Here, $6 \times B$ is a number, whose unit’s digit is $8.$
Therefore, the possible values of $B$ are $3 \& 8$ If $B = 3,$
then $A \times 6 + 1 = C6$ which is not possible for any value of A between $0$ to $9.$
$B = 8$ and then $A = 7$
The values of ? and ?
also satisfies the given condition i.e. $8 - 7 = 1$ If $A = 7$
then $7 \times 6 + 4 = C6 46 = C6 C = 4$
Hence, $A = 7, B = 8$ and $C = 4$
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