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1
CHAPTER 28
Network Security
Exercises
1.
Substitute the character that is 4 characters down. For example, X is 4 characters
down from T and L is 4 characters down from H. The encrypted message is
3.
Using statistics, we can find
5.
a.
65
66
67
65
68
69
70
71
72
b.
0
1
2
0
3
4
5
6
7
c.
00000 00001 00010 00000 00011 00100 00101 00110 00111
7.
N
= 19
×
23 = 437
(
p
−
1)(
q
−
1) = 396
Select an
e
such that it is relatively prime to 396. An
e
= 5 satisfies the condition.
Now we need a
d
such that (5
×
d
) mod 396 = 1. Solve the equation 396
y
= 5
d
−
1,
with
y
an integer. We need a y that when multiplied by 396, the ones’ digit is 4 or 9
so that it is evenly divisible by 5. If y is 4, then d = 317.
XLMW MW E KSSH IBEQTPI
ENCRYPTION IS LIKE ENCLOSING A SECRET IN AN ENVELOPE
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2
9.
( 1
×
227) mod 100
= 27
(27
×
227) mod 100
= 29
(29
×
227) mod 100
= 83
(83
×
227) mod 100
= 41
(41
×
227) mod 100
=
7
( 7
×
227) mod 100
= 89
(89
×
227) mod 100
=
3
( 3
×
227) mod 100
= 81
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 Winter '10
 Dmathur

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