SolutionCh1-10v2.

qxd 11/26/07 3:45 PM Page 1

Solutions Manual

CHAPTER 1
1.1 Base-10: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
Octal: 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 40
Hex: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20
Base-13 A B C 10 11 12 13 14 15 16 17 18 19 23 24 25 26
1.2 (a) 32,768 (b) 6,871,947,674
1.3 (4310)5 = 4 * 5 + 3 * 52 + 1 * 51 = 58010
3

(735)8 = 7 * 82 + 3 * 81 + 5 * 80 = 47710
1.4 14-bit binary: 11_1111_1111_1111
Decimal: 214 - 1 = 16,38310
Hexadecimal: 3FFF16
1.5 Let b = base
(a) 14/2 = (b + 4)/2 = 5, so b = 6
(b) (2 * b + 4) + (b + 7) = 4b, so b = 11
1.6 (x - 3)(x - 6) = x 2 - (6 + 3)x + 6 * 3 = x 2 - 11x + 22
Therefore: 6 + 3 = b + 1m so b = 8
Also, 6 * 3 = (18)10 = (22)8
1.7 68BE = 0110_1000_1011_1110 = 110_100_010_111_110 = (64276)8

1
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1.8 (a) Results of repeated division by 2 (quotients are followed by remainders):

43110 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
Answer: 1111_10102 = FA 16
(b) Results of repeated division by 16:ß
43110 = 26(15); 1(10) (Faster)
Answer: FA = 1111_1010
1.9 (a) 10110.01012 = 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
(c) 1010.10102 = 8 + 2 + .5 + .125 = 10.625
1.10 (a) 1.100102 = 0001.10012 = 1.916 = 1 + 9/16 = 1.56310
(b) 110.0102 = 0110.01002 = 6.416 = 6 + 4/16 = 6.2510
Reason: 110.0102 is the same as 1.100102 shifted to the left by two places.

1011.11
101ƒ111011.0000
1.11 101
01001
101
1001
101
1000
101
0110
The quotient is carried to two decimal places, giving 1011.11
Checking: 1110112/1012 = 5910/510
1011.112 = 58.7510
1.12 (a) 10000 and 110111
1011
+101 1011
10000 = 1610 *101
1011
1011
110111 = 5510
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1.13 (a) Convert 27.315 to binary:

Integer Remainder Coefficient
Quotient
1
27/2 = 13 + 2 a0 = 1
1
13/2 6 + 2 a1 = 1
6/2 3 + 0 a2 = 0
1
3/2 1 + 2 a3 = 1
1 1
2 0 + 2 a4 = 1
2710 = 110112
Integer Fraction Coefficient
.315 * 2 = 0 + .630 a -1 = 0
.630 * 2 = 1 + .26 a -2 = 1
.26 * 2 = 0 + .52 a -3 = 0
.52 * 2 = 1 + .04 a -4 = 1
.31510
.01012 = .25 + .0625 = .3125
27.315
11011.01012
(b) 2/3
.6666666667
Integer Fraction Coefficient
.6666_6666_67 * 2 = 1 + .3333_3333_34 a -1 = 1
.3333333334 * 2 = 0 + .6666666668 a -2 = 0
.6666666668 * 2 = 1 + .3333333336 a -3 = 1
.3333333336 * 2 = 0 + .6666666672 a -4 = 0
.6666666672 * 2 = 1 + .3333333344 a -5 = 1
.3333333344 * 2 = 0 + .6666666688 a -6 = 0
.6666666688 * 2 = 1 + .3333333376 a -7 = 1
.3333333376 * 2 = 0 + .6666666752 a -8 = 0
.666666666710
.101010102 = .5 + .125 + .0313 + .0078 = .664110
.101010102 = .1010_10102 = .AA 16 = 10/16 + 10/256 = .664110 (Same as (b)).
1.14 (a) 1000_0000 (b) 1101_1010 (c) 1000_0101
1s comp: 0111_1111 1s comp: 0010_0101 1s comp: 0111_1010
2s comp: 1000_0000 2s comp: 0010_0110 2s comp: 0111_1011
1.15 (a) 52,784,630 (b) 25,000,000
9s comp: 47,215,369 9s comp: 74,999,999
10s comp: 47,215,370 10s comp: 75,000,000
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4 Solutions Manual

1.16 B2FA B2FA: 1011_0010_1111_1010

15s comp: 4D05 1s comp: 0100_1101_0000_0101
1.17 (a) 3409 : 03409 : 96590 (9s comp) : 96591 (10s comp)
06428 - 03409 = 06428 + 96591 = 03019
(b) 6152 : 06152 : 93847 (9s comp) : 93848 (10s comp)
2043 - 6152 = 02043 + 93848 = 95891 (Negative)
Magnitude: 4109
Result: 2043 - 6152 = -4109
1.18 Note: Consider sign extension with 2s complement arithmetic.

(a) 10001
1s comp: 01110
2s comp: 01111
10011
Diff: 00010
101000
(b) 1s comp: 1010111
2s comp: 1011000
001001
Diff: 1100001 (negative)
0011111 (2s comp)
-011111 (diff is -31)
1.19 +9286 : 009286; +801 : 000801; -9286 : 990714; -801 : 999199
(a) (+9286) + (-801) = 009286 + 000801 = 010087
(b) (-9286) + (+801) = 990714 + 000801 = 991515
1.20 +49 : 0_110001 (Needs leading zero indicate + value); +29 : 0_011101
Leading 0 indicates + value) -49 : 1_001111; -29 : 1_100011
(a) (+29) + (-49) = 0_011101 + 1_001111 = 1_101100 (1 indicates negative value.)
Magnitude = 0_010100; Result (+29) + (-49) = -20
(b) (-29) + (+49) = 1_100011 + 0_110001 = 0_010100 (0 indicates positive value)
(-29) + (+49) = +20
(c) Must increase word size by 1 (sign extension) to accomodate overflow of values:
(-29) + (-49) = 11_100011 + 11_001111 = 10_110010 (1 indicates negative
result)
Magnitude: 1_001110 = 7810
Result: (-29) + (-49) = -78
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Solutions Manual 5

1.21 +9742 : 009742 : 990257 (9’s comp) : 990258 (10s) comp

+641 : 000641 : 999358 (9’s comp) : 999359 (10s) comp
(a) (+9742) + (+641) : 010383
(b) (-9742) + (+641) = 990258 + 000641 = 990899 (negative)
Magnitude: 009101
Result: (-9742) + (641) = -9101
1.22 8,723
BCD: 1000_0111_0010_0011
ASCII: 0_011_1000_011_0111_011_0010_011_0001
1.23 1000 0100 0010 (842)
0101 0011 0111 (+537)
1101 0111 1001
0110
0001 0011 0111 0101 (1,379)
1.24
(a)
6 3 1 1 Decimal
0 0 0 0 0
0 0 0 1 1
0 0 1 0 2
0 1 0 0 3
0 1 1 0 4 (or 0101)
0 1 1 1 5
1 0 0 0 6
1 0 1 0 7 (or 1001)
1 0 1 1 8
1 1 0 0 9
1.25 (a) 5,13710 BCD: 0101_0011_0111
(b) Excess-3: 1000_0100_0110_1010
(c) 2421: 1011_0001_0011_0111
(d) 6311: 0111_0001_0100_1001
1.26 5,137 9s Comp: 4,862
2421 code: 0100_1110_1100_1000
1s comp: 1011_0001_0011_0111 same as (c) in 1.25
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1.27 For a deck with 52 cards, we need 6 bits (32 6 52 6 64). Let the msb’s select the suit
(e.g., diamonds, hearts, clubs, spades are encoded respectively as 00, 01, 10, and 11. The
remaining four bits select the “number” of the card. Example: 0001 (ace) through 1011
(9), plus 101 through 1100 (jack, queen, king). This a jack of spades might be coded as
11_1010. (Note: only 52 out of 64 patterns are used.)
1.28 G (dot) (space) B o o l e
01000111_11101111_01101000_01101110_00100000_11000100_11101111_11100101
1.29 Bill Gates
1.30 73 F4 E5 76 E5 4A EF 62 73
73: 0_111_0011 s
F4: 1_111_0100 t
E5: 1_110_0101 e
76: 0_111_0110 v
E5: 1_110_0101 e
4A: 0_100_1010 j
EF: 1_110_1111 o
62: 0_110_0010 b
73: 0_111_0011 s
1.31 62 + 32 = 94 printing characters
1.32 bit 6 from the right
1.33 (a) 897 (b) 871
1.34 ASCII for decimal digits with odd parity:
(0): 10110000 (1): 00110001 (2): 00110010 (3): 10110011
(4): 00110100 (5): 10110101 (6): 10110110 (7): 00110111
(8): 00111000 (9): 10111001
1.35 (a)

a b c
a
f
b

c
g
f

g
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Solutions Manual 7

1.36

a b
a
f
b

g
f

CHAPTER 2
2.1
(a)

xyz x + y + z 1x + y + z2¿ x¿ y¿ z¿ x¿y¿z¿ xyz (xyz) 1xyz2¿ x¿ y¿ z¿ x¿ + y¿ + z¿

000 0 1 1 1 1 1 000 0 1 1 1 1 1
001 1 0 1 1 0 0 001 0 1 1 1 0 1
010 1 0 1 0 1 0 010 0 1 1 0 1 1
011 1 0 1 0 0 0 011 0 1 1 0 0 1
100 1 0 0 1 1 0 100 0 1 0 1 1 1
101 1 0 0 1 0 0 101 0 1 0 1 0 1
110 1 0 0 0 1 0 110 0 1 0 0 1 1
111 1 0 0 0 0 0 111 1 0 0 0 0 0

(b) (c)

xyz x1y + z2 xy xz xy + xz xyz yz x(yz) xy (xy)z

000 0 0 0 0 000 0 0 0 0
001 0 0 0 0 001 0 0 0 0
010 0 0 0 0 010 0 0 0 0
011 0 0 0 0 011 1 0 0 0
100 0 0 0 0 100 0 0 0 0
101 1 0 1 1 101 0 0 0 0
110 1 1 0 1 110 0 0 1 0
111 1 1 1 1 111 1 1 1 1