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100
Solutions to Exercises
(c) Suppose the digits
at
and
aj
of al
...
ak
are transposed. (Without loss of generality, assume
i
<
j.)
Let a
=
(al,""
ai, .
.. , aj,'" ak)
and a'
=
aj,"" ai,'" ak).
Ifw· a'
=
w· a,
then
Wiaj
+
Wjai
=
Wiai
+
wjaj,
so
(Wi

wj)(aj

ai)
==
0
(mod
n).
The assumption tells us
that
==
ai
n)
and, since
n
>
9, we conclude that
=
ai.
18. (a) [BB] 1
=
(4)5+3(7), so
x
=
4(4)(5) +3(3)(7)
=
17
==
18 (mod 35). Therefore,
x
=
18.
(b) 1
=
(2)4 + 1(9), so
x
=
8(2)(4) + 1(1)(9)
=
55
==
17 (mod 36). Therefore,
x
=
17.
(c) 1
=
(3)5 + 2(8), so
x
=
7( 3)(5) + 3(2)(8)
=
57
==
23 (mod 40). Therefore,
x
=
23.
(d) [BB] 1
=
(3)8 + 1(25), so
x
=
17( 3)(8) + 6(1)(25)
=
258
==
142 (mod 200). Therefore,
x
=
142.
(e) 1
=
48(1917) 
239(385), so
x
=
75(48)(1917) 
3(239)(385)
=
6,625,155
==
720,795
(mod 738,045). Therefore,
x
=
720,795.
(f)
1
=
2647(17369) + 8402(5472), so
x
=
2974(2647)(17369)"+ 1003(8402)(5472)
=
46,113,671,232 
136,731,859,682
=
90,618,188,450
Then, since 90,618,188,450
=
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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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