To divide the binary number \(111011101_2\) by \(1001_2\), we can use long division in binary.

Let's walk through the steps:

1. **Write the numbers**:
   \[
   111011101_2 \div 1001_2
   \]
   
2. **Perform binary long division**:
   - Start with the first four digits of \(111011101_2\), which is \(1110_2\). Compare it to \(1001_2\):
     \[
     1110_2 > 1001_2
     \]
     So, subtract \(1001_2\) from \(1110_2\):
     \[
     1110_2 - 1001_2 = 0111_2
     \]
     Write down a \(1\) in the quotient.

   - Now, bring down the next digit of \(111011101_2\), which gives us \(01111_2\). Compare it to \(1001_2\):
     \[
     01111_2 > 1001_2
     \]
     Subtract \(1001_2\) from \(01111_2\):
     \[
     01111_2 - 1001_2 = 00110_2
     \]
     Write down a \(1\) in the quotient.

   - Bring down the next digit, giving \(001101_2\). Compare it to \(1001_2\):
     \[
     001101_2 < 1001_2
     \]
     Bring down another digit, making it \(0011011_2\), which is still less than \(1001_2\), so add a \(0\) to the quotient.

   - Continue the division by bringing down the next digits, subtracting \(1001_2\) wherever possible, and writing the quotient digits accordingly.

After completing the division, the result is:

\[
111011101_2 \div 1001_2 = 1110_2 \text{ with a remainder of } 101_2
\]

In decimal terms:
- \(1110_2 = 14_{10}\)
- \(101_2 = 5_{10}\)

Thus, the quotient is \(1110_2\) and the remainder is \(101_2\).

---

Would you like more details on any step, or do you have any other questions?

Here are five questions to expand on this:

1. How is binary long division different from decimal division?
2. Can you divide binary numbers with larger divisors?
3. What are common uses for binary arithmetic in computer science?
4. How do you convert binary results back to decimal efficiently?
5. What are binary shifts and how are they used in division?

**Tip**: Always verify your binary division by multiplying the quotient by the divisor and adding the remainder.

Learn how to perform binary division with a detailed example of 111011101 ÷ 1001. This solution walks through binary long division, showing the quotient and remainder. Key topics include binary arithmetic and subtraction, suitable for Grades 9-12.

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