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92
Solutions to Exercises
8. [BB] Observe that 5°
==
1 (mod 7), 51
==
5 (mod 7), 52
==
4 (mod 7), 53
==
6 (mod 7), 54
==
2
(mod 7) and 55
==
3 (mod 7). Thus, each of the integers 1,2,3,4,5,6 is congruent mod 7 to some
power of 5. By Proposition 4.4.5, any integer
a
is congruent mod 7 to one of 0,1,2,3,4,5,6. So the
result follows.
9. (a) [BB] No
x
exists. The values of
3x
mod 6 are 0 and 3.
(b)
x
=
2,
x
=
5;
(c)
x
=
6.
(d) No
x
exists. The values of
4x
mod 6 are 0, 4 and 2.
(e) [BB] If 50 1
(2x
18), then
2x
18
=
50k,
so
x
=
9
+
25k
for some integer
k.
We obtain
x
=
9,
x
=
34.
(f) [BB] By trial and error,
x
=
9, or using the method of Problem 25, write
112(5)
=
1, deducing
that 2(5)
==
1 (mod 11) and so
x
=
2(1)
==
9 (mod 11).
(g)
If
25 1
(5x
 5), then
5x
 5
=
25k
so
x
=
1
+
5k
for some integer
k.
So
x
=
1,6,11,16,21.
(h) No
x
exists.
If
5921
(4x
 301), then
4x
 301 is even. This is impossible since, for any
x, 4x
is
even.
(i) No
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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, Integers

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