Kingdom of Saudi Arabia
Ministry of Higher Education
Majma’ah University
College of Engineering
EE & CEN
Logic Design (CE 207, CE 213)
Chapter No. 1 – Part No. 1
Problem 1.1: List the Octal and Hexadecimal numbers from 16 to 32.
Sol:
Decimal (Base – 10):
16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32
Octal (Base – 8):
20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 37, 40
Hexadecimal (Base – 16):
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F, 20
Problem 1.2: What is the exact number of bytes in a system that contains?
a) 32 Kbytes
Sol:
As 1 Kilo (K) is referred as 210
32 Kbytes = 32 x 210 = 32 x 1024 = 32768 Bytes
b) 44 Mbytes
Sol:
As 1 Mega (M) is referred as 220
44 Mbytes= 64 x 220 = 32 x 1048576 = 67108864 Bytes
c) 6.4 Gbytes
Sol:
As 1 Giga (G) is referred as 230
6.4 Gbytes = 6.4 x 230 = 32 x 1073741824 = 6871947673.6 Bytes
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Problem 1.3: Convert the following numbers with the indicated base to decimal.
a) (4310)5
Sol:
(4310)5 = 4 x 53 + 3 x 52 + 1 x 51 + 0 x 50
(4310)5 = 4 x 125 + 3 x 25 + 1 x 5 + 0 x 1
(4310)5 = 500 + 75 + 5 + 0
(4310)5 = (580)10
b) (198)12
Sol:
(198)12 = 1 x 122 + 9 x121 + 8 x 120
(198)12 = 1 x 144 + 9 x12 + 8 x 1
(198)12 = 144 + 108 + 8
(198)12 = (260)10
c) (735)8
Sol:
(735)8 = 7 x 82 + 7 x 81 + 7 x 80
(735)8 = 7 x 64 + 7 x 8 + 7 x 1
(735)8 = 448 + 56 + 7
(735)8 = (511)10
d) (525)6
Sol:
(525)6 = 5 x 62 + 2 x 61 + 5 x 60
(525)6 = 5 x 36 + 2 x 6 + 5 x 1
(525)6 = 180 + 12 + 5
(525)6 = (197)10
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Problem 1.4: What is the largest number that can be expressed with 14 – bits?
What are the equivalent decimal and hexadecimal numbers?
Sol:
Largest 14 – bits binary number = 11111111111111
Decimal equivalent number = 214 – 1 = 16384 – 1 = (16383)10
Hexadecimal equivalent number = (11111111111111)2
= 0011 1111 1111 1111
= (3FFF)16
Problem 1.7: Convert the Hexadecimal number 68BE to Binary and then convert it
from Binary to Octal.
Sol:
Hexadecimal to Binary:
(68BE)16 =
6
8
B
E
= 0110 1000 1011 1110
= (110100010111110)2
Binary to Octal:
(110100010111110)2 = 110 100 010 111 110
= 6
4
2
7
6
= (64276)8
Problem 1.8: Convert the decimal number 431 to binary number in two ways.
a) Convert directly to Binary:
Sol:
Therefore (431)10 = (110101111)2
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b) Convert first to Hexadecimal and then from Hexadecimal to Binary:
Sol: Decimal to Hexadecimal:
Therefore (431)10 = (1AF)16
Hexadecimal to Binary:
(1AF)16 =
1
A
F
= 0001 1010 1111
= (10101111)2
Problem 1.9: Express the following numbers in decimals.
a) (10110.0101)2
Sol:
(10110.0101)2 = 1 x 24 + 0 x 23 + 1 x 22 + 1 x 21 + 0 x 20 + 0 x 2-1 + 1 x 2-2
+ 0 x 2-3 + 1 x 2-4
(10110.0101)2 = 1 x 16 + 0 + 1 x 4 + 1 x 2 + 0 + 0 + 1/4 + 0 + 1/16
(10110.0101)2 = 16 + 4 +2 + 0.25 + 0.0625
(10110.0101)2 = (22.3125)10
b) (16.5)16
Sol:
(16.5)16 = 1 x 161 + 6 x 160 + 5 x 16-1
(16.5)16 = 1 x 16 + 6 x 1 + 5 x 1/16
(16.5)16 = 16 + 6 + 0.3125
(16.5)16 = (22.3125)10
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c) (26.24)8
Sol:
(26.24)8 = 2 x 81 + 6 x 80 + 2 x 8-1 + 4 x 8-2
(26.24)8 = 2 x 8 + 6 x 1 + 2 x 1/8 + 4 x 1/64
(26.24)8 = 16 + 6 + 0.25 + 0.0625
(26.24)8 = (22.3125)10
d) (FAFA.B)16
Sol:
(FAFA.B)16 = 15 x 163 + 10 x 162 + 15 x 161 + 10 + 160 + 11 x 16-1
(FAFA.B)16 = 15 x 4096 + 10 x 256 + 15 x 16 + 10 + 1 + 11 x 1/16
(FAFA.B)16 = 61440 + 2560 + 240 + 10 + 0.6875
(FAFA.B)16 = (64250.6875)10
Problem 1.10: Convert the following numbers from Binary to Hexadecimal and
then from Hexadecimal to Decimal.
A. (1.10010)2
Sol: Binary to Hexadecimal:
(1.10010)2 = 0001.1001
(1.10010)2 = 1 . 9
(1.10010)2 = (1.9)16
Hexadecimal to Decimal:
(1.9)16 = 1 X 160 + 9 X 16-1 = 1 X 1 + 9 X 1/16 = 1 + 0.5625
(1.9)16 = (1.5625)10
B. (110.0101)2
Sol: Binary to Hexadecimal:
(110.0101)2 = 0110.0101
(110.0101)2 =
6 . 5
(110.0101)2 = (6.5)16
Hexadecimal to Decimal:
(6.5)16 = 6 X 160 + 5 X 16-1 = 6 X 1 + 5 X 1/16 = 6 + 0.3125
(6.4)16 = (6.3125)10
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Problem 1.13: Do the following conversions.
a) Convert Decimal 27.315 to Binary
Sol: Convert 27 and 0.315 separately from Decimal to Binary:
0.315 x 2 = 0.63  0
0.63 x 2 = 1.26  1
0.26 x 2 = 0.52  0
0.52 x 2 = 1.04  1
(27)10 = (11011)2
(27.315)10 = (11011.0101)2
(0.315)10 = (0.0101)2
b) Calculate the Binary equivalent of 2/3 up to eight places. Then convert from
Binary to Decimal. How close is the result to 2/3?
Sol: Decimal to Binary:
(2/3)10 = (0.66666666)10
0.66666666 x 2 = 1.33333333 1
0.33333333 x 2 = 0.66666666  0
0.66666666 x 2 = 1.33333333 1
0.33333333 x 2 = 0.66666666  0
0.66666666 x 2 = 1.33333333 1
0.33333333 x 2 = 0.66666666  0
0.66666666 x 2 = 1.33333333 1
0.33333333 x 2 = 0.66666666  0
(2/3)10 = (0.66666666)10 = (0.10101010)2
Binary to Decimal:
(0.10101010)2 = 1 x 2-1 + 0 x 2-2 + 1 x 2-3 + 0 x 2-4 + 1 x 2-5 + 0 x 2-6 + 1 x 2-7
+ 0 x 2-8
(0.10101010)2 = 1/2 + 0 + 1/8 + 0 + 1/32 + 0 + 1/128 + 0
(0.10101010)2 = 0.5 + 0.125 + 0.03125 + 0.0078125
(0.10101010)2 = (0.6640625)10
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c) Convert the Binary result in part (b) to Hexadecimal and then convert the
result to Decimal. Is the answer the same?
Sol: Binary to Hexadecimal:
(0.10101010)2 = 0.1010 1010
(0.10101010)2 =
A
A
(0.10101010)2 = (0.AA)16
Hexadecimal to Decimal:
(0.AA)16 = 10 x 16-1 + 10 x 16-2
(0.AA)16 = 10 x 1/16 + 10 x 1/256
(0.AA)16 = 10 x 0.0625 + 10 x 0.0039062
(0.AA)16 = 0.625 + 0.0390625
(0.AA)16 = (0.6640625)10
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