To divide the binary number \( 1101101_2 \) by \( 11_2 \), we will perform binary long division.
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Set Up the Division: \[ \begin{array}{r|l} & 1101101 \ 11 & \ \end{array} \]
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Perform the Division:
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Divide the first two digits \( 11 \) by \( 11 \):
- \( 11 \) goes into \( 11 \) exactly \( 1 \) time.
- Write down \( 1 \) above the division line.
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Subtract \( 11 \) from \( 11 \): \[ 11 - 11 = 00 \quad \text{(bring down the next digit, which is 0, to make it 000)} \]
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Now we take the next group, which is \( 000 \).
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\( 11 \) does not go into \( 00 \) (0 times).
- Write down \( 0 \) above the division line.
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Bring down the next \( 1 \) (making it \( 001 \)).
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Again, \( 11 \) does not go into \( 01 \). So we write down another \( 0 \).
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Bring down the next \( 0 \) (which makes it \( 0010 \)).
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Still, \( 11 \) does not go into that, so we write another \( 0 \) again.
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Now we bring down the final \( 1 \) (making it \( 101 \)).
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\( 11 \) goes into \( 10 \) (the first two digits of \( 101 \)) \( 0 \) times.
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Then we take the last \( 1 \) (making it \( 101 \) now).
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\( 11 \) goes into \( 101 \) \( 1 \) time.
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Write down \( 1 \) above the line.
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Then subtract \( 11 \) from \( 101 \): \[ 101 - 11 = 10 \]
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At this point, we have no more digits to bring down. So the remainder is \( 10 \).
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Final Result:
- The quotient is \( 10001_2 \) and the remainder is \( 10_2 \).
Putting it all together, we find: \[ 1101101_2 \div 11_2 = 10001_2 \text{ with a remainder of } 10_2 \]