University of British Columbia. Math 312, Midterm, 6th of June 2017

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1 University of British Columbia Math 312, Midterm, 6th of June 2017 Name (please be legible) Signature Student number Duration: 90 minutes INSTRUCTIONS This test has 7 problems for a total of 100 points. This test has 8 pages including this one. Read all the questions carefully before starting to work. For problems with several parts indicate clearly which part of it you are answering. You should give complete arguments and explanations for all your claims and calculations; answers without justifications will not be marked. You may write on the backs of pages if you run out of space. Attempt to answer all questions for partial credit. This is a closed-book examination. None of the following are allowed: documents, cheat sheets or electronic devices of any kind (including calculators, cell phones, etc.) Question: Total Points: Score: 1

2 2 PROBLEM 1 (15 points) Find all positive integers x satisfying the conditions 7x 2 (mod 60) and x 180. You have to use the Euclidean algorithm and back substitution in your answer. Answer: We have (7, 60) = 1, so 7x 1 (mod 60) has a solution which is the inverse 7 1 (mod 60). We apply Euclidean algorithm 60 = , 7 = , 4 = followed by back substitution (7, 60) = 1 = = 4 (7 4 1) 1 = ( 1) = ( 1) 7 + (60 7 8) 2 = ( 17) 7, hence (mod 60). Therefore, 7x 2 (mod 60) 43 7x 43 2 (mod 60) x 26 (mod 60). Finally, the integers of the form x = k such that x 180 are x = 26, x = 86 and x = 146.

3 Math 312, Midterm 3 PROBLEM 2 (15 points) (a) (3pts) State the Chinese Reminder Theorem. Answer: Let m 1, m 2,..., m k Z >0 be pairwise coprime. Let b 1, b 2,..., b k Z. Then, the system of congruences x b 1 (mod m 1 ) x b 2 (mod m 2 ).. x b k (mod m k ) has a unique solution modulo m 1 m 2... m k. (b) (12pts) Solve the following ancient Chinese problem. A band of 17 pirates stole a sack of gold coins. When they tried to divide the fortune into equal portions, 3 coins remained. In the ensuing brawl over who should get the extra coins, one pirate was killed. The wealth was redistributed, but this time an equal division left 10 coins. Again an argument developed in which another pirate was killed. But now the total fortune was evenly distributed among the survivors. What was the least number of coins that could have been stolen? Hint: Note that = Answer: Let x be the number of gold coins. The problem is equivalent to finding the least positive solution to the following system of congruences x 3 (mod 17), x 10 (mod 16), x 0 (mod 15). As 17, 16, and 15 are pairwise coprime, we can use the CRT to find the unique solution modulo M = = Thus the solution is given by the formula x = 3 M 1 y M 2 y 2 +0 M 3 y 3 3 M 1 y M 2 y 2 (mod M), where M 1 = 15 16, M 2 = 15 17, y 1 is a solution to the congruence (15 16)y 1 (mod 17) ( 2) ( 1)y 2y 1 (mod 17) and y 2 is a solution to (15 17)y 1 (mod 16) ( 1) 1y y 1 (mod 16). Thus, we can take y 1 = 9 and y 2 = 1, obtaining x = ( 1) = 3930 (mod 4080). In other words, the number of coins can be n where n is a non-negative integer. The smallest such number is 3930.

4 4 PROBLEM 3 (15 points) Compute the number r such that 0 r < 30 and r 9 23 (mod 31). Hint: Note that (mod 31) and (mod 31). Answer: We will use fast modular exponentiation. First, we write the exponent in base 2, that is 23 = Second, we compute 9 2k (mod 31) for 0 k 4 by sucessive squaring and reduction: 9 9 (mod 31), (mod 31), (mod 31) (mod 31), 9 16 ( 3) 2 9 (mod 31). Finally, we compute the product (mod 31).

5 Math 312, Midterm 5 PROBLEM 4 (15 points) (a) (3pts) State Fermat s Little Theorem. Answer: Let p be a prime. If a Z satisfies (a, p) = 1, then a p 1 1 (mod p). (b) (6pts) Let p, a Z with p a prime and (a, p) = 1. Consider the set of integers S = {a, 2a,..., (p 1)a}. Show that, when considered mod p, the elements of S are all different and not congruent to zero. Answer: Suppose that two elements in S are equal mod p, that is, ka k a (mod p) with 1 k, k p 1. Note that a 1 exists since (a, p) = 1. Then, multiplying by a 1 cancels the a, obtaining ka k a (mod p) k k (mod p) = k = k, where the implication follows because 1 k, k p 1. Finally, suppose ka 0 (mod p); since p is prime we have p a or p k. The first is impossible, because (a, p) = 1 and the second because 1 k p 1, giving a contradiction. We conclude ka 0 (mod p) for all ka S. (c) (6pts) Use (b) to prove Fermat s Little Theorem. Hint: Start by considering the product of the elements of S. Answer: It follows from (b) that, the p 1 elements in S, when considered mod p must be the integers 1, 2,..., p 1 in some order. Therefore, taking the product of the elements in S gives a (2a) (3a) (p 1)a (p 1) (mod p) Since we also have = (p 1)! (mod p) a(2a)(3a) (p 1)a a p 1 (1 2 3 p 1) = a p 1 (p 1)! we conclude a p 1 (p 1)! (p 1)! Now, by Wilson s Theorem, we have a p 1 ( 1) 1 (mod p) a p 1 1 (mod p). (mod p),

6 6 PROBLEM 5 (15 points) An old receipt has faded. It reads 88 chickens cost a total of $x4.2y, where x and y are unreadable digits. How much did the 88 chickens cost? Answer: We know that the total cost being x42y cents is divisible by 88 = 8 11 and so is divisible by both 11 and 2 3 = 8. Thus 42y is divisible by 2 3 = 8, and 2y is divisible by 2 2 = 4 and y is divisible by 2. The only number 0 y < 10 satisfying this is y = 4. As x424 is divisible by 11 we require that x = x 6 is divisible by 11. The only number 0 x < 10 satisfying this is x = 6. Thus the total cost was $64.24.

7 Math 312, Midterm 7 PROBLEM 6 (15 points) Let φ be the Euler φ-function. Find all positive integers n such that φ(n) = 2. Answer: Let n Z >0. If n 1 then n = p a 1 1 p a p a k k where a k 1 and p i are distinct primes. We have from class φ(n) = k i=1 p a i 1 i (p i 1). Suppose now φ(n) = 2; thus n 1 because φ(n) = 1. By the formula p i 1 2 for all i; thus only the primes 2 and 3 can divide n. Write n = 2 a 1 3 a 2 ; if a 2 0 from the formula we have 3 a2 1 2 thus a 2 = 1. We conclude that a 2 = 0 or a 2 = 1. We now divide into two cases: (i) Suppose a 2 = 1, i.e. n = 2 a1 3. If a 1 2 then the formula shows that φ(n) = 2 is divisible by 4, a contradiction. We conlude a 1 1, that is n = 3 or n = 6. Both are solutions because φ(3) = φ(6) = 2. (ii) Suppose a 2 = 0, i.e n = 2 a 1 with a 1 1. Then φ(n) = 2 a1 1 = 2 implies a 1 = 2, that is n = 4. Thus φ(n) = 2 if and only if n = 3, n = 4 or n = 6.

8 8 PROBLEM 7 (10 points) Prove there are infinitely many primes of the form 6k + 5. Hint: Suppose not, i.e. that all such primes are p 1, p 2,..., p k and consider the number N = 6p 1... p k 1. Then analyze the shape of the prime factors of N using the division algorithm. Answer: Suppose there are only finitely many primes of the form 6k + 5 and denote them p 1, p 2,..., p k. Let N = 6p 1... p k 1. We have N > 0 because p 1 = 5 = Let q be a prime divisor of N. By the division algorithm we have q = 6a + r, where a Z, r {0, 1, 2, 3, 4, 5}. We now divide into cases: (1) If r = 0, 2, 4 then q is even, that is q = 2, but 2 N since 2 N + 1, a contradiction. (2) If r = 3 then q = 6k + 3 is prime only if q = 3. But 3 N since 3 N + 1, a contradiction. (3) If r = 5 then q = 6k + 5 and by hypothesis we have q = p i for some i. Since q N + 1 then q N, a contradiction. (4) From the previous cases it follows that any prime q dividing N is of the form 6a + 1, that is q 1 (mod 6). Since the product of two such primes q 1, q 2 (not necessatily distinct) also satisfies q 1 q 2 1 (mod 6) we conclude that N 1 (mod 6) which is a contradiction with N 1 5 (mod 6).

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