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MATH 11200/20 Final exam DECEMBER 9, 2016 ALAN CHANG Please present your solutions clearly and in an organized way Answer the questions in the space provided on the question sheets If you run out of room for an answer, continue on the back of the page Please note that use of a calculator is not allowed Good luck!! Full Name: Question Points Score 1 1 2 20 3 20 4 20 1 6 1 7 20 8 1 9 10 10 0 Total: 10 This exam has 10 questions, for a total of 10 points The maximum possible score for each problem is given on the right side of the problem Please do not open this exam packet unless you are instructed to do so

Math 11200/20 Final exam, Page 2 of 11 1 (a) Recall axiom M4: M4 (Multiplicative inverse) If a is any nonzero element of the set, then there is a unique corresponding element a 1 such that a a 1 = 1 and a 1 a = 1 Circle the sets below which satisfy axiom M4 (No justification needed) 2 3 4 6 7 8 9 10 (b) Recall the following theorem: 10 Let a, b, c If a bc and gcd(a, b) = 1, then a c Please use the theorem above to give a proof of Euclid s lemma, which is the following: Let p, x, y If p is a prime and p x y, then p x or p y

Math 11200/20 Final exam, Page 3 of 11 2 (a) Circle the numbers below that are relatively prime to 20: 0 1 2 3 4 6 7 8 9 10 11 12 13 14 1 16 17 18 19 Let φ(20) be the number of numbers you circled What is φ(20)? (b) Euler s theorem says that if gcd(x, 20) = 1, then x φ(20) 1 (mod 20) Check that 7 φ(20) 1 10 (mod 20) and 11 φ(20) 1 (mod 20) (c) What is the remainder when 7 1000 is divided by 20?

Math 11200/20 Final exam, Page 4 of 11 3 (a) Write down three different positive numbers which satisfy x 4 (mod 20) 2 (b) Write down a negative number which satisfies x 4 (mod 20) 2 (c) What are all x which satisfy both x 4 (mod 20) and x (mod 14)? 8 (d) What are all x which satisfy both x 4 (mod 20) and x (mod 13)? 8

Math 11200/20 Final exam, Page of 11 4 Let A = 1,120,021 (a) What is the remainder when A is divided by 4? (No justification needed) (b) What is the remainder when A is divided by 9? (No justification needed) Let B = 4 6 6 + 2 6 + 1 6 3 + 2 6 2 + 6 1 + 3 (c) What is the remainder when B is divided by 36? (d) What is the remainder when B is divided by?

Math 11200/20 Final exam, Page 6 of 11 (a) How many positive divisors does 100 have? 3 (b) How many positive divisors does 1,000,000,000 (one billion) have? 7 (c) How many positive divisors does 3,000,000,000 (three billion) have?

Math 11200/20 Final exam, Page 7 of 11 6 For this problem, use the following letter-number pairing letter A B C D E F G H I J K L M number 0 1 2 3 4 6 7 8 9 10 11 12 letter N O P Q R S T U V W X Y Z number 13 14 1 16 17 18 19 20 21 22 23 24 2 (a) I encrypted a message using the function f (x) = x + 12 (mod 26) The encrypted message is NKQ What is the original message? (b) Encrypt HI using the encryption function f (x) = x + 1 (mod 26) What is the decryption 10 function g? (Hint: The following calculations may be useful: 4 26 = 104 and 21 = 10)

Math 11200/20 Final exam, Page 8 of 11 7 Recall the RSA algorithm: Step 1: Bob chooses 2 distinct primes p and q He computes n = pq Step 2: Bob chooses e with gcd(e, (p 1)(q 1)) = 1 Step 3: Bob finds d with de 1 (mod (p 1)(q 1)) Step 4: Bob makes the two following numbers public: n and e (He keeps p, q, d secret) Step : The encryption function is f (x) = x e (mod n) Step 6: The decryption function is g(x) = x d (mod n) (a) In step 2, why does e need to satisfy gcd(e, (p 1)(q 1)) = 1? (Why can t Bob choose any e?) (b) In one short sentence, what makes RSA secure (in present times, at least)? (c) Now, suppose we do RSA with n = 77 and e = 11 What is the decryption function? 10

Math 11200/20 Final exam, Page 9 of 11 8 (a) Suppose a, b 8 Write what a b (in 8 ) means 2 (b) In 8, the statement if a 3 b 3, then a b is not true Please find a counterexample to the 6 statement (Hint: Choose b 8 so that b 3 = 0) (c) In, the statement if a 3 b 3, then a b is true What theorem from class can we use to 3 prove this? (Just state the name of the theorem) (d) Recall that using the axioms A1--A4, M1--M3, D, we can prove statements like if a b, then 4 a 2 b 2 Why is there no proof of the statement in (c) that uses only these axioms? (Recall that these axioms are: commutativity of addition, associativity of addition, additive identity, additive inverse, commutativity of multiplication, associativity of multiplication, multiplicative identity, distributive property)

Math 11200/20 Final exam, Page 10 of 11 9 Recall that 10 = {(x, y) x and y } That is, the set consists of all ordered pairs of natural numbers You can view this set as a grid of squares extending infinitely upwards and to the right: (1, 4) (2, 4) (3, 4) (4, 4) (1, 3) (2, 3) (3, 3) (4, 3) (1, 2) (2, 2) (3, 2) (4, 2) (1, 1) (2, 1) (3, 1) (4, 1) Is the set countable? Please justify

Math 11200/20 Final exam, Page 11 of 11 10 The set {1, 2, 3} has 8 subsets: 10 (bonus) {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} How many subsets does have? Let s show that it has uncountably many! Help me complete the following proof, which uses a variation of Cantor s diagonalization argument Step 1: Suppose for contradiction that has only countably many subsets Then we can list out all subsets of in a sequence S 1, S 2, S 3, (Each S i is a subset of ) For example: S 1 {3,, 7} S 2 S 3 {} (the empty set) S 4 {1, 4, 9, 16, 2, 36, } (the perfect squares) S {2, 4, 6, 8, 10, } (the even numbers) Step 2: Given a sequence S 1, S 2, S 3,, we define a new set T by T = {n n S n } In words: for each natural number n, we check to see if n is in S n If it is, then we include n in the set T If it is not, then we do not include n in the set T In our example above, 1 S 1, 2 S 2, 3 S 3, 4 S 4, S, so T contains 1, 3, but not 2, 4 Step 3: Using our set T, we get a contradiction How?