Find the missing digit in the number 23104*791 
Find the missing digit in the number $23104*791$ if
(i) it is divisible by $11$,
(ii) it is divisible by $13$,
(iii) it is divisible by $63$.

(i) $23104*791=231 (990+10)^2+4*(990+10)+11\times 72-1 \implies 11|4*0-1 \implies 11|400+*0-1 \implies 11|3+*0 \implies 11|*3 \implies *=3$
(ii) & (iii) Unable to adjust. Please help me to solve.

For the second, since we have 
$$10^3\equiv (-3)^3\equiv -27\equiv -1\pmod{13},$$
we have
$$(231)\cdot (10^3)^2+(40+*)\cdot 10^3+791\equiv 0\pmod{13}$$
$$\Rightarrow 10\cdot 1+(1+*)\cdot (-1)+11\equiv 0\pmod{13}$$
$$\Rightarrow *\equiv 10-1+11\equiv 7\pmod{13}$$
Hence, we have $*=\color{red}{7}$. (This is sufficient.)
For the third, it is divisible by $9$, so we have
$$2+3+1+0+4+*+7+9+1\equiv 0\pmod 9\Rightarrow *\equiv 0\pmod{9}$$
Also, it is divisible by $7$. Since we have 
$$10^3\equiv 3^3\equiv 27\equiv -1\pmod{7},$$
we have
$$(231)\cdot (10^3)^2+(40+*)\cdot 10^3+791\equiv 0\pmod{7}$$
$$\Rightarrow 0+(5+*)\cdot (-1)+0\equiv 0\pmod 7$$
$$\Rightarrow *\equiv 2\pmod 7$$
Hence, we have $*=\color{red}{9}$. (This is sufficient.)
