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Divide the following numbers:

(a) [tex]$1001_2 \div 11_2$[/tex]

(b) [tex]$110102 \div 10$[/tex]

Solution:

For (a) [tex][tex]$11_2 \sqrt{1001}$[/tex][/tex]:

[tex]\[

\begin{array}{c}

11_2 \, | \, 1001_2 \\

\hline

= 11 \, \downarrow \\

111 \\

-1 \\

= 0

\end{array}

\][/tex]

1. [tex]$11_2 \div 11_2$[/tex]

2. [tex]$110012 \div 1012$[/tex]

[tex]\[

\begin{array}{l}

1 \\

\text{1} \quad 1211 = \\

-112 \\

10110112 \div 1112

\end{array}

\][/tex]

3. [tex]$1_2 \div 1_2$[/tex]

Answer :

Let's divide the given binary numbers step by step:

1. Divide [tex]\( 1001_2 \)[/tex] by [tex]\( 11_2 \)[/tex]:

- Convert the binary numbers to decimals:
- [tex]\( 1001_2 \)[/tex] is [tex]\( 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 0 + 0 + 1 = 9 \)[/tex].
- [tex]\( 11_2 \)[/tex] is [tex]\( 1 \times 2^1 + 1 \times 2^0 = 2 + 1 = 3 \)[/tex].
- Now, divide 9 by 3:
- [tex]\( 9 \div 3 = 3 \)[/tex].

2. Divide [tex]\( 11010_2 \)[/tex] by [tex]\( 10_2 \)[/tex]:

- Convert the binary numbers to decimals:
- [tex]\( 11010_2 \)[/tex] is [tex]\( 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 16 + 8 + 0 + 2 + 0 = 26 \)[/tex].
- [tex]\( 10_2 \)[/tex] is [tex]\( 1 \times 2^1 + 0 \times 2^0 = 2 \)[/tex].
- Now, divide 26 by 2:
- [tex]\( 26 \div 2 = 13 \)[/tex].

3. Divide [tex]\( 11001_2 \)[/tex] by [tex]\( 101_2 \)[/tex]:

- Convert the binary numbers to decimals:
- [tex]\( 11001_2 \)[/tex] is [tex]\( 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 16 + 8 + 0 + 0 + 1 = 25 \)[/tex].
- [tex]\( 101_2 \)[/tex] is [tex]\( 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5 \)[/tex].
- Now, divide 25 by 5:
- [tex]\( 25 \div 5 = 5 \)[/tex].

4. Divide [tex]\( 1_2 \)[/tex] by [tex]\( 1_2 \)[/tex]:

- Convert the binary numbers to decimals:
- [tex]\( 1_2 \)[/tex] is [tex]\( 1 \)[/tex].
- Now, divide 1 by 1:
- [tex]\( 1 \div 1 = 1 \)[/tex].

Therefore, the results of the divisions are: [tex]\(3, 13, 5, \)[/tex] and [tex]\(1\)[/tex].

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