Launchpad Maths. Arithmetic II

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1 Launchpad Maths. Arithmetic II

2 LAW OF DISTRIBUTION The Law of Distribution exploits the symmetries 1 of addition and multiplication to tell of how those operations behave when working together. Consider the expression: Graphically, we could represent it as 2 lots of (3 + 4); that is, 2 x (3 + 4): Keeping the objects fixed, but now shifting the boundaries, we have an alternative, but equivalent, way of considering the situation: The diagram above represents 2 lots of 3 added to 2 lots of 4, or (2 x 3) + (2 x 4). 1 Those symmetries being (i) associativity and (ii) commutativity. More will be said of these properties in Algebra II. Page 2 of 31

3 Consequently, we see that 2 (3 + 4) = (2 3) + (2 4) We say that the 2 distributes through the brackets as indicated above. Really, though, the term Law of Distribution refers to the two, equivalent ways of expressing the distribution of a single set of objects. We can also express the result notationally as follows: = Multiplication of (3+4) by 2 means to make two copies of (3+4) and add together. = Removal of brackets since order of addition makes no difference. = Addition is commutative, so we can switch the order of the middle 3+4 to become 4+3. = Brackets inserted to indicate our focus. = By definition of multiplication (two groups of 3 and two groups of 4). Page 3 of 31

4 What about the situation where subtraction occurs in the brackets, such as 2 5 3? We have, = and we see that we are adding 5 twice, giving 2 5, and subtracting 3 twice; that is, subtracting 2 3, to give = Note that the subtraction in the brackets here will give a positive number, i.e. 2. It won t always be the case that distribution will be applied to situations where numbers are positive when evaluated in the brackets (e.g. the case of (3 5)). The Law of Distribution will apply to such cases, but their development requires an understanding of algebra and negative numbers (and the equivalency between subtraction and the addition of a negative numbers), all of which will be covered I later topics. For now, we will state the Law of Distribution as follows: Law of Distribution Given three (non- negative) numbers, represented here by a, b and c, we have that a (b + c) = (a b) + (a c) & a (b c) = (a b) (a c) Note that the description above mentions non- negative numbers. Although the law applies generally to all numbers that you ll encounter at the secondary level, we re restricting application, for now, to positive whole numbers, or zero.

5 Also, given that multiplication is commutative (that is, order of (standard) multiplication doesn t matter (e.g. 2 3 = 3 2)), if we have an expression that doesn t match the law explicitly, such as we need only swap the order of the inputs into the multiplication, and then apply the distributive law; that is, (i) = (simple swapping of the order of (3+5) and 7) (ii) now apply the Law of Distribution to give Moreover, once again, because of the commutativity of multiplication, we have that and so = = We see that the 7 distributes as expected. Example Use the distributive law to complete the following. (i) = (ii) = 10 7 (iii) = Solutions (i) = (ii) = (iii) = Page 5 of 31

6 Applying the Distributive Law to Calculations Some worked examples will serve to demonstrate how the distribution law can be used in calculations. Example Calculate: (i) (ii) (iii) Solutions (i) = = = = 3531 (ii) = = = = = = 960 (iii) = = = = = 4500 In the following exercise, if you get stuck, go back to the description of the Law of Distribution, decide upon which of your expressions will be a, b and c, and apply the pattern strictly. Page 6 of 31

7 Exercise 1 1. Use the distributive law to complete the following. (a) (5 + ) 12 = (b) (7 + 3) = (c) 2 (21 9) = (d) (5 ) 17 = Calculate the following by grouping first, using the distributive law. (a) (b) (c) (d) Page 7 of 31

8 3. Use the distributive law to calculate the following by first expanding one of the two numbers of the multiplication into a sum (that s convenient ). Refer to the second example above if you have trouble. (a) (b) 83 9 (c) (d) Use the distributive law to simplify the following. You should apply the law twice by choosing appropriate pairs of expressions, simplify and then use the law again. See part (iii) in the second example to help if needed. (a) (b) Page 8 of 31

9 5. Apply the distributive law to the expression to show that = by (i) letting a = (18 12), b = 73 and c = 11, and expanding, and then (ii) applying the distributive law again to your result. TESTS FOR DIVISIBILITY To say that one whole number is divisible by another whole number means that the result of the division is yet another whole number. For example, 6 is divisible by 3 because six objects can be distributed amongst three groups equally; that is 6 3 = 2, and 2 is a whole number. Page 9 of 31

10 On the contrary, 6 is not divisible by 4 since we cannot divide six objects into four equal groups, without breaking some of the wholes before distributing. Divisibility Tests A divisibility test is something that allows for us to determine whether a whole number is divisible by another whole number, without having to perform the actual division (using, say, the long division algorithm). For example, you probably already know that if a number ends in an even digit, then the number is divisible by 2. Also, if a number ends in a zero, then you know that the number is divisible by 10. Here, we will give divisibility tests for a small set of whole numbers. Technically, if we construct divisibility tests for all prime numbers, we can build divisibility tests for any number by combining the individual prime number tests (this will be explored when discussing divisibility by six or 12). Divisor Test Example 1 Every whole number is divisible by 1. 2 Last digit of number is even (0,2,4,6,8) 120 has last digit 0, and zero is an even number, so 120 is an even number. 3 The sum of the digits forms a number that is divisible by three. For 123, the sum of the digits is = 6, and since 6 is divisible by 3, it follows that 123 is divisible by 3. Page 10 of 31

11 4 The last two digits of the number forms a number that is itself divisible by 4. For 4536, the last two digits (3 and 6) form the number, 36, and since 36 is divisible by 4, it follows that 4536 is divisible by 4. 5 The last digit of the number is either 0 or is divisible by 5 since the number ends in either a 5 or 0. 6 If the number is divisible by 2 and 3, then the number is divisible by 6. That is, to see if a number is divisible by 6, test for divisibility by 2 and 3. 7 (i) Split off the last digit of the number, giving two fragments (e.g. ABC à AB C) (ii) Subtract twice the split digit from the leftover fragment (e.g. AB 2 x C) (iii) If the resulting number is divisible by 7, then the original number is divisible by is divisible by 6 because (i) the last digit of 252 is 2 and 2 is even, and (ii) the sum of the digits of 252 is 2+5+2=9, and 9 is divisible by 3. (i) For 231, split as 23, 1 (ii) 23 (2x1) = 21 (iii) As 21 is divisible by 7, the test allows us to say that 231 is divisible by 7. 8 If the last three digits of the number forms a number that is divisible by 8, then the original number is divisible by 8. To test the divisibility of 513,715,712 by 8, we need only see if 712 is divisible by 8 (use short or long division). Here, _ Page 11 of 31

12 9 If the sum of the digits of the number forms a number that is divisible by 9, then the original number is divisible by If the number s last digit is 0, then the number is divisible by Form the alternating sum of the digits of the number. If the result is a number divisible by 11, then the original number is divisible by 11. The alternating in alternating sum refers to the +/- pattern that is demonstrated to the right. 12 If the number is divisible by 3 and 4, then the number is divisible by 12. Since 712 is divisible by 8, it follows that 513,715,712 is divisible by 8. For 5886, we have that =27, and since 27 is divisible by 9, 5886 is divisible by 9. Easy: 100 is divisible by 10 because 100 ends in a zero. Consider the number 71,973. We form the alternating sum as follows: = 11 and since 11 is divisible by 11, it follows that 71,973 is also divisible by 11. For the number 1044: (i) the sum of the digits is =9, and since 9 is divisible by 3, 1044 is divisible by 3; (ii) the last two digits form a number (44) that is divisible by 4. Hence 1044 is divisible by 4. (iii) Since 1044 is divisible by 3 and by 4, it is divisible by 12. Page 12 of 31

13 25 If the number is ends in the digits, 00, 25, 50, or 75, then the number is divisible by is divisible by 25 because the last two digits form the number 75. Applying the Tests Recursively Note that the divisibility tests can be used once again on the results of your tests. For example, if we wanted to see if 4515 were divisible by 7, we would apply the test as follows: 4515 à x 5 = = 441 Now, if 441 is divisible by 7, then 4515 is divisible by 7 but we don t know, off the top of our heads, whether 441 is divisible by 7. But we have a test to determine divisibility by 7 so we apply the test again, this time to 441: 441 à 44 2 x 1 = 42 We now have something that we can use: we know 42 is divisible by 7, which implies 441 is divisible by 7, which implies further that 4515 is divisible by 7. Note, if you didn t know that 42 was divisible by 7, you could apply the test to again: 42 à 4 2 x 2 = 0, and 0 is divisible by 7 since 0 7 = 0, a whole number. We say that we can apply these tests recursively; that is, again and again, on the newer objects/numbers, until we find an answer to the problem. Outline of a Proof for the Divisibility Test for 9 We will look at the rationale behind the divisibility test for 9 by examining the number First, express the number in expanded form 2376 = and then express the powers of 10 in the following form: Page 13 of 31

14 2376 = We now expand, using the distributive law, and collect to get two expressions: one having 9 as a factor, and one being the sum of the digits of 2376: 2376 = = = Now, we see that the first term in the last expression is divisible by 9 since it contains 9 as a factor, so whether 2376 is divisible by 9 will depend then on whether the sum of the digits is also divisible by 9. In this case, the sum is 18, which is divisible by 9; that is 2376 = = and so, we see that 9 is a factor of Hence, 2376 is divisible by 9. The proof of divisibility by 9 for this specific number can be generalized to any number. The trick is to (i) write the power of 10 in the expansion as (ii) (iii) (iv) (v) expand using distribution collect all terms with common factor of 9; the remaining sum will be a sum of the digits of the number if the sum of those digits is divisible by 9, then 9 will be a common factor of the two main terms, and so can be factored out using distribution (in reverse), and so 9 will be a factor of the original number. Page 14 of 31

15 Exercise 2 1. Apply the divisibility tests to determine the following. Only use a division algorithm to assist, as part of the test, in checking for divisibility by 4 or 8. (a) Which of the following numbers are divisible by 25? (b) Which of the following numbers are divisible by 4? (c) Which of the following numbers are divisible by 7? (d) Which of the following numbers are divisible by 9? Page 15 of 31

16 2. Apply the divisibility test for 8 to determine whether 8 divides Apply the divisibility test for 11 to determine if 11 divides Without using a division algorithm, show that is divisible by 105. Page 16 of 31

17 Factors FACTORS AND MULTIPLES OF A NUMBER If a whole number can be written as a product of other whole numbers, then the numbers so forming the product are termed factors. For example, the number 28 can be written as a product of the whole numbers, 7 and 4, namely 28 = 7 4 Hence 7 and 4 are factors of 28. But so too are 14 and 2, since we have 28 = We can compose a list of all the factors of 28 in a methodical manner by checking the divisibility of 28 against all whole numbers from 1 to 28 (there are more efficient methods, but this will do for now). So, to determine methodically all the factors of 28, we can work as follows: Test factor: Complementary factor: No 7 No No Yes, but no need to continue because 7 was already discovered to be a factor, and any higher factor from here will also have been discovered because of the symmetry in multiplication. 28 = So, the factors of 28 are 1, 2, 4, 7, 14, 28. Page 17 of 31

18 Note that the divisibility tests can be used to help in the discovery of the factors of a number. Multiples The multiples of a positive whole number is the set of all positive products that may be formed when using that number as a factor. This is more easily understood through examples. Below is the beginning of the list of multiples for the numbers 5, 12 and 113: 5 1 = 5, 5 2 = 10, 5 3 = 15, 5 4 = 20, 12 1 = 12, 12 2 = 24, 12 3 = 36, 12 4 = 48, = 113, = 226, = 339, = 452, The lists show how to generate the multiples of a number. They also indicate that the lists are infinite, or unending. PRIME NUMBERS AND COMPOSITE NUMBERS Prime Numbers A prime number is any whole number that is greater than 1, and whose only factors are 1 and itself. For example, 11 is a prime number since (i) it is greater than 1, and (ii) (ii) its only factors are 1 and 11. Similarly, 61 is prime since it is (i) greater than 1, and (ii) (ii) has only the factors of 1 and 61. Page 18 of 31

19 Technically, to show that a number is prime requires that we go about checking to ensure that it has no factors other than 1 and itself. There is no theorem in Mathematics that allows us to check whether a number is prime by means of a simple test. Prime numbers play a role in the mathematics of numbers similar to that played by fundamental particles in nature: they are the building blocks of whole numbers, and a study of those building blocks extends to a study of anything that those building blocks can make. Composite Numbers A composite number is any whole number that is greater than 1 and that is not prime. Another way of stating the above is that composite numbers are numbers that may be expressed as a product of primes. For example, 6 is a composite number since it has factors other than 1 and 6. So too is 60 composite, since it has factors other than 1 and 60. Moreover, both 6 and 60 may be written as a product of primes. We have, 6 = 2 3, where both 2 and 3 are prime, and 60 = , where each of the factors, again, are prime. A theorem in Mathematics called The Unique Factorisation Theorem states that any whole number may be decomposed into a product of primes, unique up to order of factorisation (that is, although 2 3 and 3 2 are technically different, we consider them equivalent because they yield the same answer. We say they are equivalent up to factorisation ). Page 19 of 31

20 An Algorithm to Determine Prime Numbers: The Sieve of Eratosthenes The Sieve of Eratosthenes is an algorithm (procedure) devised by the Greek mathematician, Eratosthenes of Cyrene, to find the set of all prime numbers up to a given limit. A proper formulation of the algorithm would require some knowledge of algebra and recursion, both of which you re not required to know yet. Below, we present the first few steps in the application of the algorithm. Hopefully, the pattern will be clear. Algorithm Outline To find all prime numbers less than or equal to a given whole number, do the following: Setup List all consecutive whole numbers from 2 to the final number wanted. Application I. Label the number 2 as prime. II. Delete all multiples of 2, other than 2. III. Select the smallest remaining number greater than 2 (here it will be 3) and label it as prime. IV. Delete all multiples of 3, other than 3. V. Select the smallest remaining number greater than 3 (here it will be 5) and label it as prime. VI. Delete all multiples of 5, other than 5. VII. Select the smallest remaining number greater than 5 (here it will be 7) and label it as prime. VIII. Delete all multiples of 7, other than 7. etc. An example follows on the next page. Page 20 of 31

21 Launchpad Mathematics Arithmetic II TALENT 100 Example Find all primes from 2 to 30 using the Sieve of Eratosthenes. Solution We first list: Fix 2 as our first prime and delete all other multiples of 2: Next fix the smallest number remaining that is greater than 2 (i.e. 3) and delete all other multiples of 3: TALENT 100: HSC SUCCESS. SIMPLIFIED. Page 21 of 31

22 Next fix the smallest number remaining that is greater than 3 (i.e. 5) and delete all other multiples of 5: Continuing would show that there are no more numbers that the algorithm would delete. The algorithm deletes, methodically, all numbers that are composites, hence leaving behind only those numbers that are prime. The set of prime numbers from 2 to 30 is therefore: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Prime Factorisation and Factors of a Number Although we won t prove the unique factorization theorem, we will still make use of it. The theorem says that we may factorise any whole number into a product of primes. One way of doing this is by use of factorization trees. We will demonstrate this by factorising the number 60. Decomposition will make use of the divisibility tests, especially when decomposing larger numbers. We start by breaking 60 into any two factors (other than 1 and 60). Since 60 ends in the digit 0, we know that it s divisible by 10. The complementary factor to 10 is 6 here Page 22 of 31

23 We now expand the tree by factorising the leaves (the lowest set of numbers). There is only one possibility for each number: 6 = 2 3 and 10 = 2 5. We continue then as follows: We stop once every number generated is a prime (i.e. no more composites to factorise). We see then that 60 has the prime factorisation, 60 = We can use the prime factorisation of a number to investigate properties of that number. For example, knowing that the primes of 60 are 2, 2, 3, 5, we can generate all the factors of 60 by considering all possible combinations of its primes (and remembering to include the trivial factor, 1). To do this: (i) Consider all combinations of primes taken one at- a- time: 2, 3, 5 (ii) Consider all combinations of primes taken two at- a- time: 2 2 (start with 2 and apply all other possible primes) Page 23 of 31

24 3 5 (move to 3 and apply all other possible primes, excluding any multiplications that have already been determined (such as 2 3)) 5? (no need to continue since all two- prime products including 5 have already been listed) (iii) Consider all combinations of primes taken three at- a- time: (start by fixing 2 and 2, and apply all other possible primes) (tick over the right- most fixed number (i.e. the second 2) to 3, and apply all other possible primes, excluding any multiplications that have already been listed) 2 5? (no need to continue since all three- prime products including 5 have already been listed) (iv) Consider all combinations of primes taken four at- a- time: (start by fixing 2, 2 and 3, and apply all other possible primes. There is only one possible combination taking all four primes at once) Hence, the factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Page 24 of 31

25 Exercise 3 1. (a) Show that 72 = : by decomposing 72 using a factor tree. (b) Hence, using the prime factorisation or otherwise, list the factors of 72 (there are twelve factors). 2. Use the Sieve of Eratosthenes to determine the prime numbers from 2 to 120 (use the grid below and be efficient by noticing that entire lines may be deleted as you work through the algorithm). Page 25 of 31

26 HIGHEST COMMON FACTOR (HCF) & LOWEST COMMON MULTIPLE (LCM) Recall that to factorise a whole number means to write it as a product of whole numbers. So, 4 9 is a factorisation of 36, but so too is The point is, factorisations are not unique. An equivalent way of thinking about a factor is that it is a smaller number that divides into a bigger number, leaving no remainder. Highest Common Factor Consider the following list of factors of 48 and 60. Factors of: The boxed factors are factors in common, the highest of which being 12; that is, the highest common factor (HCF) of 48 and 60 is 12. Clearly, then, one method by which to find the HCF of two numbers is to list the factors of each number select the greatest factor in common The method is simple, but relies upon you being able to determine all the factors of a number. For numbers with many prime factors, this can be a tedious task. Page 26 of 31

27 Lowest Common Multiple Now consider the following list of multiples of 8 and 12: Multiples of: The boxed multiples are multiples in common, the lowest of which being 30; that is, the lowest common multiple (LCM) of 6 and 15 is 30. Clearly, then, one method by which to find the LCM of two numbers is to list the multiples of each number select the lowest multiple that the numbers have in common. The method is straightforward, but can be a little inefficient if the numbers used aren t nice, which happens if they don t share many primes as common factors. Using the Primes of a Number to Determine HCF and LCM NOTE that the following is not necessary for your understanding of HCF and LCM. It s presented here for further development and allows for a more efficient means of calculating HCF and LCM, particularly in cases concerning more than two numbers or where numbers are large. Below we show two sets, the left containing the prime factors 36, and the right containing the prime factors of Page 27 of 31

28 Combining the two sets, we can represent those prime factors that are shared as an overlap between the two sets This is an example of a Venn diagram, a method for representing sets, or collections, of objects. Venn diagrams will be covered more formally in the latter part of the course. For now, to understand what s to follow, you need only understand what the diagram represents: all prime factors contained in the blue circle are those of 36, and all prime factors contained in the black circle are those of 60. All those contained in the overlap (the intersection) are primes that the two numbers share. Using the diagram to find the HCF If primes build factors, then common primes build common factors. The greater the number of common primes that you multiply, the greater the common factor you create. Hence multiplying the highest number of common primes will give you the highest common factor. The HCF of a set of numbers can be found by multiplying all primes held in common between those numbers. Here, the HCF of 36 and 60 is 2 x 2 x 3 = 12 (which can be verified by listing). Page 28 of 31

29 Using the diagram to find the LCM This is a little harder to justify without some algebraic knowledge and some experience in number theory. Also, some work with equations is required too (for proofs that are the most straightforward). So, for now, we ll give a statement of fact, and leave a proof for some time later in the course. The LCM of two or more whole numbers can be found by multiplying each of the primes in the set diagram; that is, multiply (i) (ii) all shared prime factors with (i.e. the HCF) each of the unshared prime factors. OR 6 : m = 6 10n The LCM of two or more whole numbers can be found by multiplying those numbers and then dividing the result by the HCF of those numbers. So, to find the LCM of 36 and 60 using the diagram, we have: LCM = [product of all shared primes] x [product of all unshared primes] = [HCF] x [product of all unshared primes] = [2 x 2 x 3] x [3 x 5] = [12] x [15] = 180 To find the LCM using the second method: LCM = [product of numbers] [HCF of those numbers] = [36 x 60] [2 x 2 x 3] = [2160] [12] = 180 Page 29 of 31

30 Exercise 4 1. (a) List the factors of 12 and 32. (b) Hence write down the HCF of 12 and (a) List the first six multiples of 8 and 12. (b) Hence write down the LCM of 8 and Find the HCF of 12 and 18 by either listing their factors, or by using the method involving their primes. Page 30 of 31

31 4. Find the LCM of 6 and 8 by either listing their factors, or by using the method involving their primes. Note: if you choose to list multiples, do so by listing multiples of the larger number first (since it takes bigger jumps ) and allow the smaller number s multiples to catch up. 5. Apply the idea of lowest common multiple to answer the following question. A pair of athletes begin racing on a closed- loop racetrack. If athlete A completes one circuit in 45 seconds, and athlete B completes one circuit in 50 seconds, at what time (that is, number of seconds from the start) will they first meet again? Page 31 of 31

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