Transcript Preface

Answer to exercises(2)
P74, 2.1
(a) 11010112=6B16
(b) 1740038=1 111 100 000 000 0112
(d) 67.248=110 111.010 12
(e) 10100.11012=14.D16
(f) F3A58=1111 0011 1010 01012
(i) 101111.01112=57.348
(j) 15C.3816=1 0101 1100.0011 12
Answer to exercises(2)
P74, 2.2
(e) 5436.158=101 100 011 110.001 1012 =B1E.3416
(f) 13705.2078=1 011 111 000 101.010 000 1112
=17C5.43816
P74, 2.3
(d) C35016=1100 0011 0101 00002 =1415208
(e) 9E36.7A16=1001 1110 0011 0110 .0111 1012
=117066.3648
Answer to exercises(2)
P75, 2.4
123456701238
= 1 010 011 100 101 110 111 000 001 010 0112
= 01 010 011 10 010 111 01 110 000 01 010 0112
 The octal values of the four 8-bit bytes are:
1238 2278 1608 1238
P75, 2.5
(d) 67.248=681+780+28-1+48 –2=55.312510
(e) 10100.11012= 124+122.12-1+ 12-2+ 12-4
= 20.812510
(j) 15C.3816= 1162+5161+C160.316-1+816-2
= 348.2187510
Answer to exercises(2)
P75, 2.6
(a) 12510=?2
125
2
62
2
31
2
15
2
7
2
2
3
2 1
0
(b) 348910=?8
(g) 72710=?5
727 (2
(1 8
3489 (1 5
145 (0
436 (4 5
(0 8
54 (6
(1
5 29 (4
8
(1
5 5 (0
8 6 (6
5 1 (1
(1
0
0
(1
348910=66418
(1
72710=104025
12510=11111012
Answer to exercises(2)
P76, 2.15
16 61453 (D
16 3840 (0
16 240 (0
16 15 (F
0
6145310 =F00D16
P73, 2.31
n  C68
8!

 28
6!(8  6)!
There are 28 different
3-bit binary encodings
So the result “FOOD” are possible for the
traffic-light
controller
can whet your appetite.
of Table 2-12.