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Chapter 4
(c) [BB] AIR corresponds to the number 10918.
E
=
M
5
(mod 391)
=
259.
(d) BYE corresponds to the number 22505.
E
=
M
5
(mod 391) = 97.
(e)
NOW
corresponds to the number 141523.
E
=
M5
(mod 391) = 104.
24.
Since 1 = 1(5)
1(4) and 1 = 5(5)  4(6) we have
a
=
1,
b
= 5 in the notation of the text.
103
(a) [BB]
E
a
(mod
p)
is 31
1
(mod 5) = 1;
Eb
(mod
q)
is 31
5
(mod 7) = 5. To solve
M
==
1
(mod 5),
M
==
5 (mod 7), we write 1 = 10(5) 
7(7) and get
M
= 5(10)(5)  1(7)(7) =
201
==
26 (mod 35). So the message was the letter
Z.
(b)
Ea
(mod
p)
is 241 (mod 5) = 4;
Eb
(mod
q)
is 24
5
(mod 7) = 5. To solve
M
==
4 (mod 5),
M
==
5 (mod 7), we write 1 = 10(5) 
7(7) and get
M
= 5(10)(5)  4(7)(7) = 54 == 19
(mod 35). So the message was the letter S.
(c) [BB]
E
a
(mod
p)
is 7
1
(mod 5) = 2;
Eb
(mod
q)
is 7
5
(mod 7) = O. To solve
M
==
2
(mod 5),
M
==
0 (mod 7), we write 1 = 10(5) 
7(7) and get
M
= 0(10)(5)  2(7)(7) =
98
==
7 (mod 35). So the message was the letter G.
(d)
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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