hw3sol - Fall 2010 Math 55 Homework 3 Solutions September...

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Unformatted text preview: Fall 2010 Math 55 Homework 3 Solutions September 22, 2010 3.4 9. Find the quotient and remainders (a) 19 = 2 * 7 + 5, so q = 2 ,r = 5 (b)- 111 =- 11 * 11 + 10, so q =- 11 ,r = 10 (c) 789 = 34 * 23 + 7, so q = 34 ,r = 7 (d) 1001 = 77 * 13 + 0, so q = 77 ,r = 0 (e) 0 = 0 * 19 + 0, so q = 0 ,r = 0 (f) 3 = 0 * 5 + 3, so q = 0 ,r = 3 (g)- 1 =- 1 * 3 + 2, so q =- 1 ,r = 2 (h) 4 = 4 * 1 + 0, so q = 4 ,r = 0 12. Show that a mod m = b mod m if a b ( mod m ) Were asked to show that a b (mod m ) a mod m = b mod m . Well do this with a direct proof. Assume a b (mod m ). Then by the definition of modular equivalence, we know m | ( a- b ). By the definition of divides, we can find an integer k such that a- b = mk . By applying the mod m operator to both sides and using Corollary 2 from p. 205, we get: a- b = mk a = b + mk a mod m = ( b + mk ) mod m a mod m = (( b mod m ) + ( mk mod m )) mod m a mod m = ( b mod m + 0) mod m a mod m = b mod m And thus we have the desired result. 16. Evaluate each quantity (a)- 17 mod 2 = 1 (- 17 =- 9 * 2 + 1) (b) 144 mod 7 = 4 (144 = 20 * 7 + 4) (c)- 101 mod 13 = 3 (- 101 =- 8 * 13 + 3) (d) 199 mod 19 = 9 (199 = 10 * 19 + 9) 21. Show that if n | m , then a b ( mod m ) a b ( mod n ) Well do this with a direct proof: Let n,m be integers such that n | m . Assume also that a b (mod m ). By the definition of modular equivalence, we know m | ( a- b ). Then since n | m , we can use theorem 1(iii) on p.202 to conclude n | ( a- b ). By definition, this means a b (mod n ). 31. Encrypt DO NOT PASS GO using f(p) = a) p + 3 mod 26 , b) p + 13 mod 26 , c) 3 p + 7 mod 26 This is essentially the same as example 9, but with different numbers. Replacing letters with numbers in DO NOT PASS GO gives 3 14 13 14 19 15 0 18 18 6 14. Applying each encryption function gives the following results: (a) 6 17 16 17 22 18 3 21 21 9 17 which is GR QRW SDVV JR (b) 16 1 0 1 6 2 13 5 5 19 1 which is QB ABG CNFF TB (c) 16 23 20 23 12 0 7 9 9 25 23 which is QX UXM AHJJ ZX 32. Decrypt each of the following messages To decrypt, you just need to apply the rule g ( p ) = p- 3 mod 26. This gives: (a) BLUE JEANS (b) TEST TODAY (c) EAT DIM SUM 3.5 2. Determine if each of the following numbers are prime One key here is that to test if a number n is prime, you only need to test if it has any prime factors less than n . (a) Yes (19 is not divisible by 2, 3) (b) No (27 = 3 3 ) (c) No (93 = 3 31) (d) Yes (101 is not divisble by 2, 3, 5, 7) (e) Yes (107 is not divisible by 2, 3, 5, 7) (f) Yes (113 is not divisible by 2, 3, 5, 7) 4. (a) 39 = 3 13 (b) 81 = 3 4 (c) 101 = 101 (it is prime, and thus its prime factorization is just itself) (d) 143 = 11 13 (e) 289 = 17 2 (f) 899 = 29 31 8. Prove that for every positive integer n , there are n consecutive composite numbers Let n be given. As per the hint, consider the n consecutive numbers beginning with ( n + 1)! + 2: (...
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hw3sol - Fall 2010 Math 55 Homework 3 Solutions September...

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