If $B$ is $5$,
then $1$ will be carried.
Then $A \times 3 + 1 = A$ will not be possible for any number.
$\therefore B = 0$
$A \times 3 = A$ is possible for either $0$ or $5.$
If we take $A = 0,$ then all number will become $0.$
however this not a possible.
$\therefore A = 5$
Then, $1$ will be carried.
$\therefore C = 1$
$\therefore A = 5, B = 0$ and $C = 1$
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