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KS4 Mathematics N1 Integers

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**N1 Integers Contents A N1.1 Classifying numbers A**

N1.2 Calculating with integers A N1.3 Multiples, factors and primes A N1.4 Prime factor decomposition A N1.5 LCM and HCF

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**Classifying numbers Natural numbers**

Positive whole numbers 0, 1, 2, 3, 4 … Integers Positive and negative whole numbers … –3, –2, 1, 0, 1, 2, 3, … Rational numbers Numbers that can be expressed in the form n/m, where n and m are integers. All fractions and all terminating and recurring decimals are rational numbers, for example, ¾, –0.63, 0.2. . This slide gives the sets of numbers that pupils have met so far. All off these sets belong to the set of real numbers. Pupils are not required to know about complex numbers at this level. Explain that natural numbers are also called counting numbers. This is the most basic set of numbers. Rational numbers (from the word ratio) can be written as one integer divided by another, as a terminating decimal or as a recurring decimal. Remind pupils that all terminating and recurring decimals can be written as fractions. Irrational numbers cannot be written exactly as fractions or decimals. As fractions they would consist of an infinitely long string of non-repeating digits. See N4.2 Terminating and recurring decimals Irrational numbers Numbers that cannot be expressed in the form n/m, where n and m are integers. Examples of irrational numbers are and 2.

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**Even numbers Even numbers are numbers that are exactly divisible by 2.**

For example, 48 is an even number. It can be written as 48 = 2 × 24. All even numbers end in 0, 2, 4, 6 or 8. Even numbers can be illustrated using dots or counters arranged as follows: E(1) = 2 E(2) = 4 E(3) = 6 E(4) = 8 E(5) = 10 Numbers that that can be illustrated using patterns of evenly spaced dots or counters are called figurate numbers. Pupils no not need to remember this term, however. Examples of figurate numbers include even numbers, odd numbers, triangular numbers, square numbers and cube numbers. Illustrating numbers using patterns of dots help us to deduce and justify their properties. Explain that we can use the notation E(1) to mean the first even number, E(2) to mean the second even number, E(3) to mean the third even number, and so on. We can also write E1, E2, E3, … Of course, although (half of) even numbers are negative, these cannot be shown as a pattern of dots. The nth even number can be written as E(n) = 2n.

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**Odd numbers Odd numbers leave a remainder of 1 when divided by 2.**

For example, 17 is an odd number. It can be written as 17 = 2 × 8 + 1 All odd numbers end in 1, 3, 5, 7 or 9. Odd numbers can be illustrated using dots or counters arranged as follows: U(2) = 3 U(3) = 5 U(4) = 7 U(5) = 9 Explain that odd numbers can also be negative even though they cannot be shown figuratively in this way. U(1) = 1 The nth odd number can be written as U(n) = 2n – 1.

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Triangular numbers Triangular numbers are numbers that can be written as the sum of consecutive whole numbers starting with 1. For example, 15 is a triangular number. It can be written as 15 = Triangular numbers can be illustrated using dots or counters arranged in triangles: T(5) = 15 T(4) = 10 T(3) = 6 T(2) = 3 T(1) = 1

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Triangular numbers Suppose we want to know the value of T(50), the 50th triangular number. We could either add together all the numbers from 1 to 50 or we could find a rule for the nth term, T(n). If we double the number of counters in each triangular arrangement we can make rectangular arrangements: 2T(5) = 30 2T(4) = 20 T(5) = 15 2T(3) = 12 T(4) = 10 Explain that adding all the numbers from 1 to 50 would be a long task. If we knew the formula for the nth term them we could simply substitute this value The following slides show the derivation of the nth triangular number using the triangular arrangements and doubling them. 2T(2) = 6 T(3) = 6 2T(1) = 2 T(2) = 3 T(1) = 1

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Triangular numbers Any rectangular arrangement of counters can be written as the product of two whole numbers: T(5) = 15 2T(5) = 30 T(4) = 10 2T(4) = 20 T(3) = 6 2T(3) = 12 T(2) = 3 2T(2) = 6 T(1) = 1 2T(1) = 2 2T(1) = 1 × 2 2T(2) = 2 × 3 2T(3) = 3 × 4 2T(4) = 4 × 5 2T(5) = 5 × 6 Use the rectangular arrangements to help pupils see that two times a triangular number is equal to the product of the term number and the term number plus 1. Each triangular number is therefore equal to half of the product of the term number and the term number plus 1. From these arrangements we can see that 2T(n) = n(n + 1) So, for any triangular number T(n) T(n) = n(n + 1) 2

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Triangular numbers We can now use this rule to find the value of the 50th triangular number. T(n) = n(n + 1) 2 T(50) = 50(50 + 1) 2 T(50) = 50 × 51 2 T(50) = 2550 2 T(50) = 1275

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**Gauss’ method for adding consecutive numbers**

There is a story that when the famous mathematician Karl Friedrich Gauss was a young boy at school, his teacher asked the class to add up the numbers from one to a hundred. Gauss was only six years old when he worked this out. Ask the pupils what kind of number 5050 is. How did Gauss do the calculation? The teacher expected this activity to keep the class quiet for some time and so he was amazed when Gauss put up his hand and gave the answer, 5050, almost immediately!

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**Gauss’ method for adding consecutive numbers**

Gauss worked the answer out by noticing that you can quickly add together consecutive numbers by writing the numbers once in order and once in reverse order and adding them together. For example, to add the numbers from 1 to 10: 1 + 2 3 4 5 6 7 8 9 10 10 + 9 8 7 6 5 4 3 2 1 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 = 110 Sum of the numbers from 1 to 10 = 110 ÷ 2 = 55 Use this method to show that the nth triangular number is: T(n) = n(n + 1) 2

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Square numbers Square numbers are obtained when a whole number is multiplied by itself. They are sometimes called perfect squares. For example, 49 is a square number. It can be written as 49 = 7 × 7 or 49 = 72. Square numbers can be illustrated using dots or counters arranged in squares: S(5) = 25 S(4) = 16 S(3) = 9 Pupils should be able to recognise square numbers up to 152 = 225 S(2) = 4 S(1) = 1

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Making square numbers The nth square number S(n) can be written as S(n) = n2. There are several ways to generate a sequence of square numbers. We can multiply a whole number by itself. For example, 25 = 5 × 5 or 25 = 52. We can add consecutive odd numbers starting from 1. For example, 25 = We can add together two consecutive triangular numbers. For example, 25 = As a challenge ask pupils to demonstrate the last three methods for generating square numbers either algebraically or by using diagrams of dots. We can find the product of two consecutive even or odd numbers and add 1. For example, 25 = 4 ×

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**Difference between consecutive squares**

Show that the difference between two consecutive square numbers is always an odd number. If we use the general form for a square number n2, where n is a whole number, we can write the square number following it as (n + 1)2. The difference between two consecutive square numbers can therefore be written as (n + 1)2 – n2 = (n + 1)(n + 1) – n2 Explain that we can use the general forms of special numbers, such as square numbers and odd numbers as shown here, to prove certain results. = n2 + n + n + 1 – n2 = 2n + 1 2n + 1 is always an odd number for any whole number n.

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Cube numbers Cube numbers are obtained when a whole number is multiplied by itself and then by itself again. For example, 64 is a cube number. It can be written as 64 = 4 × 4 × 4 or 64 = 43. Cube numbers can be illustrated using spheres arranged in cubes: Count the number of spheres in each arrangement by multiplying the number of cubes in each layer by the total number of layers. Link to volume = length × width × height. C(1) = 1 C(2) = 8 C(3) = 27 C(4) = 64 C(5) = 125 The nth cube number C(n) can be written as C(n) = n3.

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**Squares, triangles and primes**

Discuss how the highest possible score can be achieved.

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**N1.2 Calculating with integers**

Contents N1 Integers A N1.1 Classifying numbers A N1.2 Calculating with integers A N1.3 Multiples, factors and primes A N1.4 Prime factor decomposition A N1.5 LCM and HCF

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Negative numbers A positive or negative whole number, including zero, is called an integer. For example, –3 is an integer. –3 is read as ‘negative three’. This can also be written as –3. It is 3 less than 0. Explain that in maths the ‘–’ sign has two uses. One is to denote subtraction, as in 0 – 3 which means ‘take away three’ or ‘subtract three’, and the other is to denote a negative number, as in –3 ‘negative three’ which means ‘three less than zero’. In the first case, the – sign is being used as an operation and in the second case the – sign is being used to tell us that the number is negative. We could write a + sign in front of positive numbers although this is not usually necessary. 0 – 3 = –3 Here the ‘–’ sign means minus 3 or subtract 3. Here the ‘–’ sign means negative 3. We say, ‘zero minus three equals negative three’.

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**Integers on a number line**

Positive and negative integers can be shown on a number line. –8 –3 Negative integers Positive integers We can use the number line to compare integers. Discuss what we mean when we use the terms ‘greater than’ and ‘less than’ with negative numbers. We say that –3 is ‘greater than’ –8 because it is further along the line in the positive direction. For example, –3 > –8 –3 ‘is greater than’ –8

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Adding integers We can use a number line to help us add positive and negative integers. –2 + 5 = = 3 -2 3 How can we use the number line to work out –2 + 5? Start at –2 (click to highlight the –2 on the number line) and count forwards 5. On a number line we move to the right for forwards (in a positive direction) and to the left for backwards (in a negative direction). Explain that when we add or subtract integers the signs tell us whether we move up or down the number line. Unlike multiplication and division, they do not generally tell us what sign the answer will have. This depends on the starting point and the size of the numbers. It is important to stress this fact because when pupils learn the rules for multiplying and dividing negative numbers they often confuse these rules with the rules for addition and subtraction. To add a positive integer we move forwards up the number line.

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**–3 + –4 = = –7 Adding integers**

We can use a number line to help us add positive and negative integers. –3 + –4 = = –7 -7 -3 How can we use the number line to work out –3 + –4? Explain that when we add a negative number we have to move backwards down the number line. The answer can be positive or negative depending on the starting point. In this example, we start at –3 and move 4 back down the number line. Stress that adding a negative number is equivalent to subtracting the positive value of that number. To add a negative integer we move backwards down the number line. –3 + –4 is the same as –3 – 4

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**5 – 8 = = –3 Subtracting integers**

We can use a number line to help us subtract positive and negative integers. 5 – 8 = = –3 -3 5 How can we use the number line to work out 5 – 8? Explain that this could also be written as 5 – +8 (five minus positive eight), but that we don’t usually need to write + in front of a number to show that it is positive. In this example, we start at 5 (click to make the 5 orange) and count backwards 8. To subtract a positive integer we move backwards down the number line. 5 – 8 is the same as 5 – +8

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**3 – –6 = = 9 Subtracting integers**

We can use a number line to help us subtract positive and negative integers. 3 – –6 = = 9 3 9 How can we use the number line to work out 3 – –6? Many pupils may find this difficult to understand. Explain what is happening. When we subtract 6 we move backwards down the number line, so, when we subtract -6 we we need to move forwards up the number line. “3 – –6 is equivalent to 3 + 6” You may wish to give further examples of double negatives being equivalent to positives. To subtract a negative integer we move forwards up the number line. 3 – –6 is the same as 3 + 6

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**–4 – –7 = = 3 Subtracting integers**

We can use a number line to help us subtract positive and negative integers. –4 – –7 = = 3 -4 3 Here is another example of subtracting a negative number. Again, explain carefully that subtracting a negative number is equivalent to adding. In this example, –4 is the starting point and then we move 7 forwards up the number line. Ask pupils to tell you the equivalent calculation before revealing it on the board. To subtract a negative integer we move forwards up the number line. –4 – –7 is the same as –4 + 7

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**Adding and subtracting integers**

To add a positive integer we move forwards up the number line. To add a negative integer we move backwards down the number line. a + –b is the same as a – b. To subtract a positive integer we move backwards down the number line. Stress that adding a negative integer is the same as subtracting and that subtracting a negative integer is the same as adding. Unlike multiplying and dividing integers, when adding and subtracting integers the answer is positive or negative depending on the start number and how much we move up or down the number line. To subtract a negative integer we move forwards up the number line. a – –b is the same as a + b.

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Integer circle sums Select a volunteer to come to the board and drag numbers into the intersections so the the total inside each circle is the same. Members of the class should call out suggestions. Simplify the problem by revealing the circle sum.

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**Rules for multiplying and dividing**

When multiplying negative numbers remember: + × = – + × = – + × = – + × = Dividing is the inverse operation to multiplying. When we are dividing negative numbers similar rules apply: These rules have been drawn graphically to make it easier for pupils to spot the pattern. As each rule appear read it as, for example, A positive number multiplied by a positive number always equals a positive number. Remind pupils of the meaning of ‘inverse operation’ – one ‘undoes’ the other. For example, if 4 × –3 = –12, then –12 ÷ –3 must equal 4. Tell pupils that easiest way to remember these rules is that when we multiply together (or divide) two numbers with different signs (a positive number times a negative number or a negative number times a positive number) the answer will always be negative. If we multiply together (or divide) two numbers with a different sign (a positive number times a positive number or a negative number times a negative number) the answer will always be positive. Encourage pupils to first work out whether their answers will be positive or negative and then multiply or divide. Ask pupils to write down rules for multiplying (or dividing) three numbers. For example, negative × positive × negative = positive and negative × negative × negative = negative. + ÷ = – + ÷ = – + ÷ = – + ÷ =

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**Multiplying and dividing integers**

Complete the following: –3 × 8 = –24 –36 ÷ = –4 9 42 ÷ = –6 –7 ÷ –90 = –6 540 × –8 = 96 –12 –7 × = 175 –25 For each example ask pupils what sign the missing number will have and then what the number is. 47 × = –141 –3 –4 × –5 × –8 = –160 –72 ÷ –6 = 12 3 × –8 ÷ = 1.5 –16

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Using a calculator We can enter negative numbers into a calculator by using the sign change key: (–) For example: –456 ÷ –6 can be entered as: (–) 4 5 6 ÷ = The answer will be displayed as 76. Ask pupils to locate the sign change key on their calculator. Note that on some non-scientific calculators this key may be shown as +/– and must be pressed after the number is entered to make it negative. Discuss ways that we can check that the answer given by the calculator is correct. First check that the sign is correct. In this example, we are dividing a negative number by another negative number and so the answer must be positive. 456 ÷ 6 is approximately equal to 480 ÷ 6 = 80. Always make sure that answers given by a calculator are sensible.

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Sums and products What two integers have a sum of 2 and a product of –8? Start by writing down all of the pairs of numbers that multiply together to make –8. Since –8 is negative, one of the numbers must be positive and one of the numbers must be negative. We can have: –1 × 8 = –8 Pupils will need to solve this type of problem when factorizing quadratic expressions. Explain that we start by looking for products because there are less of these. There are infinitely many pairs of integers with a sum of 2. Link: A1.4 Factorization 1 × –8 = –8 –2 × 4 = –8 or 2 × –4 = –8 –1 + 8 = 7 1 + –8 = –7 –2 + 4 = 2 2 + –4 = –2 The two integers are –2 and 4.

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Sums and products

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**N1.3 Multiples, factors and primes**

Contents N1 Integers A N1.1 Classifying numbers A N1.2 Calculating with integers A N1.3 Multiples, factors and primes A N1.4 Prime factor decomposition A N1.5 LCM and HCF

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Multiples A multiple of a number is found by multiplying the number by any whole number. What are the first six multiples of 7? To find the first six multiples of 7 multiply 7 by 1, 2, 3, 4, 5 and 6 in turn to get: Discuss the fact that any given number has infinitely many multiples. We can check whether a number is a multiple of another number by using divisibility tests. 7, 14, 21, 28, 35 and 42. Any given number has infinitely many multiples.

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Factors A factor (or divisor) of a number is a whole number that divides into it exactly. Factors come in pairs. For example, What are the factors of 30? 1 and 30, 2 and 15, 3 and 10, 5 and 6. Ask pupils to tell you what a factor is and reveal the definition on the board. Remind pupils that factors always go in pairs (in the example of rectangular arrangements these are given by the length and the width of the rectangle). The pairs multiply together to give the number. Ask pupils if numbers always have an even number of factors. They may argue that they will because factors can always be written in pairs. Establish, however, that when a number is multiplied by itself the numbers in that factor pair are repeated. That number will therefore have an odd number of factors. Pupils could investigate this individually. Establish that it follows that if a number has an odd number of factors it must be a square number. So, in order, the factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30.

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Prime numbers If a whole number has two, and only two, factors it is called a prime number. For example, the number 17 has only two factors, 1 and 17. Therefore, 17 is a prime number. The number 1 has only one factor, 1. Therefore, 1 is not a prime number. Establish the 2 is the only even prime number. Ask pupils to name all the prime numbers less than 20. There is only one even prime number. What is it? 2 is the only even prime number.

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**Prime numbers There are 25 prime numbers less than 100. These are: 2 3**

7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 Establish the 2 is the only even prime number. Ask pupils to learn the first 10 prime numbers. 73 79 83 89 97 What if we go above 100? Around 400 BC the Greek mathematician, Euclid, proved that there are infinitely many prime numbers.

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**N1.4 Prime factor decomposition**

Contents N1 Integers A N1.1 Classifying numbers A N1.2 Calculating with integers A N1.3 Multiples, factors and primes A N1.4 Prime factor decomposition A N1.5 LCM and HCF

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**Prime factors A prime factor is a factor that is a prime number.**

For example, What are the prime factors of 70? The factors of 70 are: 1 Ask for the factors of 70 before revealing them. Then ask which of these factors are prime numbers. 2 5 7 10 14 35 70 The prime factors of 70 are 2, 5, and 7.

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**Products of prime factors**

70 = 2 × 5 × 7 56 = 2 × 2 × 2 × 7 This can be written as 56 = 23 × 7 99 = 3 × 3 × 11 This can be written as 99 = 32 × 11 First of all, explain that one of the reasons prime numbers are so important is that by multiplying together prime numbers you can make any whole number bigger than one. Go through each of the products. Remind pupils of index notation and how to read it. For example, 22 is read as “2 squared” or “2 to the power of 2”. 23 is read as “2 cubed” or “2 to the power of 3” Tell pupils that every whole number greater than 1 is either a prime number or can be written as a product of two or more prime numbers. This is called the Fundamental Theorem of Arithmetic. Pupils are not expected to know this term, but it is an important idea. It is a good reason for defining prime numbers to exclude 1. If 1 were a prime, then the prime factor decomposition would lose its uniqueness. This is because we could multiply by 1 as many times as we liked in the decomposition. Number that are not prime are often called composite numbers because they are made up of products of prime numbers. Every whole number greater than 1 is either a prime number or can be written as a product of two or more prime numbers.

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**The prime factor decomposition**

When we write a number as a product of prime factors it is called the prime factor decomposition or prime factor form. For example, The prime factor decomposition of 100 is: 100 = 2 × 2 × 5 × 5 Verify that 2 × 2 × 5 × 5 = 100 = 22 × 52 There are two methods of finding the prime factor decomposition of a number.

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Factor trees 36 4 9 2 2 3 3 Explain that to write 36 as a product or prime factors we start by writing 36 at the top (of the tree). Next, we need to think of two numbers which multiply together to give 36. Ask pupils to give examples. Explain that it doesn’t matter whether we use, 2 × 18, 3 × 12, 4 × 9, or 6 × 6, the end result will be the same. Let’s use 4 × 9 this time. Next, we must find two numbers that multiply together to make 4. Click to reveal two 2s. 2 is a prime number so we draw a circle around it. Now find 2 numbers which multiply together to make 9. Click to reveal two 3s. 3 is a prime number so draw a circle around it. State that when every number at the bottom of each branch is circled we can write down the prime factor decomposition of the number writing the prime numbers in order from smallest to biggest. Ask pupil how we can we write this using index notation (powers) before revealing this. 36 = 2 × 2 × 3 × 3 = 22 × 32

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Factor trees 36 3 12 4 3 2 2 Show this alternative factor tree for 36. The prime factor decomposition is the same. 36 = 2 × 2 × 3 × 3 = 22 × 32

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Factor trees 2100 30 70 6 5 10 7 2 3 2 5 Again, explain that there are many ways to draw the factor tree for 2100 but the final factor decomposition will be the same. The prime factors are written in order of size and then simplified using index notation. 2100 = 2 × 2 × 3 × 5 × 5 × 7 = 22 × 3 × 52 × 7

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Factor trees 780 78 10 2 39 5 2 3 13 Here is another example. 780 = 2 × 2 × 3 × 5 × 13 = 22 × 3 × 5 × 13

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**Dividing by prime numbers**

2 96 2 3 2 48 96 = 2 × 2 × 2 × 2 × 2 × 3 2 24 2 12 = 25 × 3 2 6 Explain that another method to find the prime factor decomposition is to divide repeatedly by prime factors putting the answers in a table as follows: To find the prime factor decomposition of 96 start by writing 96. Click to reveal 96. Now, what is the lowest prime number that divides into 96? Establish that this is 2. Remind pupils of tests for divisibility if necessary. Any number ending in 0, 2, 4, 6, or 8 is divisible by 2. Write the 2 to the left of the 96 and then divide 96 by 2. Click to reveal the 2. This may be divided mentally. Discuss strategies such as halving 90 to get 45 and halving 6 to get 3 and adding 45 and 3 together to get 48. We write this under the 96. Now, what is the lowest prime number that divides into 48? Establish that this is 2 again. Continue dividing by the lowest prime number possible until you get to 1. When you get to 1 at the bottom, stop. The prime factor decomposition is found by multiplying together all the numbers in the left hand column. 3 3 1

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**Dividing by prime numbers**

3 5 7 3 315 315 = 3 × 3 × 5 × 7 3 105 5 35 = 32 × 5 × 7 7 7 Talk through this example as before. Remind pupils that to test for divisibility by 3 we must add together the digits and check whether the result is divisible by 3. 1

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**Dividing by prime numbers**

2 702 2 3 13 3 351 702 = 2 × 3 × 3 × 3 × 13 3 117 3 39 = 2 × 33 × 13 13 13 Here is another example. Ask pupils to find the prime factor decomposition of given numbers. 1

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**Using the prime factor decomposition**

Use the prime factor form of 324 to show that it is a square number. 2 324 2 3 324 = 2 × 2 × 3 × 3 × 3 × 3 2 162 = 22 × 34 3 81 This can be written as: (2 × 32) × (2 × 32) 3 27 or (2 × 32)2 If necessary break this down further to show that 324 = (2 × 3 × 3) × (2 × 3 × 3). Ask pupils to use the information on the board to tell you the square root of 324. 324 = 2 × 32 = 2 × 9 = 18. 3 9 If all the indices in the prime factor decomposition of a number are even, then the number is a square number. 3 3 1

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**Using the prime factor decomposition**

Use the prime factor form of 3375 to show that it is a cube number. 3 3375 3375 = 3 × 3 × 3 × 5 × 5 × 5 3 5 = 33 × 53 3 1125 This can be written as: 3 375 (3 × 5) × (3 × 5) × (3 × 5) 5 125 or (3 × 5)3 Ask pupils to use the information on the board to tell you the cube root of 3375. 33375 = 3 × 5 = 15 5 25 If all the indices in the prime factor decomposition of a number are multiples of 3, then the number is a cube number. 5 5 1

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**Using the prime factor decomposition**

168 = 23 × 3 × 7 4116 = 22 × 3 × 73 294 = 2 × 3 × 72 Use the prime factor decompositions of the numbers given above to answer the following questions. 1) What is 168 × 294 as a product of prime factors? 168 × 294 = (23 × 3 × 7) × (2 × 3 × 72 ) Recall that to multiply two numbers with the same base written in index form we add the indices. If required, add another two lines to write 168 × 294 = 2 × 2 × 2 × 3 × 7 × 2 × 3 × 7 × 7 and then rearranging to write 168 × 294 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7. Link: N2.3 Index notation = 23 × 2 × 3 × 3 × 7 × 72 = 24 × 32 × 73

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**Using the prime factor decomposition**

168 = 23 × 3 × 7 4116 = 22 × 3 × 73 294 = 2 × 3 × 72 Use the prime factor decompositions of the numbers given above to answer the following questions. 2) What is 4116 ÷ 294? 4116 ÷ 294 = 22 × 3 × 73 2 × 3 × 72 Recall that to divide two numbers with the same base written in index form we subtract the indices. Link: N2.3 Index notation = 2 × 2 × 3 × 7 × 7 × 7 2 × 3 × 7 × 7 = 2 × 7 = 14

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**Using the prime factor decomposition**

168 = 23 × 3 × 7 4116 = 22 × 3 × 73 294 = 2 × 3 × 72 Use the prime factor decompositions of the numbers given above to answer the following questions. 3) Is 4116 divisible by 168? If we divide 4116 by 168 we have: 4116 ÷ 168 = 22 × 3 × 73 23 × 3 × 7 If 4116 were exactly divisible by 168 then the prime factor decomposition of 168 would not contain prime numbers raised to higher powers than those in the prime factor decomposition of is not divisible by 168 because the prime factor decomposition contains 23, while the prime factor decomposition of 4116 contains 22. = 2 × 2 × 3 × 7 × 7 × 7 2 × 2 × 2 × 3 × 7 There is a 2 left in the denominator No, 4116 is not divisible by 168.

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**Using the prime factor decomposition**

168 = 23 × 3 × 7 4116 = 22 × 3 × 73 294 = 2 × 3 × 72 Use the prime factor decompositions of the numbers given above to answer the following questions. 4) Show that 294 × 6 is a square number. We can write 6 as 2 × 3 294 × 6 = 2 × 3 × 72 × 2 × 3 Reveal the solution on the slide and ask pupils to use this to tell you the square root of 294 × 6. Rearranging, 294 × 6 = 2 × 2 × 3 × 3 × 72 = 22 × 32 × 72 = (2 × 3 × 7)2

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**Using the prime factor decomposition**

168 = 23 × 3 × 7 4116 = 22 × 3 × 73 294 = 2 × 3 × 72 Use the prime factor decompositions of the numbers given above to answer the following questions. 168 294 5) Write the fraction in its simplest form. 168 294 = 23 × 3 × 7 2 × 3 × 72 2 × 2 × 2 × 3 × 7 2 × 3 × 7 × 7 = = 4 7

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**N1 Integers Contents A N1.1 Classifying numbers A**

N1.2 Calculating with integers A N1.3 Multiples, factors and primes A N1.4 Prime factor decomposition A N1.5 LCM and HCF

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**The lowest common multiple**

The lowest common multiple (or LCM) of two numbers is the smallest number that is a multiple of both the numbers. For small numbers we can find this by writing down the first few multiples for both numbers until we find a number that is in both lists. For example, Multiples of 20 are : 20, 40, 60, 80, 100, 120, . . . You may like to add that if the two numbers have no common factors (except 1) then the lowest common multiple of the two numbers will be the product of the two numbers. For example, 4 and 5 have no common factors and so the lowest common multiple of 4 and 5 is 4 × 5, 20. Pupils could also investigate this themselves later in the lesson. Multiples of 25 are : 25, 50, 75, 100, 125, . . . The LCM of 20 and 25 is 100.

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**The lowest common multiple**

We use the lowest common multiple when adding and subtracting fractions. For example, Add together 4 9 5 12 and The LCM of 9 and 12 is 36. × 4 × 3 Establish by asking for multiples that the lowest common multiple of 9 and 12 is 36. Remind pupils that to add two fractions together they must have the same denominator. The LCM is the lowest number that both 9 and 12 will divide into. Ask pupils how many 9s ‘go into’ 36. Establish that we must multiply 9 by 4 to get 36 before revealing the first arrow. Now, we’ve multiplied the bottom by 4 so we must multiply the top by 4. Remember, if you multiply the top and the bottom of a fraction by the same number, you do not change its value. 16/36 is just another way of writing 4/9. Repeat this explanation as you convert 5/12 to 15/36. 16/36 plus 15/36 equals 31/36. Can this fraction be simplified? Establish that it cannot. + 4 9 5 12 = 16 15 31 36 + 36 = 36 × 4 × 3

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**The highest common factor**

The highest common factor (or HCF) of two numbers is the highest number that is a factor of both numbers. We can find the highest common factor of two numbers by writing down all their factors and finding the largest factor in both lists. For example, Factors of 36 are : 1, 2, 3, 4, 6, 9, 12, 18, 36. Point out that 3 is a common factor of 36 and 45. So is 1. But 9 is the highest common factor. Factors of 45 are : 1, 3, 5, 9, 15, 45. The HCF of 36 and 45 is 9.

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**The highest common factor**

We use the highest common factor when cancelling fractions. For example, Cancel the fraction 36 48 The HCF of 36 and 48 is 12, so we need to divide the numerator and the denominator by 12. Talk through the use of the highest common factor to cancel fractions in one step. ÷12 36 48 3 = 4 ÷12

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**Using prime factors to find the HCF and LCM**

We can use the prime factor decomposition to find the HCF and LCM of larger numbers. For example, Find the HCF and the LCM of 60 and 125. 2 60 2 294 2 30 3 147 Recap on the method of dividing by prime numbers introduced in the previous section. 3 15 7 49 5 5 7 7 1 1 60 = 2 × 2 × 3 × 5 294 = 2 × 3 × 7 × 7

61
**Using prime factors to find the HCF and LCM**

60 = 2 × 2 × 3 × 5 294 = 2 × 3 × 7 × 7 60 294 2 7 2 3 5 7 We can find the HCF and LCM by using a Venn diagram. We put the prime factors of 60 in the first circle. Any factors that are common to both 60 and 294 go into the overlapping section. Click to demonstrate this. Point out that we can cross out the prime factors that we have included from 294 in the overlapping section to avoid adding then twice. We put the prime factors of 294 in the second circle. The prime factors which are common to both 60 and 294 will be in the section where the two circles overlap. To find the highest common factor of 60 and 294 we need to multiply together the numbers in the overlapping section. The lowest common multiple is found by multiplying together all the prime numbers in the diagram. HCF of 60 and 294 = 2 × 3 = 6 LCM of 60 and 294 = 2 × 5 × 2 × 3 × 7 × 7 = 2940

62
**Using prime factors to find the HCF and LCM**

Ask volunteers to take turns to find the LCM and HCF using the Venn diagram.

63
**The LCM of co-prime numbers**

If two numbers have a highest common factor (or HCF) of 1 then they are called co-prime or relatively prime numbers. For two whole numbers a and b we can write: a and b are co-prime if HCF(a, b) = 1 If two whole numbers a and b are co-prime then: LCM(a, b) = ab Explain that if two numbers are co-prime, that is they do not share any common factors other than 1, their LCM is equal to the product of the two numbers. This rule is also true if more than two numbers are co-prime. For example, the LCM of 3, 5 and 8 is 3 × 5 × 8 = 120. For example, the numbers 8 and 9 do not share any common multiples other than 1. They are co-prime. Therefore, LCM(8, 9) = 8 × 9 = 72

64
**The LCM of numbers that are not co-prime**

If two numbers are not co-prime then their highest common factor is greater than 1. If two numbers a and b are not co-prime then their lowest common multiple is equal to the product of the two numbers divided by their highest common factor. We can write this as: LCM(a, b) = ab HCF(a, b) Repeat the activity on slide 62 verifying using this rule to check each solution. Rearranging this formula, we can also say that the LCM(a,b) × HCM(a,b) = ab. In other words, for any two whole numbers a and b, the LCM of a and b multiplied by the HCF of a and b equals the product of a and b. This rule is true for any two whole numbers. If the two numbers are co-prime we have LCM (a,b) = ab/1, as on the previous slide. For example, 8 × 12 HCF(8, 12) = 96 4 LCM(8, 12) = = 24

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