# hw4sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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Math 55: Discrete Mathematics UC Berkeley, Spring 2009 Solutions to Homework # 4 (due February 18) § 3.6, #4 a) 1 1011 2 = 1 B 16 = 1 · 16 + 11 = 27 b) 10 1011 0101 2 = 2 B 5 16 = 2 · 16 2 + 11 · 16 + 5 = 693 c) 11 1011 1110 2 = 3 BE 16 = 3 · 16 2 + 11 · 16 + 14 = 958 d) 111 1100 0001 1111 2 = 7 C 1 F 16 = 7 · 16 3 +12 · 16 2 +1 · 16+15 = 31775 § 3.6 #5 a) 80 E 16 = 100 0000 1110 2 b) 135 AB 16 = 1 0011 0101 1010 1011 2 c) ABBA 16 = 1010 1011 1011 1010 2 d) DEFACED 16 = 1101 1110 1111 1010 1100 1110 1101 2 § 3.6 #8 a) 1111 0111 2 = F 5 16 b) 1010 1010 1010 2 = AAA 16 c) 111 0111 0111 0111 2 = 7777 16 § 3.6 #19 644 = 1010000100 2 so: i=0: a 0 = 0, x = 1, power = 7 2 MOD 645 = 49 i=1: a 1 = 0, x = 1, power = 49 2 MOD 645 = 466 i=2: a 2 = 1, x = 466, power = 466 2 MOD 645 = 436 i=3: a 3 = 0, x = 466, power = 436 2 MOD 645 = 466 i=4: a 4 = 0, x = 466, power = 466 2 MOD 645 = 436 i=5: a 5 = 0, x = 466, power = 436 2 MOD 645 = 466 i=6: a 6 = 0, x = 466, power = 466 2 MOD 645 = 436 i=7: a 7 = 1, x = 466 · 436 MOD 645 = 1, power = 436 2 MOD 645 = 466 1

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i=8: a 8 = 0, x = 1, power = 466 2 MOD 645 = 436 i=9: a 9 = 1, x = 436 § 3.6 #21 2003 = 11111010011 2 so: i=0: a 0 = 1, x = 3, power = 3 2 MOD 99 = 9 i=1: a 1 = 1, x = 3 · 9 MOD 99 = 27, power = 9 2 MOD 99 = 81 i=2: a 2 = 0, x = 27, power = 81 2 MOD 99 = 27 i=3: a 3 = 0, x = 27, power = 27 2 MOD 99 = 36 i=4: a 4 = 1, x = 27 · 36 MOD 99 = 81, power = 36 2 MOD 99 = 9 i=5: a 5 = 0, x = 81, power = 9 2 MOD 99 = 81 i=6: a 6 = 1, x = 81 · 81 MOD 99 = 27, power = 81 2 MOD 99 = 27 i=7: a 7 = 1, x = 27 · 27 MOD 99 = 36, power = 27 2 MOD 99 = 36 i=8: a 8 = 1, x = 36 · 36 MOD 99 = 9, power = 36 2 MOD 99 = 9 i=9: a 9 = 1, x = 9 · 9 MOD 36 = 81, power = 9 2 MOD 99 = 81 i=10: a 10 = 1, x = 81 · 81 MOD 99 = 27 § 3.6 #24 a) gcd (1 , 5) = gcd (1 , 5 MOD 1 = 0) = 1 b) gcd (101 , 100) = gcd (101 MOD 100 = 1 , 100) = gcd (1 , 100 MOD 1 = 0) = 1 c) gcd (277 , 123) = gcd (277 MOD 123 = 31 , 123) = gcd (31 , 123 MOD 31 = 30) = gcd (31 MOD 30 = 1 , 30) = gcd (1 , 30 MOD 1 = 0) = 1 d) gcd (14039 , 1529) = gcd (14039 MOD 1529 = 278 , 1529) = gcd (278 , 1529 MOD 278 = 139) = gcd (278 MOD 139 = 0 , 139) = 139 e) gcd (14038 , 1529) = gcd (14038
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## This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.

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hw4sol - Math 55: Discrete Mathematics UC Berkeley, Spring...

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