Question
In: Computer Science
Using the below given ASCII table (lowercase letters) convert the sentence “welcome to cci college” into...
Using the below given ASCII table (lowercase letters) convert the sentence “welcome to cci college” into binary values.
|
a - 97 |
b - 98 |
c - 99 |
d - 100 |
e - 101 |
f - 102 |
g - 103 |
h - 104 |
i - 105 |
|
j - 106 |
k - 107 |
l - 108 |
m - 109 |
n - 110 |
o - 111 |
p - 112 |
q - 113 |
r -114 |
|
s -115 |
t - 116 |
u - 117 |
v - 118 |
w - 119 |
x - 120 |
y - 121 |
z - 122 |
Space - 32 |
Solutions
Expert Solution
Given Sentence is
“welcome to cci college”
Based on the given text we have to find the binary values for
w,e,l,c,o,m,t,i,g,space
Converting “w” to Binary using its ASCII value – 119
|
Division |
Quotient |
Remainder |
Bit position |
|
119/2 |
59 |
1 |
0 |
|
59/2 |
29 |
1 |
1 |
|
29/2 |
14 |
1 |
2 |
|
14/2 |
7 |
0 |
3 |
|
7/2 |
3 |
1 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(119)10 = (01110111)2
Converting “e” to Binary using its ASCII value – 101
|
Division |
Quotient |
Remainder |
Bit position |
|
101/2 |
50 |
1 |
0 |
|
50/2 |
25 |
0 |
1 |
|
25/2 |
12 |
1 |
2 |
|
12/2 |
6 |
0 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(101)10 = (01100101)2
Converting “l” to Binary using its ASCII value – 108
|
Division |
Quotient |
Remainder |
Bit position |
|
108/2 |
54 |
0 |
0 |
|
54/2 |
27 |
0 |
1 |
|
27/2 |
13 |
1 |
2 |
|
13/2 |
6 |
1 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(108)10 = (01101100)2
Converting “c” to Binary using its ASCII value – 99
|
Division |
Quotient |
Remainder |
Bit position |
|
99/2 |
49 |
1 |
0 |
|
49/2 |
24 |
1 |
1 |
|
24/2 |
12 |
0 |
2 |
|
12/2 |
6 |
0 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(99)10 = (01100011)2
Converting “o” to Binary using its ASCII value – 111
|
Division |
Quotient |
Remainder |
Bit position |
|
111/2 |
55 |
1 |
0 |
|
55/2 |
27 |
1 |
1 |
|
27/2 |
13 |
1 |
2 |
|
13/2 |
6 |
1 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(111)10 = (01101111)2
Converting “m” to Binary using its ASCII value – 109
|
Division |
Quotient |
Remainder |
Bit position |
|
109/2 |
54 |
1 |
0 |
|
54/2 |
27 |
0 |
1 |
|
27/2 |
13 |
1 |
2 |
|
13/2 |
6 |
1 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(109)10 = (01101101)2
Converting “t” to Binary using its ASCII value – 116
|
Division |
Quotient |
Remainder |
Bit position |
|
116/2 |
58 |
0 |
0 |
|
58/2 |
29 |
0 |
1 |
|
29/2 |
14 |
1 |
2 |
|
14/2 |
7 |
0 |
3 |
|
7/2 |
3 |
1 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(116)10 = (01110100)2
Converting “i” to Binary using its ASCII value – 105
|
Division |
Quotient |
Remainder |
Bit position |
|
105/2 |
52 |
1 |
0 |
|
52/2 |
26 |
0 |
1 |
|
26/2 |
13 |
0 |
2 |
|
13/2 |
6 |
1 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(105)10 = (01101001)2
Converting “g” to Binary using its ASCII value – 103
|
Division |
Quotient |
Remainder |
Bit position |
|
103/2 |
51 |
1 |
0 |
|
51/2 |
25 |
1 |
1 |
|
25/2 |
12 |
1 |
2 |
|
12/2 |
6 |
0 |
3 |
|
6/2 |
3 |
0 |
4 |
|
3/2 |
1 |
1 |
5 |
|
1/2 |
0 |
1 |
6 |
(103)10 = (01100111)2
Converting “space” to Binary using its ASCII value – 32
|
Division |
Quotient |
Remainder |
Bit position |
|
32/2 |
16 |
0 |
0 |
|
16/2 |
8 |
0 |
1 |
|
8/2 |
4 |
0 |
2 |
|
4/2 |
2 |
0 |
3 |
|
2/2 |
1 |
0 |
4 |
|
1/2 |
0 |
1 |
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the
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a - 97
b - 98
c - 99
d - 100
e - 101
f - 102
g - 103
h - 104
i - 105
j - 106
k - 107
l - 108
m - 109
n - 110
o - 111
p - 112
q - 113
r -114
s -115
t - 116
u - 117...
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