Discrete Mathematics with Graph Theory (3rd Edition) 106

Discrete Mathematics with Graph Theory (3rd Edition) 106 -...

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104 Solutions to Review Exercises 3. Inbaseb, 31 = 1+3b, 156 = 6+5b+b 2 and 220 = 2b+2b 2 . The equation (1+3b)+(6+5b+b 2 ) = 2b + 2b 2 implies 7 + 8b + b 2 = 2b + 2b 2 , so b 2 - 6b - 7 = 0 = (b - 7)(b + 1). Buddy is correct in base b = 7. 4. The integer n has prime divisors 2, 3 and 7, so n = 2(3)(7)k for some integer k. Since 2(3)(7) has eight factors-I, 2, 3, 7, 6, 14, 21, 42-k = 1 and n = 42. 5. The given integer is 1 + 2b + 3b 2 + 4b 3 + 5b 4 + 6b 5 + 7b 6 + 8b 7 + 9b B = (1 + 2b + 4b 3 + 6b 5 + 8b 7 ) + (3b 2 + 5b 4 + 7b 6 + 9b B ), the sum of an odd and an even integer. The given integer is never even. 6. (a) Since 8215 = 5 + 1(10) + 2(100) + 8(1000), the stated "observation"-popularly called "casting out 9's"-follows from Proposition 4.4.7 and the fact that any power of 10 is == 1 (mod 9). This holds for any integer, because (an -la n -2 ... alaoho is just ao+ lOal + 10 2 a2+· . ·+10 n - l an_l. 7.
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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.

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