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Section 5.1
107
26.
341
1
0
Thus 1 = gcd(341, 189) = 46(341) + 83(189). So
189
0
1
(5)(83)(189) + 11( 46)(341) = 94111. Since
152
1
1 (341)(189) = 64449, the answer is
37
1 2 94111 + 2(64449) = 34787.
4
5
9
1
46 83
27. We use the method described in Problem 29. First, we solve the first two congruences. We have
3(5)  2(7) = 1, so
x
= 5(3)5  3(2)7 = 33. Now we solve
x
==
33 (mod 35) and
x
==
7 (mod 11).
We have 5(35) + 16(11)
=
1, so
x
= 7( 5)(35) + 33(16)(11) = 4583. Since 5(7)(11) = 385, the
smallest positive integer is 4583 (mod 385) = 348.
Exercises 5.1
1. (a) [BB]
2:~=1
i
2
=
12 + 22 + 3
2
+ 4
2
+ 52 = 1 + 4 + 9 + 16 + 25 = 55
(b) [BB]
2:t=l
2i
=
21 + 22 + 2
3
+ 24 = 2 + 4 + 8 + 16 = 30
(c) [BB]
2:i=l
sin nt = sin
n(l)
= sin n = 0
(d)
2:~=0
3)+2 = 30+2 + 31+2 + 32+2 = 3
2
+ 3
3
+ 3
4
= 9 + 27 +
81
= 117
(e)
2:!=_1(2k
2

k
+ 1) = [2(1)2  (1) + 1] + [(2(0
2
)  0 + 1]
+[2(12)
1 + 1] + [2(22)  2 + 1] + [2(3
2
)  3 + 1] + [2(4
2
)  4 + 1]
=
4 + 1 + 2 + 7 + 16 + 29
=
59
(0
2:~=o(_I)k
= (_1)0 + (_1)1 + (1)2 + (_1)3 +
... +
(_I)n
= 1  1 + 1  1 +
... + ( 1)
n
= 0 if
n
is odd and 1 if
n
is even.
2. (a) [BB] The values
of2:~=l
(_I)i forn = 0, 1, 2, 3 are
1,11 = 0,
11+1 = 1 and
11+11 =
O. The answer is 0, 1.
(b) The values of
2:~1
2i for
n
= 1,2,3,4,5 are 2,2 + 4 = 6,2 + 4 + 8 = 14,2 + 4 + 8 + 16 =
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 Summer '10
 any
 Graph Theory

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