1. Answer to exercises(2) P74, 2.1 (a) 1101011 2 =6B 16 (b) 174003 8 =1 111 100 000 000 011 2 (d) 67.24 8 =110 111.010 1 2 (f) F3A5 8 =1111 0011 1010 0101 2 (i) 101111.0111 2 =57.34 8 (j) 15C.38 16 =1 0101 1100.0011 1 2 (e) 10100.1101 2 =14.D 16

2. Answer to exercises(2) P74, 2.2 (e) 5436.15 8 =101 100 011 110.001 101 2 =B1E.34 16 (f) 13705.207 8 =1 011 111 000 101.010 000 111 2 =17C5.438 16 (e) 9E36.7A 16 =1001 1110 0011 0110.0111 101 2 =117066.364 8 P74, 2.3 (d) C350 16 =1100 0011 0101 0000 2 =141520 8

3. Answer to exercises(2) P75, 2.4  12345670123 8 = 1 010 011 100 101 110 111 000 001 010 011 2 = 01 010 011 10 010 111 01 110 000 01 010 011 2  The octal values of the four 8-bit bytes are: 123 8 227 8 160 8 123 8 P75, 2.5 (d) 67.24 8 =6  8 1 +7  8 0 +2  8 -1 +4  8 –2 =55.3125 10 (e) 10100.1101 2 = 1  2 4 +1  2 2.1  2 -1 + 1  2 -2 + 1  2 -4 = 20.8125 10 (j) 15C.38 16 = 1  16 2 +5  16 1 +C  16 0.3  16 -1 +8  16 -2 = 348.21875 10

4. Answer to exercises(2) P75, 2.6 (a) 125 10 =? 2 125 2 62 (1 2 31 (0 2 15 (1 2 7 2 3 1 2 0 2 125 10 =1111101 2 (b) 3489 10 =? 8 3489 8 436 (1 8 54 (4 8 6 (6 8 0 3489 10 =6641 8 (g) 727 10 =? 5 727 10 =10402 5 727 5 145 (2 5 29 (0 5 5 (4 5 1 (0 5 0 (1

5. Answer to exercises(2) P76, 2.15 61453 10 =F00D 16 So the result “FOOD” can whet your appetite. 0 61453 16 3840 (D 16 240 (0 16 15 (0 16 (F P73, 2.31 There are 28 different 3-bit binary encodings are possible for the traffic-light controller of Table 2-12.