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A brief look at Number Theory A problem that has occasionally been used in Math 300 as a Habits of Mind problem is: Example 1: An Open and Shut Case: In a certain school there are 100 lockers lining a long hallway. The lockers are numbered 1, 2, 3,..., 99, 100. All are closed. Suppose that 100 students walk down the hall in single file, one after another. Suppose the first student (who we will call “Student #1” for obvious reasons) opens every locker. The second student (i.e. Student #2) comes along and closes every 2nd locker beginning with locker #2. (i.e. lockers #2, 4, 6,…, 98, 100). Along comes Student #3 who changes the position of every third locker; if it is open, this student closes it; if it is closed, this student opens it. (I.e., (s)he closes locker #3, opens locker #6, closes locker #9, etc.) Student #4 changes the “open or shut” position of every fourth locker, and so forth, until the 100th student changes the position of locker #100. Which lockers are open at the end of this event? To solve the problem, you had to think about the factors of each number between 1 and 100 and have the insight that the perfect squares were the only numbers that had an odd number of factors. Here are two other problems that are in the same vein as our problem about the lockers. Example 2: A Rare Occurrence: In May of 1998, Missouri and many other parts of the U.S. were inundated by two species of cicadas. One species had a 13-year life cycle and others had a 17-year life cycle. To appreciate that this was a rare occurrence, we ask two questions: a) When was the most recent time before 1998 that these 13-year cicadas were heard in Missouri? What about the 17-year cicadas? b) When was the most recent time before 1998 that both the 13-year and the 17-year cicadas were heard in Missouri? 2 Example 3: Riding the Ferris Wheel: You and your little sister go to a carnival that has both a large and a small Ferris wheel. You get on the large Ferris wheel at the same time your sister gets on the small Ferris wheel. The rides begin as soon as you are both buckled into your seats. Determine the number of seconds that will pass before you and your sister are both at the bottom again. Note : The Ferris Wheel problem is taken from a 6th grade unit in Connected Mathematics. a) Assume the large wheel makes one revolution in 60 seconds and the small wheel makes one revolution in 20 seconds. b) Assume the large wheel makes one revolution in 50 seconds and the small wheel makes one revolution in 30 seconds. These last two problems call for us to investigate multiples of a number. Note: Our basic assumption is that in discussing these concepts, we are talking about integers. Definition: In particular, if a, b are integers (and a 0), then we say: a is a divisor of b (factor), and b is a multiple of a if there exists an integer c such that ac = b. One exception involves zero. Since 0 * 3 = 0, we would say 0 is a factor of 0, but we would never say that 0 is a divisor of 0. Our notation is a|b. 2 3 Here are some key divisibility facts about integers a, b, c, d. 1) If a|b, then it is also true that a|bc. 2) Transitivity: If a|b and b|c, then it is also true that a|c. 3) The 2 out of 3 Idea: Given a nonzero integer d, it is possible to find examples of numbers a, b, c such that a + b = c, and d factors none of the three numbers, exactly one of the three numbers, or all three of the numbers, but it is impossible to find an example where d factors exactly two of the three numbers. That is, a. If a + b = c, and d|a and d|b, then d|c; b. If a + b = c, and d|a and d|c, then d|b; and c. If a + b = c, and d|b and d|c, then d|a. Given two integers, a, b, we might make a list of the divisors of each number and ask “what divisors do they have in common?” Or, “does there exist a greatest common divisor?” Similarly, we might look at the multiples of each number and ask “do they have a least common multiple?” Greatest common divisor: Let a, b be integers (not both 0). Then the gcd(a, b) = d if 1. d|a and d|b; and 2. If g|a and g|b, for some integer g, then g < d. Practice Problems: a) gcd(15,60) b) gcd(17,21) Least common multiple: Let a, b be nonzero integers. Then the lcm(a, b) = m if 1. m is positive; 2. a|m and b|m; and 3. If a|c and b|c, for some positive integer c, then m < c. Practice Problems: a) lcm(15,60) b) lcm(17,21) 3 4 One of our goals is to gain insight into how many divisors a number might have and into how we might find the greatest common divisor or least common multiple of two integers. One approach calls for us to have a good understanding of prime numbers. Definition: A positive integer p is a prime number if it has exactly two distinct positive factors. A positive integer that has more than two distinct factors is said to be a composite number. We treat 1 as a very special number that we say is a unit. In particular, 1 is not a prime. Note that we have chosen our definitions so that 1 is not a prime. Note that 0 and 1 are neither prime nor composite. Why do you think this is a good idea? One of our first tasks is to investigate the prime numbers. How many primes are there under 100, 200, 1000? How are primes arranged among the positive integers? Sieve of Eratosthenes Eratosthenes, c. 275 – 195 BC, was a Greek scholar who made a map of the world, devised a system of chronology, and accurately estimated the circumference of the Earth and the distance to the moon. One can create a list of the prime numbers by writing down the whole numbers 2,3,4,5,… and successively checking each one. Of course, the even numbers larger than 2 are not prime, so there is no need to check them. So, we can circle 2 and then cross out all larger multiples of 2. Move to the next number that is not crossed out (it should be 3) and circle it. Now cross out all multiples of 3. Move to the next number that is not crossed out (it should be 5) and circle it. Now cross out all multiples of 5. Continue this procedure, at each step circling the first number that is not circled or crossed out, and then crossing out all its multiples. The circled numbers are the primes (less than the size you chose) and the numbers crossed out are composites. Question: Why is 2 the only even prime? Example 4: Find the first 50 prime numbers using the Sieve of Eratosthenes. (See next page) 4 5 11 21 31 41 51 61 71 81 91 101 111 121 131 141 151 161 171 181 191 201 211 221 231 241 2 12 22 32 42 52 62 72 82 92 102 112 122 132 142 152 162 172 182 192 202 212 222 232 242 3 13 23 33 43 53 63 73 83 93 103 113 123 133 143 153 163 173 183 193 203 213 223 233 243 4 14 24 34 44 54 64 74 84 94 104 114 124 134 144 154 164 174 184 194 204 214 224 234 244 5 15 25 35 45 55 65 75 85 95 105 115 125 135 145 155 165 175 185 195 205 215 225 235 245 6 16 26 36 46 56 66 76 86 96 106 116 126 136 146 156 166 176 186 196 206 216 226 236 246 7 17 27 37 47 57 67 77 87 97 107 117 127 137 147 157 167 177 187 197 207 217 227 237 247 8 18 28 38 48 58 68 78 88 98 108 118 128 138 148 158 168 178 188 198 208 218 228 238 248 9 19 29 39 49 59 69 79 89 99 109 119 129 139 149 159 169 179 189 199 209 219 229 239 249 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 The first 50 primes are: ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ 5 6 Having learned an algorithm for finding the primes less than a given number, we next turn our attention to asking, given a number n, how can we tell whether n is prime or composite? Or more generally, if a number is not prime, what process enables us to find factors of that number other than 1 and the number itself? If we have lots of patience, we could start checking numbers one at a time. Is 2 a 2 factor of the number? Is 3 a factor of the number? What about 4, 5, 6, 7, 8, etc.? Actually, this process would be horribly inefficient, because we are able to argue: Prime number test: An integer n > 1 is either prime or it has a prime divisor d such that 1 < d < n . Thus to test whether n is prime one need only check divisibility by the primes n. Example 5: a) Is 221 prime? b) Is 1067 prime? Lemma: Every whole number greater than 1 is a multiple of a prime. Encouraged by what we have already learned about prime numbers, we set our sights on the Fundamental Theorem of Arithmetic. Fundamental Theorem of Arithmetic: Every whole number greater than 1 can be written as a finite product of primes, and this factorization is unique except for the order of the primes. (In other words, every number has a unique prime factorization.) Example 6: Find the factorizations for the following numbers: a) 180 = 2*2*3*3*5 (Note that some primes occur more than once in this factorization.) 2 2 = 235 b) 1430 c) 80 d) 591600 e) 7 = 7 (It bothers some people to call this a factorization into a product of one prime. They restate the fundamental theorem as below.) 6 7 Fundamental Theorem of Arithmetic (2). Every whole number greater than 1 is either prime or it can be written as a finite product of primes that is unique except for the order of the primes. One approach that can be used to keep track of our efforts to factor a number into a product of primes is called a factor tree. At the “top” of the factor tree is the number. If the product of two numbers is that number, we write them under the number and “connect” the two numbers to their product with “limbs” of the factor tree. We repeat this process until finally every number at the bottom of any set of branches is a prime. (Think of these as the “low hanging fruit.”) Example 7: What is the prime factorization of 5040? 10584? Most curricula include work on factorizations and primes sometime in Grades 5-7. This work solidifies understanding of multiplication, division, and divisibility tests, and many students find the topic interesting. The Fundamental Theorem of Arithmetic shows that each whole number is built out of primes, much as molecules are built out of atoms (in both cases everything is assembled from a collection of indivisible ‘particles’.) This same idea shows up again in the context of factoring polynomials such as x 2 5 x 6 . Binomials such as ( x 3) and ( x 12) are simple polynomials. The Fundamental Theorem of Algebra states that, if we consistently use complex numbers, every polynomial p (x ) can be written as a constant times a product of binomials. These binomials can be considered as ‘prime factors’ of p (x ) . Question: Does the list of primes 2, 3, 5, 7, 11, 13, 17, 19, … end? Well, either this list ends or it never ends. Theorem: There are infinitely many primes. The list of primes never ends. We will show the list of primes never ends by using an indirect argument or sometimes called a proof by contradiction. Let’s suppose the list of primes does end. In fact, we could write them in order: 2,3,5, 7,11, , P , where P is the largest prime number. We could then multiply all the primes together, add 1, and call the resulting number N: N (2*3*5*7 *11* * P) 1 If we divide N by any prime on the list, the remainder is 1. Thus, N is not a multiple any prime. But that contradicts that every whole number is a multiple of a prime! (by the Lemma preceding the Fundamental Theorem of Arithmetic). This contradiction means that our assumption that there is a largest prime P is NOT true. Thus, the list of primes never ends. 7 8 Now let us return to a fairly basic question. How many factors does a given number have? One approach is to check numbers, one at a time, up to the square root of the number. Note: if ab = c, then either a or b is smaller than the square root of c. Why? Another approach is to find the prime factorization of the number and then use this to determine the answer. Example 8: How many factors does a given number have? a) 12 22 *3 b) 63 32 *7 c) 180 22 *32 *5 d) a 25 *36 *57 e) b pi q j r k , p, q, r are prime numbers and i, j, k are whole numbers. Example 9: Find a number T with exactly 20 factors such that a) T has exactly 2 distinct prime factors b) T has exactly 3 distinct prime factors Example 10: Explain your answer to each of the following: a) Is 340 a factor of 8130 ? b) Is 340 a factor of 6330 ? 8 9 Example 11: What is the smallest positive integer with exactly 9 proper divisors? Note: In 1982 Jim’s oldest son was an 8th grader at Pound. He (and his teacher) encountered this problem as a practice problem when they were preparing for the Math Counts competition. Divisibility Tests: If someone tells you that there are 63 shoes in the room, then you would know something is peculiar since every even number ends with a 0, 2, 4, 6, or 8. That familiar fact is one example of a divisibility test. We will develop divisibility tests for 2, 3, 4, 5, 6, 8, 9, and 10. When you look at the number a = 412763895, you immediately know that a is divisible by 5. Similarly, you know that b = 37475064 is divisible by 2 and that a is not divisible by 2. Why? Because you know a few “divisibility tests” for whether a number is divisible by 5 (the last digit is divisible by 5) or by 2 (the last divisible by 2). But do you know whether a or b is divisible by 9? Or, can you tell whether b is divisible by 8? Because we often want a quick answer to whether a number is divisible one or more small positive integers, there are divisibility tests (some call them divisibility tricks) that tell us whether a number is divisible by 2, 3, 4, 5, 6, 8, 9, 10, or 11. There are divisibility tests for 7, but are more complicated and we will not discuss divisibility tests for 7 at this time. Divisibility Tests: A number is divisible by 2 if and only if its last digit is 0, 2, 4, 6, or 8 by 3 if and only if the sum of the digits is divisible by 3. by 4 if and only if its last two digits are a number divisible by 4 by 5 if and only if its last digit is 0 or 5. by 6 if and only if the number is divisible by 2 and 3. by 8 if and only if its last three digits are a number divisible by 8. by 9 if and only if the sum of the digits is divisible by 9. by 10 if and only if its last digit is 0 9 10 Example of the divisibility test for 2: a) 144 = 10(14) + 4 Reason why the divisibility test for 2 is true: Using expanded form, any whole number N can be written as N 10a b where b is the last digit. But 10a 2(5a) is divisible by 2, so by the 2 out of 3 rule, N is divisible by 2 if and only if b is divisible by 2, i.e., b is 0,2,4,6, or 8. Example of the divisibility test for 3: a) 345 = 100(3) + 10(4) + 5 = 99(3) + 9(4) + (3 + 4 + 5) b) 3822 = 1000(3) + 100(8) + 10(2) + 2 = 999(3) + 99(8) + 9(2) + (3 + 8 + 2 + 2) Reason why the divisibility test for 3 is true: Consider a three digit number N which digits abc. In expanded form N 100a 10b c . Note that 100a 99a a and 10b 9b b . So, N 100a 10b c (99a 9b) (a b c) 9(11a b) (a b c) . Now, 9(11a b) is divisible by 3. Thus, N is divisible by 3 if and only if (a b c) is divisible by 3. But, (a b c) is the sum of the digits. [Note: This reason is for a 3 digit number, however, a similar argument shows why a number with more than 3 digits is divisible by 3.] Example of the divisibility test for 4: a) 144 = 100(1) + 44 b) 36 = 100(0) + 36 Reason why the divisibility test for 4 is true: Using expanded form, any whole number N can be written as N 100a b , where b is the last 2 digits. But 100a 4(25a) is divisible by 4. So by the 2 out of 3 rule, N is divisible by 4 if and only if b (the last two digits) is divisible by 4. Example of the divisibility test for 5: a) 345 = 10(34) + 5 b) 670 = 10(67) + 0 Reason why the divisibility test for 5 is true: Adapt the reason for the divisibility test for 2. Example of the divisibility test for 6: a) 818 b) 2178 Reason why the divisibility test for 6 is true: Since gcd(2,3)=1 and 2 divides N and 3 divides N, then 2*3=6 divides N. Example of the divisibility test for 8: a) 1248 = 1000(1) + 248 b) 3537208 = 1000(3537) + 208 Reason why the divisibility test for 8 is true: Adapt the reason for the divisibility test for 4. 10 11 Example of the divisibility test for 9: a) 342 = 100(3) + 10(4) + 2 = 99(3) + 9(4) + (3 + 4 + 2) b) 3822 = 1000(3) + 100(8) + 10(2) + 2 = 999(3) + 99(8) + 9(2) + (3 + 8 + 2 + 2) Reason why the divisibility test for 9 is true: Consider a three digit number N which digits abc. In expanded form N 100a 10b c . Note that 100a 99a a and 10b 9b b . So, N 100a 10b c (99a 9b) (a b c) 9(11a b) (a b c) . Now, 9(11a b) is divisible by 9. Thus, N is divisible by 9 if and only if (a b c) is divisible by 9. But, (a b c) is the sum of the digits. [Note: This reason is for a 3 digit number, however, a similar argument shows why a number with more than 3 digits is divisible by 3.] Casting out nines. Here is a simple trick that makes checking divisibility by 3 and 9 very fast and easy: When checking divisibility by 9, it is not actually necessary to find the sum of the digits; we need only determine whether the sum is a multiple of 9. Thus, we can ignore any digits that are 9, and pairs of digits that sum to 9. Question: Is 429761 divisible by 9? Question: Is 429761 divisible by 3? 11