ELEC 2200 HW 1 Solutions Spring 2018 Solution: A + B: A -­‐ B: A × B: A ÷ B: 10101 10101 10101 1 + 1011 -­‐ 1011 × 1011 1011 10101 100000 01010 10101 1011 10101 1010 remainder 00000 10101 11100111 Solution: A + B: A -­‐ B: A × B: A ÷ B: 372 372 372 2 + 156 -­‐ 156 × 156 156 372 550 214 2734 334 2342 36 remainder 372 65554 Solution: A + B: A -­‐ B: A × B: A ÷ B: 2CF3 2CF3 2CF3 10B + 2B -­‐ 2B × 2B 2B 2CF3 2D1E 2CC8 1EE71 2B 59E6 1F3 78CD1 1D9 1A remainder Solution: Binary: Octal: Hexadecimal: 2 174 (010101110.010)₂ (10101110.0100)₂ 2 87 0 2 5 6. 2 A E . 4 2 43 1 2 21 1 2 10 1 2 5 0 2 2 1 1 0 0.25 × 2=0.5 0.5 × 2=1 (174.25)₁₀= (10101110.01)₂ Solution: 1 5. C Hexadecimal 00010 101. 1100 Binary 2 5. 6 Octal 15.C = 1×161＋ 5×16⁰ ＋ C×16-­‐1=1×16 + 5×1 + 12×0.0625 = (21.75)₁₀ Decimal Solution: F 8. A 7 01111 1000. 1010 01110 Binary 3 7 0. 5 1 6 Octal F8.A7 =F×161 ＋ 8×16⁰ ＋ A×16-­‐1 ﹢ 7×16-­‐2 = 15×16 + 8×1 + 10×0.0625 + 7×0.00390625 = (248.65234375)₁₀ Decimal