1. **Problem 1: Binary Addition** We need to add the binary numbers $11111101_2$ and $10101_2$. 2. **Align the numbers by their least significant bit:** $$\begin{array}{r} 11111101 \\ +00010101 \\ \hline \end{array}$$ 3. **Add bit by bit from right to left, carrying over when sum exceeds 1:** - $1 + 1 = 10_2$ (write 0, carry 1) - $0 + 0 + 1_{carry} = 1$ - $1 + 1 = 10_2$ (write 0, carry 1) - $1 + 0 + 1_{carry} = 10_2$ (write 0, carry 1) - $1 + 1 + 1_{carry} = 11_2$ (write 1, carry 1) - $1 + 0 + 1_{carry} = 10_2$ (write 0, carry 1) - $1 + 0 + 1_{carry} = 10_2$ (write 0, carry 1) - $1 + 0 + 1_{carry} = 10_2$ (write 0, carry 1) 4. **Write down the final carry:** The carry out is 1, so the result has an extra bit. 5. **Final sum:** $$11111101_2 + 10101_2 = 100010010_2$$ --- 6. **Problem 2: Binary Long Division** Divide $11111011_2$ (dividend) by $11110_2$ (divisor). 7. **Convert to decimal for clarity:** - $11111011_2 = 251_{10}$ - $11110_2 = 30_{10}$ 8. **Perform division in decimal:** $251 \div 30 = 8$ remainder $11$ 9. **Convert quotient and remainder back to binary:** - Quotient: $8_{10} = 1000_2$ - Remainder: $11_{10} = 1011_2$ 10. **Therefore:** $$11111011_2 \div 11110_2 = 1000_2 \text{ remainder } 1011_2$$