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Chapter 4
105
97
1
0
17
0
1
10. (a) We use the Euclidean algorithm:
12
1
5
5
1 6
Thus 1
= 7(97)
+
40(17).
2
3
17
1
7 40
(b) By part (a), 10
= 70(97)
+
400(17). Mod 97, this says 400(17)
(mod 97).
10, so
x
=
400
==
11. (a) We know the solution is unique since gcd 8,27 =
1.
A patient person might use trial and error to
get
x
= 17. Alternatively, using the method suggested by Problem 25, writing 1 = 3(27)  10(8)
shows that
x
= 10 ==
17 (mod 27).
(b) As in part (a), we know the solution is unique. Since 3(17)
==
1 (mod 27), 3(15)(17)
==
15
(mod 27), so
x
=
3(15)
=
45
==
18 (mod 27).
(c) We want 16
I
(4x
 8), so 4
I
(x
 2); that is,
x
 2
=
4k
for some
k.
The values of
x
(mod 16)
are
x
=
2,6, 14.
(d) There is no solution because the values of
4x
(mod 16) are 0, 4, and 8.
(e) We want 380
I
(6x

7), so
6x

7
=
380k
for some integer
k.
This says 7
=

is even,
which is not true. There is no solution.
(f)
We want 380
I

8), so

8
=
for some integer
k.
This is equivalent to
3x =
190k
+
4 = 4, 194, 384, 574, 764, 954,
... so
x
= 128 or
x
= 318.
12. (a) Since gcd(6, 25)
=
1, we know there exist integers
a
and
b
so that
6a
+
25b
=
In fact, we find
that 6( 4)
+
25(1) =
Multiplying by 13 gives 6( 52)
+
25(13) = 13, so take
Xo
= 52 and
Yo
=
13.
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 Summer '10
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 Graph Theory

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