To divide the binary number \(10010001\) by \(101\), we can perform binary long division.
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Begin by setting up the division: Divide \(10010001\) (which is the dividend) by \(101\) (which is the divisor).
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Align the numbers:
- \(100\) is the first 3 bits of the dividend. Since \(100\) is less than \(101\) (which is \(5\) in decimal), we move to the next bit, which gives us \(1001\).
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Repeat the process:
- \(1001\) is the next section of the dividend. Compare \(1001\) (which is \(9\) in decimal) with \(101\).
- \(101\) goes into \(1001\) once. Write down \(1\).
- Subtract \(101\) from \(1001\):
\[ 1001 - 101 = 1000 \]
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Bring down the next bit:
- Bring down the next bit from \(10010001\)—this gives you \(10000\).
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Divide \(10000\) by \(101\):
- \(101\) (which is \(5\) in decimal) fits into \(10000\) (which is \(16\) in decimal) three times since \(3 \times 5 = 15\).
- Write down \(3\) (which is \(11\) in binary).
- Subtract:
\[ 10000 - 111 = 0011 \]
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Continue:
- Bring down the next bit, which results in \(00111\).
- \(101\) fits into \(00111\) (which is \(7\) in decimal) one time. Write down \(1\).
- Subtract \(101\) from \(00111\):
\[ 00111 - 00101 = 00010 \]
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Final bit:
- Bring down the last bit to get \(000100\).
- \(101\) cannot fit, so it's \(0\). Write down \(0\).
- Finally, we don't need any more bits, so the remainder is \(10\).
Now we can compile the quotient and remainder:
The final quotient is \(110011\) with a remainder of \(10\).
So,
\[ 10010001 \div 101 = 110011 \quad \text{(Quotient)}, \quad \text{Remainder: } 10 \]