Finding Nth value

Question

If $1$ is $A$ and $2$ is $B$ and $3$ is $C$ and $4$ is $A$ again and $5$ is $B$ against and $6$ is $C$ again, what will be $16863$ is ?

Could some one explain logic to attain the answer ??

Answer 1

If I an not wrong, $16863$ is $C$.

Look $A$ occurs for $1, 4, 7, 10, . . . $

$B$ occurs for $2, 5, 8, 11, . . . $

$C$ occurs for $3, 6, 9, 12, . . . $

Now $16863$ can be written as $16863 = 3 \times 5621 $.

Her all the three digits $A, B, C$ occur in common difference $3$. But for $C$ it is also the multiple of $3$.

So $16863$ is $C$.