New Mathematics Connection P.1-P.6 sample chapter (update)

From 8 = 1 × 8, we can say that 8 can be divided exactly by 1 and 8.
Thus, 1, 2, 4 and 8 are exact divisors of 8. They are called the factors of 8.

Example 1

Find all the factors of 4.

Solution:
4 ÷ 1 = 4 with no remainder. Therefore, 1 is a factor of 4.
4 ÷ 2 = 2 with no remainder. Therefore, 2 is a factor of 4.
4 ÷ 3 = 1 with a remainder of 1. Therefore, 3 is not a factor of 4.
4 ÷ 4 = 1 with no remainder. Therefore, 4 is a factor of 4.

Answer: The factors of 4 are 1, 2 and 4.

Example 2 Example 3

Write all the factors of 12. Write all the factors of 36.

Solution: Solution:

12 = 1 × 12 36 = 1 × 36
12 = 2 × 6 36 = 2 × 18
12 = 3 × 4 36 = 3 × 12
We can stop here because 3 and 4 are 36 = 4 × 9
consecutive numbers. 36 = 6 × 6
Answer: The factors of 12 are 1, 2, 3, 4, We can stop here because both
6 and 12. factors are the same.
Answer: The factors of 36 are 1, 2, 3, 4, 6, 9, 12,
18 and 36.

We can observe the following facts about factors:

New Mathematics Connection 3

Prime Numbers
The following table shows the factors and number of factors of the numbers from 1 to 12.

Numbers Factors Number of Factors
1 1 1
2 1, 2 2
3 1, 3 2
4 3
5 1, 2, 4 2
6 1, 5 4
7 1, 2, 3, 6 2
8 1, 7 4
9 1, 2, 4, 8 3
10 1, 3, 9 4
11 1, 2, 5, 10 2
12 1,11 6
1, 2, 3, 4, 6, 12

We find that

(a) Some numbers have exactly two factors.The numbers which have only two factors,
1 and the number itself, are called ‘prime numbers’.

For example: 2, 3, 5, 7, 11

(b) The numbers which have more than two factors are called ‘composite numbers’.
For example: 4, 6, 8, 10

(c) The number 1 has only one factor. So, it is neither a prime number nor a composite
number.

Example 4

Determine whether 13 is a prime number.

Solution:
Let us find the factors of 13.
13 can only be divided by 13 and 1. Therefore, 1 and 13 are the only factors of 13.
Answer: 13 is a prime number.

4 Chapter 1 Factors and Multiples

Prime Factorisation

When a number is expressed as a product of its factors, we say that the number has been
prime factorised.
Consider the number 24.
24 = 2 × 12
We say that 24 has been factorised. This is one of the factorisations of 24.
The others are: 24 = 3 × 8; 24 = 4 × 6

Now, let us learn to find the prime factors of 24.

Method 1: 24 = 3 × 8 24 = 4 × 6
24 = 2 × 12

=2× 2 × 6 =3 × 2 × 4 =2×2×2×3

=2×2×2×3 =3×2×2×2

Method 2:

2 24
2 12
26
33

1

In both methods, we finally reach at only one factorisation 2 × 2 × 2 × 3. In this factorisation, the
only factors 2 and 3 are prime numbers.This is called the prime factorisation of a number.

Example 5

Find the prime factorisation of 40.

Solution:

Method 1: Method 2:

40 = 4 × 10 2 40
2×2×2×5 2 20
2 10

5

Thus, the prime factorisation of 40 is 2 × 2 × 2 × 5.

New Mathematics Connection 5

Try It

1. Find the factors of the following numbers.

(a) 28 (b) 54 (c) 35

(d) 26 (e) 44 (f) 60

2. Circle the prime numbers.

(a) 2, 4, 6 (b) 12, 9, 7 (c) 13, 15, 19
(f) 41, 57, 81
(d) 31, 38, 42 (e) 48, 54, 43

3. Write all the prime numbers less than 50.

4. Find the prime factorisation of the following numbers.

(a) 25 (b) 18 (c) 30

(d) 45 (e) 27 (f) 72

Go to Exercise 1

B Greatest Common Divisor (GCD)

Common Prime Factors
Let us find the common prime factors of 6 and 8.

The factors of 6 are 1, 2, 3 and 6.

The factors of 8 are 1, 2, 4 and 8.
Therefore, 2 is the common prime factor of 6 and 8.

6 Chapter 1 Factors and Multiples

Example 6
What is the common prime factors of 9 and 18?

Solution:

The factors of 9 are 1, 3 and 9.

The factors of 18 are 1, 2, 3, 6, 9 and 18.
1, 3 and 9 are present in both the factors of 9 and 18. Therefore, they are the common factors
of 9 and 18.
Answer: The common prime factor of 9 and 18 is 3.

The Greatest Common Divisor or GCD, of two or more
numbers is the greatest of their common factors. It is
also known as the Highest Common Factor or HCF.

Fun Facts

A number is divisible by 2 if its last digit is 0 or even.

New Mathematics Connection 7

Finding GCD Using Prime Factorisation Method
Let us find the GCD of 15 and 25.

15 = 3 × 5
25 = 5 × 5
Therefore, the GCD of 15 and 25 is 5.

Example 7
Find the GCD of 10, 14 and 18.

Solution:
10 = 2 × 5
14 = 2 × 7
18 = 2 × 3 × 3
Answer: The GCD of 10, 14 and 18 is 2.

Example 8
Find the GCD of 28, 36 and 48.

Solution:
28 = 2 × 2 × 7
36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
The GCD is 2 × 2 = 4.
Answer: The GCD of 28, 36 and 48 is 4.

Think Aloud

Is 10 a factor of 5? Is 2 a multiple of 8?

8 Chapter 1 Factors and Multiples

Finding GCD Using Short Division

We can also find GCD of numbers using short division. Divide the given number by the prime number
until there is no prime number divisor.Then, find the product of the common divisors.

Example 9

Find the GCD of 16, 28 and 44.

Solution:
Step 1: Find all the prime factors of 16, 28 and 44.

2 16 28 44 We cannot carry on further because there is
2 8 14 22 no common factors besides 1.

4 7 11

Step 2: Find the product of all the common prime factors.
GCD = 2 × 2 = 4.

Therefore, the GCD of 16, 28 and 44 is 4.

Try It

1. Find the common factors. (b) 38 and 57
(a) 6 and 16 (d) 11 and 121
(c) 12 and 64 (f) 65 and 104
(e) 5 and 15

2. Find the GCD using prime factorisation method.

(a) 16 and 20 (b) 12 and 48

(c) 15, 35 and 50 (d) 25, 45 and 90

3. Find the GCD using short division. (b) 25 and 45
(a) 18 and 30 (d) 12, 24 and 32
(c) 26, 34 and 42

Go to Exercise 2

New Mathematics Connection 9

C Lowest Common Multiple (LCM)

When we write 18 as 18 = 2 × 9, we say that 2 and 9 are factors of 18. We can also say that 18 is
a multiple of 2 and 9.
The first 10 multiples of 2 are shown in the box:

Thus, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, … are multiples of 2.
Each of these multiples is greater than or equal to 2,
and the list is endless.
We can observe the following facts about multiples:

Example 10
Find the first five multiples of 7.

Solution:
1×7= 7
2 × 7 = 14
3 × 7 = 21
4 × 7 = 28
5 × 7 = 35
The first five multiples of 7 are 7, 14, 21, 28 and 35.

10 Chapter 1 Factors and Multiples

Common Multiples of Two Numbers

Let us find the multiples of 3 and 5.

Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30,...
Multiples of 5 are 5, 10, 15, 20, 25, 30, 35,...
We can find that 15 and 30 are multiples of both 3 and 5.
15 and 30 are called the common multiples of 3 and 5.

Example 11

Find the first three common multiples of 2 and 4.
Solution:
Multiples of 2 are 2, 4, 6, 8, 10, 12, ...
Multiples of 4 are 4, 8, 12, 16, 20, 24,...

The first three common multiples of 2 and 4 are 4, 8 and 12.

Finding LCM Using the Multiples

The smallest among the common multiples of two or more
numbers is called the Lowest Common Multiple or (LCM).

Example 12

Find LCM of 12 and 20.

Solution:
Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120 ,...
Multiples of 20 are 20, 40, 60, 80, 100, 120 ,...
Common multiples are 60 and 120.
The least common multiple is 60.
The LCM of 12 and 20 is 60.

New Mathematics Connection 11

Finding LCM Using the Prime Factorisation Method
You can also use the prime factorisation method to find the lowest common multiple.

Example 13
Find LCM of 6 and 8.

Solution:
6= 2 × 3
8= 2 × 2 × 2
LCM of 6 and 8 is 2 × 2 × 2 × 3 = 24
Count the number of times each prime factor appears in both numbers.
The product of those factors is the LCM.
Answer: The LCM of 6 and 8 is 24.

Example 14
Find the LCM of 14, 21 and 9.

Solution:
14 = 2 × 7
21 = 3 × 7
9= 3 × 3
LCM of 14, 21 and 9 is 3 × 7 × 2 × 3 = 126
Answer: The L.C.M of 14, 21 and 9 is 126.

Finding LCM Using Short Division
We can also use short division to find the LCM of numbers.

12 Chapter 1 Factors and Multiples

Example 15

Find the LCM of 12, 40 and 48.

Solution:

Step 1: Divide each number by their common prime factors.

2 12 40 48
2 6 20 24

3 10 12

Step 2: Divide each pair of numbers by their common prime factors.

2 12 40 48 Stop when there is no common factors beside 1.
2 6 20 24
2 3 10 12
335 6

152

Step 3: Multiply all the prime factors to get the LCM.
LCM = 2 × 2 × 2 × 3 × 1 × 5 × 2 = 240

Therefore, the LCM of 12, 40 and 48 is 240.

Try It

1. Find the LCM for the following using any one of the methods.

(a) 21 and 24 (b) 33 and 24
(c) 15 and 40 (d) 8, 15 and 40
(e) 34, 48 and 56 (f) 16, 18 and 20

2. Write the first five multiples of the following numbers.

(a) 5 (b) 13

(c) 19 (d) 70

(e) 15 (f) 22

New Mathematics Connection 13

3. Find.
(a) The common multiples of 4 and 10.
(b) The first three common multiples of 6 and 12.
(c) The first two common multiples of 5, 6 and 10.
(d) The first two common multiples of 20 and 30.
(e) The first three common multiples of 7, 9 and 15.

4. Match the numbers in Group A with the correct answers given in Group B.

Group A Group B
(a) 15 Multiple of 8
(b) 25 Factor of 60
(c) 16 Factor of 25
(d) 35 Multiple of 17
(e) 51 Multiple of 7

5. Write true or false.
(a) The first five multiples of 3 are 3, 6, 18, 21 and 24.
(b) The first two common multiples of 5 and 10 is 20.
(c) 24 is a multiple of both 3 and 8.
(d) The LCM of 15, 20 and 25 is 80.
(e) Is 17 a factor of 34?

Go to Exercise 3

14 Chapter 1 Factors and Multiples

D Word Problems Using GCD and LCM

Example 16
A florist has 36 roses, 27 orchids and 18 daisies to make bouquets. What is the largest number of
bouquets that can be made without having any flowers left over? And how many flowers for each
kind are there in each bouquet?

Solution:

To find the largest number of bouquets that can be made without
having any flowers left, we have to find the GCD of the numbers.
The remaining factors in each number give the number of flowers of

each variety to be put in a bouquet.

Let us find the prime factors of each of the numbers.
Roses: 36 = 2 × 2 × 3 × 3
Orchids: 27 = 3 × 3 × 3
Daisies: 18 = 2 × 3 × 3

GCD = 3 × 3 = 9
Therefore, the number of bouquets that can be made is 9.

Answer: From the given flowers, 9 bouquets can be made and each bouquet will have 4 roses,
3 orchids and 2 daisies.

New Mathematics Connection 15

Example 17

Tom and his three friends decide to share a pack of 6 cookies. What is the least number of packets
that Tom would buy, so that each of them will get the same number of cookies and none are left.

Solution:

Cookies have to be shared among 4 people.
A pack contains 6 cookies.
Let us find the LCM of 4 and 6.

4= 2 ×2
6= 2 ×3
So, LCM of 4 and 6 is 2 × 2 × 3 = 12.
Therefore, 12 cookies are needed.The number of packs of cookies 12 ÷ 6 = 2.
Answer: The number of packs that Tom would buy is 2.

Try It

1. Solve the following word problems.
(a) What is the least number of pencils needed to make groups of 3, 4 or 9 pencils with none
left over?
(b) What is the smallest number that is divisible by 5, 6 and 15?
(c) Jill bought some apples to give her friends. If she gave 4 or 5 or 6 apples to each of her
friends she will be left with 100 apples. What is the least number of apples she bought?
(d) What is the least number of students needed so that they can be arranged equally in
rows of 12, 16 or 20?
(e) Two cans contain 60 litres and 165 litres of milk respectively. Find the largest capacity of a
containing which can measure the milk in the two cans without any quantity of milk left
over.
(f) Three pieces of timber 42 metres, 49 metres and 63 metres long have to be divided into
pieces of the same length. What is the greatest possible length of each piece?

Go to Exercise 4

16 Chapter 1 Factors and Multiples