10. Find the result of the following binary division operations:
(i) $(1100011)_{2} \div(100001)_{2}=$ ?
(ii) $(110001)_{2} \div(111)_{2}=$ ?
10.
(i) Solution: We first use the conventional division operation to get

Therefore, we have $(1100011)_{2} \div(100001)_{2}=(11)_{2}$.
Now we will use the method of finding complements to perform the operation. We first find the complement of the divisor $(100001)_{2}$ to get $(011111)_{2}=(11111)_{2}$, then add the complement $(11111)_{2}$ to the dividend $(1100011)_{2}$ to get
$$\begin{array}{r}
(1100011)_{2} \\
+\quad(11111)_{2} \\
\hline(10000010)_{2}
\end{array}$$

Then subtract $(1000000)_{2}$, to get the remainder after the first subtraction of the divisor, i.e.,
$$\begin{array}{r}
(1000000010)_{2} \\
=\quad \begin{aligned}
(10000000)_{2}
\end{aligned} \\
\hline
\end{array}$$

Then add the remainder $(1000010)_{2}$ to the complement $(111111)_{2}$ and subtract $(1000000)_{2}$, to get the remainder after the second subtraction of the divisor, i.e.,

Then add the remainder $(100001)$ to the complement $(11111)$ and subtract $(1000000)_{2}$, to get the remainder after the third subtraction of the divisor, i.e.,
$$\begin{array}{r}
(100001)_{2} \\
+\quad(111111)_{2} \\
\hline(1000000)_{2} \\
-(1000000)_{2} \\
\hline
\end{array}$$

It is not difficult to see that, because it was subtracted three times in total and just finished, the quotient is 3, converting 3 to a binary number is $(11)_{2}$, so
$(1100011)_{2} \div(100001)_{2}=(11)_{2}$.
(ii) Solution: We first use the conventional division operation to get
$$\begin{array}{r}
\left.(111)_{2}\right) \frac{(111)_{2}}{(110001)_{2}} \\
=\frac{(111100)_{2}}{(10101)_{2}} \\
\frac{-(11100)_{2}}{(1111)_{2}} \\
\frac{-(111)_{2}}{(0)_{2}}
\end{array}$$

Therefore, we have $(110001)_{2} \div(111)_{2}=(111)_{2}$.
Now we will use the method of finding complements to perform the operation. First, find the complement of the divisor $(111)_{2}$ to get $(1)_{2}$. Subtract the divisor from the dividend (i.e., add the complement $(1)_{2}$, then subtract $(1000)_{2}$;

It is not difficult to see that, because it was subtracted seven times in total and just finished, the quotient is 7, converting 7 to a binary number is $(111)_{2}$, so
$(110001)_{2} \div(111)_{2}=(111)_{2}$.
$$\begin{array}{l}
\left.(100001)_{2}\right) \frac{(11)_{2}}{(110001} \\
=\frac{(1000010)_{2}}{(100001)_{2}} \\
\frac{-(100001)_{2}}{(0)_{2}} \text {. }
\end{array}$$
$$\begin{array}{l}
\text { (1000010)2 } \\
+\frac{(11111)_{2}}{(1100001)_{2}} \\
-\frac{(1000000)_{2}}{(100001)_{2}} .
\end{array}$$
$$\begin{array}{l}
-\frac{(1000)_{2}}{(101010)_{2}}
\end{array}$$
$$\begin{array}{l}
\begin{array}{l}
-\quad(1000)_{2} \\
\hline(111000)_{2}
\end{array}
\end{array}$$
$$\begin{array}{l}
=\frac{(1000)_{2}}{(10101)_{2}}
\end{array}$$
$$\begin{array}{l}
\therefore \quad(1000)_{2} \\
+\quad \frac{(1)_{2}}{(10.00)_{2}} \text { + seventh subtraction of the divisor, remainder is }(0)_{2} \text { . } \\
-\quad(1000)_{2} \\
(0)_{2}
\end{array}$$