1. **Problem i:** Perform the binary subtraction $10000010_2 - 01010111_2$. 2. Convert both binary numbers to decimal to understand the operation: $$10000010_2 = 1\times2^7 + 0 + 0 + 0 + 0 + 0 + 1\times2^1 + 0 = 128 + 2 = 130$$ $$01010111_2 = 0 + 1\times2^6 + 0 + 1\times2^4 + 0 + 1\times2^2 + 1\times2^1 + 1 = 64 + 16 + 4 + 2 + 1 = 87$$ 3. Subtract in decimal: $130 - 87 = 43$. 4. Convert $43$ back to binary: $$43 = 32 + 8 + 2 + 1 = 2^5 + 2^3 + 2^1 + 2^0$$ So, $$43_{10} = 00101011_2$$ 5. **Answer for i:** $$10000010_2 - 01010111_2 = 00101011_2$$ --- 6. **Problem ii:** Divide binary number $11010101_2$ by $101_2$ using long division. 7. Convert divisor and dividend to decimal for clarity: $$11010101_2 = 213_{10}$$ $$101_2 = 5_{10}$$ 8. Perform decimal division: $$213 \div 5 = 42 \text{ remainder } 3$$ 9. Now perform binary long division step-by-step: - Divisor: $101_2$ (5 decimal) - Dividend: $11010101_2$ 10. Align divisor under the leftmost bits of dividend and subtract if possible, shifting right each time. 11. The quotient in binary is $101010_2$ (which is 42 decimal), and the remainder is $11_2$ (which is 3 decimal). 12. **Answer for ii:** Quotient = $101010_2$ Remainder = $11_2$ --- 13. **Problem iii:** Divide $1011010_2$ by $10_2$. 14. Dividing by $10_2$ (which is 2 decimal) in binary is equivalent to shifting right by 1 bit. 15. So, $$1011010_2 \div 10_2 = 101101_2$$ 16. The remainder is the last bit shifted out, which is $0$. 17. **Answer for iii:** Quotient = $101101_2$ Remainder = $0$