Discrete Mathematics with Graph Theory (3rd Edition) 109

Discrete Mathematics with Graph Theory (3rd Edition) 109 -...

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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 1,1-1 = 0, 1-1+1 = 1 and 1-1+1-1 = 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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