# 5.1. Primes, Composites, and Tests for Divisibility

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1 CHAPTER 5 Number Theory 5.1. Primes, Composites, and Tests for Divisibility Definition. A counting number with exactly two di erent factors is called a prime number or a prime. A counting number with more than two factors is called a composite number, or a composite. The Sieve of Eratosthenes is a tool used to find primes. Start by circling 2 and then crossing out every second number. Then circle 3 and cross out every third number. Do the same for 5, 7, etc. The circled numbers are primes and the crossed out numbers are composites. Note. The number 1 is neither prime nor composite. Factor trees can be used to express a composite as a product of smaller numbers and even as a product of primes. 75

2 76 5. NUMBER THEORY We see that each of the four cases above give the same prime factorizasion. This is an example of the following theorem: Theorem (Fundamental Theorem of Arithmetic). Each composite number can be expressed as the product of primes in exactly one way (except for the order of the factors). Example. (1) (2) 192 = 2 96 = = = = = 9 40 = =

3 5.1. PRIMES, COMPOSITES, AND TESTS FOR DIVISIBILITY 77 Definition. Let a and b be whole numbers with a 6= 0. We say that a divides b, and write a b, if and only if there is a whole number x such that ax = b. The symbol a - b means tha a does not divide b. In other words, a b if and only if a is a factor of b. When a b, we say that a is a divisor of b, a is a factor of b, b is a multiple of a, and b is divisible by a. Example. (1) since = 210. (2) 2 ( ) since 2 (2 5 7) = ( ). Tests for Divisibility Theorem (Tests for Divisibility by 2, 5, and 10). A number is divisible by 2 if and only if its ones digit is 0, 2, 4, 6, or 8. A number is divisible by 5 if and only if its ones digit is 0 or 5. A number is divisible by 10 if and only if its ones digit is 0. We will prove the first part of this below. Example and But we also have then that 4 ( ) and 4 (36 24). This suggests the following theorem. Theorem. Let a, m, n, and k be whole numbers with a 6= 0. (a) If a m and a n, then a (m + n). (b) If a m and a n, then a (m n) for m n. (c) If a m, then a km.

4 78 5. NUMBER THEORY Proof. (of (b)) a m =) ax = m for some whole number x. a n =) ay = n for some whole number y. m n =) m n is a whole number. Then ax ay = m n =) a(x y) = m n and x y is a whole number =) a (m n). We now can look at the proof of the test for divisibility by 2 for an arbitrary 3-digit number. The same idea used in this proof works for any number of digits. Proof. Let r = a b 10 + c be any 3-digit number (a, b, and c are the digits from L to R). Note a b 10 = 10(a 10 + b). Since 2 10, 2 10(a 10 + b) =) 2 (a b 10) for any digits a and b. Thus if 2 c, then 2 [10(a 10 + b) + c], so 2 (a b 10 + c), or 2 r. [Thus 2 c =) 2 r.] Now suppose 2 r or 2 (a b 10 + c). Since 2 (a b 10), 2 [(a b 10 + c) (a b 10)] or 2 c. [Thus 2 r =) 2 c.] We have shown that 2 r if and only if 2 c.

5 5.1. PRIMES, COMPOSITES, AND TESTS FOR DIVISIBILITY 79 Theorem (Tests for Divisibility by 4 and 8). A number is divisible by 4 if and only if the number represented by the last two digits is divisible by 4. A number is divisible by 8 if and only if the number represented by the last three digits is divisible by 8. Example is divisible by 4 since 36 is since since since Theorem (Tests for Divisibility by 3 and 9). A number is divisible by 3 if and only if the the sum of its digits is divisible by 3. A number is divisible by 9 if and only if the the sum of its digits is divisible by 9. Example and since = 18, 3 18, and and since = 13, 3-13, and Theorem (Test for Divisibility by 11). A number is divisible by 11 if and only if 11 divides the di erence of the sum of the digits whose place values are odd powers of 10 and the sum of the digits whose place values are even powers of 10, with the smaller sum subtracted from the larger.

6 80 5. NUMBER THEORY Example since = 26, = 15, = 11, and since = 5, = 5, 5 5 = 0, and since = 9, = 11, 11 9 = 2, and Theorem (Test for Divisibility by 6). A number is divisible by 6 if and only if it is divisible by both 2 and 3. Theorem. A number is divisible by the product ab of two nonzero whole numbers a and b if and only if it is divisible by both a and b, and a and b have only 1 as a common whole number factor. Example. Is 557 prime? Theorem (Prime Factor Test). To test for prime factors of a number n, one need only search for prime factors p where p 2 apple n or p apple p n. Example. By calculator, p , 23 2 = 529 and 24 2 = 576. Thus we need check only whether 2, 3, 5, 7, 11, 13, 17, 19, or 23 divide 557. Since none of these primes divide 557, 557 is prime.

7 5.1. PRIMES, COMPOSITES, AND TESTS FOR DIVISIBILITY 81 Example. Factor 5899 into primes = 30, 3 30 but 9-30, so 5889 = , , 11 - (12 7), so , so 5889 = < p 151 < 13, so 151 is prime since we have already tried each prime factor less than 12. Problem (Page 189 # 37). Ink is spilled on a bill for 36 sweatshirts. If only the first and last digits were covered, and the other three digits were, in order, 839 as in?83.9?, how much did each cost? Solution. Strategy 10 Use properties of numbers. 36 bill. Since 9 4 = 36 and 9 and 4 have only 1 as a common factor, 4 bill and 9 bill. Thus 4 9?, so? is 2 or 6 since 4 92 and Suppose we have? Since 9 sum of the digits and = 22, the first digit is 5. Thus the bill is \$ and the cost of each sweatshirt is /16= \$ Now suppose we have? Since 9 sum of the digits and = 26, the first digit is 1. Thus the bill is \$ and the cost of each sweatshirt is /16 = \$5.11. But this cost is unrealistic, so the cost per sweatshirt is \$16.22.

8 82 5. NUMBER THEORY 5.2. Counting Factors, Greatest Common Factor, and Least Common Multiple Counting Factors Example. How many factors (or divisors) does 10,800 have? First, 10, 800 = as a product of primes. Then a divisor of 10,800 can only have 2, 3, and 5 as prime factors, with no more than 4 twos, no more than 3 threes, and no more than 2 fives. There are five possibilities for the number of twos to include: 0, 1, 2, 3, 4. There are four possibilities for the number of threes to include: 0, 1, 2, 3. There are three possibilities for the number of fives to include: 0, 1, 2. Thus there are = 60 di erent factors of 10,800. Theorem. Suppose that a counting number n is expressed as a product of distinct primes with their respective exponents, say n = (p n 1 1 )(pn 2 2 ) (pn m m ). Then the number of factors of n is the product Greatest Common Factor (n 1 + 1)(n 2 + 1) (n m + 1). Definition (Greatest Common Factor). The greatest common factor (GCF) of two (or more) nonzero whole numbers is the largest whole number that is a factor of both (all) of the numbers. The GCF of a and b is written GCF(a, b).

9 5.2. COUNTING FACTORS, GREATEST COMMON FACTOR, AND LEAST COMMON MULTIPLE 83 Set Intersection Method Example. Find GCF(36, 48). A = {x x is a factor of 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36} B = {x x is a factor of 48} = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} A \ B = {x x is a factor of 36 and 48} = {1, 2, 3, 4, 6, 12} Thus GCF(36, 48) = 12, the largest element in A \ B. Prime Factorization Method Example. Find GCF(36, 60). (1) Factor each number into primes. 36 = = (2) For the GCF, take each factor the fewest number of times it is used in any one number. GCF(36, 60) = = 12. Euclidean Algorithm Method Suppose a and b are whole numbers with a b. If c a and c b, c a b, so c is a common factor of b and a b. Also, if c b and c a b, c [(a b) + b)], so c a and c is a common factor of a and b. We have proven the following theorem:

10 84 5. NUMBER THEORY Theorem. If a and b are whole numbers with a GCF(a, b) = GCF(a b, b). b, then Example. GCF(98, 28) = GCF(70, 28) = GCF(42, 28) = GCF(14, 28) = GCF(28, 14) = GCF(14, 14) = 14 Recalling that division can be viewed as repeated subtraction, and that we can shorten the above to GCF(98, 28) = GCF(14, 28) = 14 since In general,

11 5.2. COUNTING FACTORS, GREATEST COMMON FACTOR, AND LEAST COMMON MULTIPLE 85 Theorem (Euclidean Algoritm Method). If a and b are whole numbers with a Example. Find GCF(399, 102). GCF(a, b) = GCF(r, b). b and a = bq + r where r < b, then The final divisor that leads to a remainder of 0 is the GCF. This is the Euclidean Algorithm method.

12 86 5. NUMBER THEORY Least Common Multiple Definition (Least Common Multiple). The least common multiple (LCM) of two (or more) nonzero whole numbers is the smallest nonzero whole number that is a multiple of each (all) of the numbers. The LCM of a and b is written LCM(a, b). Set Intersection Method Example. Find LCM(36, 48). A = {x x is a multiple of 36} = {36, 72, 108, 144, 180, 216, 252, 288, 324, 360,... }. B = {x x is a multiple of 48} = {48, 96, 144, 192, 240, 288, 336, 384,... }. A \ B = {x x is a multiple of 36 and 48} = {144, 288,... }. LCM(36, 48) = 144, the smallest number in A \ B. Prime Factorization Method Example. Find GCF(36, 60). (1) Factor each number into primes. 36 = = (2) For the GCF, take each factor the most number of times it is used in any one number. GCF(36, 60) = = = 180.

13 5.2. COUNTING FACTORS, GREATEST COMMON FACTOR, AND LEAST COMMON MULTIPLE 87 Build-Up Method Example. Find LCM(36, 60). (1) Factor each number into primes. 36 = = (2) Start with one number s prime factorization, say 36, and compare its prime factorization to the prime factorization of the other number, 60 in this case. We begin with Comparing with the prime factors of 60, there is no need to increase the exponents of 2 or 3. We simply need to add a factor of 5. LCM(36, 60) = = 180. Example. Find GCF(90, 36, 54) and LCM(90, 36, 54). 90 = = = GCF(90, 36, 54) = = 18 LCM(90, 36, 54) = = = 540 Note. GCF(36, 60) = 12 and LCM(36, 60) = = 2160 and = This is an instance of Theorem. Let a and b be any two whole numbers. Then GCF(a, b) LCM(a, b) = ab.

14 88 5. NUMBER THEORY How many prime numbers are there? Theorem. There is an infinite number of primes. Proof. We us indirect reasoning. Suppose there is a finite number of primes, say where p is the greatest prime. Let 2, 3, 5, 7, 11,..., p, N = p + 1. N > 1 and N is greater than any prime. But when N is divided by any prime, the remainder is always 1. So N is not 1, is not prime, and is not a composite, which is impossible. Thus there must be infinitely many primes.

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