PLEASE DO ANY 6 of the following 8 problems. Work carefully

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1 Ma2201/CS2022 Discrete Mathematics D Term, 2015 FINAL EXAM PRINT NAME: SIGN : PLEASE DO ANY 6 of the following 8 problems. Work carefully and neatly. 1. (5 pts) A St. Patricks Day deal at Mike Tattoo Parlor costs $2000 and lets the customer pick either a leprechaun, a mug of beer, or a harp, together with up to 5 little green shamrocks, and up to 5 little orange shamrocks. How many different types of tattoos could be made? The O Grady brothers decide to all get tattooed. Sean goes first and has decided to definitely get a Harp, and insists that he has more green shamrocks than orange shamrocks. Sean s little brother Brian goes second and wants to have more orange shamrocks than green shamrocks. Kevin, the littlest, goes last and doesn t want his tattoo to match either of his brothers. According to the multiplicative principle, circle those of the following which are true. a) Kevin has more choices than Sean. b) Brian has 3 (6 2) choices. c) Kevin has 3 (6!) 2 2 choices. d) Kevin has choices e) Kevin must pick an equal number of green and orange shamrocks. Solution: Sean has no choice for main tattoo, and hand to pick to different numbers from {0, 1, 2, 3, 4, 5} for the orange and green shamrocks, so the multiplicative principle gives 1 (6 2) choices. Brian has three choices for large tattoo, and also needs two distinct numbers. The larger will just be applied to the other color. So by the multiplicative principle 3 (6 2) choices. Kevin only wants to avoid the choices his brothers make, which are distinct choices since Sean s shamrocks are majority green and Brian s are majority orange, so there are always exactly two to subtract. So by the multiplicative principle Kevin has = choices. Thus a) and b) are TRUE, and c), d) and e) are FALSE. 1 of 6

2 2. (5 pts) Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, and set B = P 2 (A) P 3 (A) Taking the usual ordering on A, and then placing the subsets of A in lexicographical order with the 0 th subset of A and {0} the first subset. a) What is the 25 th 29 th subsets of A? SOLUTION: In binary 25 is corresponding to the length 10 bit-vector which encodes the subset {0, 3, 4}. The next four bit vectors are , , , and , giving subsets {1, 3, 4}, {0, 1, 3, 4}, {2, 3, 4}, and {0, 2, 3, 4} respectively. b) What is the least subset of A in lexicographical order which belongs to B. SOLUTION: B consists of the subsets of A containing either exactly 2 or exactly 3 elements, so the subsets with bit vectors containing either exactly 2 or exactly 3 ones. The smallest in lexicographic order is {0, 1}, with bit vector c) What is the greatest subset of A in lexicographical order which belongs to B. SOLUTION: B consists of the subsets of A containing either exactly 2 or exactly 3 elements, so the subsets with bit vectors containing either exactly 2 or exactly 3 ones. The largest in lexicographic order is {7, 8, 9}, with bit vector of 6

3 3. (5 pts) Let A = {1, 2, 3} and B = {2, 3, 4}. Use the double inclusion method to show P(A) P(B) = P(A B). Solution: Proof. We first show P(A) P(B) P(A B): f Let X P(A) P(B), so X P(A) and X P(B). Thus X A and X B by the definition of the power set, and so X A B. Hence, by the definition of the power set, X P(A B), as required. Next we show P(A B) P(A) P(B): Let Y P(A B). By the definition of the power set, Y A B, so Y A and Y B, hence, by the definition of the power set, Y P(A) and Y P(B), so Y P(A) P(B), as required. Thus the identity is proved. 3 of 6

4 4. (5 pts) Let p i be statements for i 0. Suppose p k p k+3 for all k 0. Suppose also that p 31 is true and p 33 is false. For each of the following mark T if it is true, F if it is false, and U if it cannot be determined. p 8111 p 8112 is true. p 8111 p 8112 p 8113 p 8114 is true. p 1 is true. p 1 p 2 p 3 is true. p 991 is true. By induction we know that p 31+3j is true for all j 0, and p 3i is false for all 0 i 11. The truth of any of the other individual p k s is unknown. U p 8111 p 8112 is true. [We need to know both statements are true, but we cannot know two in a row.] T p 8111 p 8112 p 8113 p 8114 is true. [For large k, every third p k is true, so without doing any arithmetic, one of these must be true, so the or statement is true.] U p 1 is true. F p 1 p 2 p 3 is true. [Definitely false by the argument above, or by noting that p 3 is false.] T p 991 is true. [True by induction. 991 = ] 4 of 6

5 5. (5 pts) Prove by induction that (2n 1) = n 2 for all n 1. Proof. Base case: When n = 1, we have 2n 1 = 1, so the statement is 1 = 1 2, which is true, so the base case is proved. Inductive Step: Suppose (2n 1) = n 2 is true for some particular n. We must show the n + 1 st statement is true, that is, we must show (2(n + 1) 1) = (n + 1) 2 Computing (2n 1)+(2n+1) = [ (2n 1)]+(2n+1) = [n 2 ] + (2n + 1) by the inductive hypothesis, and [n 2 ] + (2n + 1) = n 2 + 2n + 1 = (n + 1) 2, so the (n + 1) st statement is true, as required. So (2n 1) = n 2 is true for all n by induction. 6. (5 pts) Use the binomial theorem to compute the following sums 50 ( ) 50 5 k 7 50 k k k=0 121 k=0 ( 121 k ) ( 1) k 5 2k 7 k The binomial theorem states (a + b) n = n k=0 ( ) n a k b n k k So for a) we take n = 50, a = 5 and b = 7 and then compute (5 + 7) 50 = So for b) we may take n = 120, a = ( 1)(5 2 )(1/7) = 25/7 and b = 1 to compute (1 25/7) 121 = ( 18/7) of 6

6 7. (5 pts) a) Find k so that 5 k 1 mod 17. SOLUTION: We use the Euclidean Algorithm to find λ and µ so that λ 5 + µ 17 = = : 2 5 = : 1 Adding gives ( 2)(17) + (7)(5) = 1. (Checking = 1) So modulo 17 this says mod 17. b) Find j so that 5 j 9 mod 17. SOLUTION: We don t need to start from scratch. From part a) we have mod 17, so mod 17, or mod 17, so we can take j = mod 16. (Check: 5 12 = 60 = 3(17) + 9. ) 8. (5 pts) a) Suppose in RSA that p = 11 and q = 7. Suppose Alice s encoding key is e = 31. What is Bob s decoding key? SOLUTION: The decoding key is the multiplicative inverse of 31 modulo (p 1)(q 1) = 10 6 = 60. [Not modulo 77]. We grab our trusty Euclidean Algorithm: and adding gives ( 29)(31) = = : = : = : 1 (check: (1 30)(1 + 30) = ( ) = 1.) So the multiplicative inverse is mod 60. (Not a good choice, to have the decoding key to be the same as the encoding key!) b) Compute mod 101 SOLUTION: You could use fast exponentiation. 101 in binary is so you would have to compute 7 squares, and 3 multiplications modulo 101, which would take about 15 minutes or diligent work and use up all your scrap paper. But it is doable. With naive exponentiation you have to do 10 times that amount of work, 102 multiplications modulo 101, so the problem is impossible even if this was the only problem on the test. Better, first note that 101 is prime. To check it is prime we only have to check divisibility of primes below 101, that is 2, 3, 5, and 7. So 101 = = = = , so 101 is indeed prime. Now, Little Fermat says mod 101, so = = mod 101. which requires only 1 multiplication modulo of 6