Chapter 2: Highest common factor and lowest common multiple

2.1 Highest common factor

Factors

A factor is any of the two or more whole numbers we multiply to get a product. For example: $6 \times 2 = 12$ . The numbers 6 and 2 are factors of 12.

A factor is also described as a whole number that divides into a larger number without a remainder. For example: $\frac{12}{6} = 2$ . The number 6 divides perfectly into 12. Therefore, the number 6 is a factor of 12.

factor A factor is a whole number that divides into a larger number without a remainder.

Worked example 2.1: Factors

Find all the factors of the number 60.

Step 1: Any number larger than 1 has the number itself and 1 as factors.
$1 \times 60=60$
Step 2: Start with 2 and see if it divides into 60 without a remainder.
$\frac{60}{2} = 30$ $2 \times 30=60$
Step 3: Try each consecutive number to see if it divides into 60 without a remainder. Stop when you get to a number that you already have as an answer.
\begin{array}{2} \frac{60}{3} = 20 \hspace{10mm} &3 \times 20 = 60 \newline \frac{60}{4} = 15 &4 \times 15 = 60 \newline \frac{60}{5} = 12 &5 \times 12 = 60 \newline \frac{60}{6} = 10 &6 \times 10 = 60 \newline \end{array}
The numbers 7, 8 and 9 do not divide into 60 without a remainder, so they are not factors of 60.

You already have the number 10, as 60 divided by 6 is 10, so you can stop there.
$\frac{60}{10} = 6$
Step 4: List all the factors from the smallest number to the largest number.

The factors of 60 are: 1; 2; 3; 4; 5; 6; 10; 12; 15; 20; 30; 60

Exercise 2.1: Find factors

Find all the factors of the following numbers:

54

\begin{array}{2} \text{1 and itself} \hspace{10mm}&1 \times 54=54 \newline \frac{54}{2} = 27 &2 \times 27 = 54 \newline \frac{54}{3} = 18 &3 \times 18 = 54 \newline \frac{54}{6} = 9 & 6 \times 9 = 54 \newline \frac{54}{9} = 6 & \text{already divided} \end{array}
The factors of 54 are: 1; 2; 3; 6; 9; 18; 27; 54
39

\begin{array}{2} \text{1 and itself} \hspace{10mm}&1 \times 39=39 \newline \frac{39}{3} = 13 & 3 \times 13 = 39 \newline \frac{39}{13} = 3 & \text{already divided} \end{array}
The factors of 39 are: 1; 3; 13; 39
45

\begin{array}{2} \text{1 and itself} \hspace{10mm}&1 \times 45=45 \newline \frac{45}{3} = 15 & 3 \times 15 = 45 \newline \frac{45}{5} = 9 & 5 \times 9 = 45 \newline \frac{45}{9} = 5 & \text{already divided} \end{array}
The factors of 45 are: 1; 3; 5; 9; 15; 45

Prime factors

The number 3 has only two factors: 1 and 3. Similarly, the number 11 has only two factors: 1 and 11. These are examples of prime numbers.

prime number A prime number is a number that has only two factors: 1 and the number itself.

The prime numbers between 1 and 100 are highlighted in the grid below.

prime factors These are the factors of a number that are also prime numbers. Any non-prime number can be broken down into prime factors. We can then write the number as a product of its prime factors.

prime factor A prime factor is a factor that is also a prime number.

Worked example 2.2: Prime factors

Find the prime factors of 80.

Step 1: Find the smallest prime number that is a factor of 80. Express 80 as a product of two factors.
$\frac{80}{2}=40$ $80=2 \times 40$
Step 2: Find the smallest prime number that is a factor of the non-prime factor from Step 1. Express 80 as a product of three factors.
$\frac{40}{2}=20$ $80=2 \times 2 \times 20$
Step 3: Repeat Step 2 until 80 is expressed as a product of only prime numbers.
$\frac{20}{2}=10$ $80=2 \times 2 \times 2 \times 10$ $\frac{10}{2}=5$ $80=2 \times 2 \times 2 \times 5$
Remember that 1 is not a prime number.

Exercise 2.2: Express numbers as products of prime factors

Express the following numbers as products of prime factors:

30

\begin{array}{2} & \frac{30}{2}=15 &\hspace{10 mm}30=2 \times 15 \newline & \frac{15}{3}=5 &\hspace{10mm} 30=2 \times 3 \times 5 \end{array}
140

\begin{array}{2} & \frac{140}{2}=70 &\hspace{10 mm}140=2 \times 70 \newline & \frac{70}{2}=35 &\hspace{10mm} 140=2 \times 2 \times 35 \newline & \frac{35}{5}=7 &\hspace{10mm} 140=2 \times 2 \times 5 \times 7 \newline \end{array}
195

\begin{array}{2} & \frac{195}{3}=65 &\hspace{10 mm}195=3 \times 65 \newline & \frac{65}{5}=13 &\hspace{10mm} 195=3 \times 5 \times 13 \newline \end{array}
252

\begin{array}{2} & \frac{252}{2}=126 &\hspace{10 mm}252=2 \times 126 \newline & \frac{126}{2}=63 &\hspace{10mm} 252=2 \times 2 \times 63 \newline & \frac{63}{3}=21 &\hspace{10mm} 252=2 \times 2 \times 3 \times 21 \newline & \frac{21}{3}=7 &\hspace{10mm} 252=2 \times 2 \times 3 \times 3 \times 7 \newline \end{array}
297

\begin{array}{2} & \frac{297}{3}=99 &\hspace{10 mm}297=3 \times 99 \newline & \frac{99}{3}=33 &\hspace{10mm} 297=3 \times 3 \times 33 \newline & \frac{33}{3}=11 &\hspace{10mm} 297=3 \times 3 \times 3 \times 11 \newline \end{array}

Index form

Finding and writing down the prime factors of large numbers can become confusing. To make the process easier, we may use continued division and write the prime factors in index form. The index or power of a number tells us how many times we multiply the number by itself. For example, $2^4$ means $2 \times 2 \times 2 \times 2$ . In this case, 4 is the power or index of 2.

index form Index form tells you how many times to use the same number in a multiplication by writing a superscript next to the number, for example $3^2$ means multiply 3 by 3.

An index of 1 means a number is not multiplied by itself. Therefore, $2$ and $2^1$ have the same meaning.

Worked example 2.3: Prime factors in index form

Find the prime factors of 180 and write them in index form.

Step 1: Divide the smallest prime factor of 180 until it cannot divide any further.
\begin{array}{r | r} 2 & 180 \newline \hline 2 & 90 \newline \hline & 45 \end{array}
Step 2: Move to the next prime factor of 180. Keep on dividing until it cannot divide any further.
\begin{array}{r | r} 2 & 180 \newline \hline 2 & 90 \newline \hline 3 & 45 \newline \hline 3 & 15 \newline \hline & 5 \end{array}
Step 3: Repeat Step 2 until you get 1 as an answer.
\begin{array}{r | r} 2 & 180 \newline \hline 2 & 90 \newline \hline 3 & 45 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array}
Step 4: Write down all the prime factors of 180. Use superscripts to write the index of each prime factor, that is, how many times the prime factor divided.
$180 = 2^2 \times 3^2 \times 5$
The index of each prime factor shows us how many times we must multiply each prime factor to get 180 again:
$180 = 2^2 \times 3^2 \times 5 = 2 \times 2 \times 3 \times 3 \times 5$

Exercise 2.3: Express numbers as products of prime factors in index form

Express the following numbers as prime factors in index form:

60

\begin{array}{r | r} 2 & 60 \newline \hline 2 & 30 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $60=2^2 \times 3 \times 5$
144

\begin{array}{r | r} 2 & 144 \newline \hline 2 & 72 \newline \hline 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $144=2^4 \times 3^2$
198

\begin{array}{r | r} 2 & 198 \newline \hline 3 & 99 \newline \hline 3 & 33 \newline \hline 11 & 11 \newline \hline & 1 \end{array} $198=2\times 3^2 \times 11$
300

\begin{array}{r | r} 2 & 300 \newline \hline 2 & 150 \newline \hline 3 & 75 \newline \hline 5 & 25 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $300=2^2 \times 3\times 5^2$
3 375

\begin{array}{r | r} 3 & 3\space 375 \newline \hline 3 & 1\space 125 \newline \hline 3 & 375 \newline \hline 5 & 125 \newline \hline 5 & 25 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $3\space375=3^3 \times 5^3$

Common factors

Two numbers have a common factor if the same number is a factor of both numbers. For example, the number 2 divides into both 10 and 12 without a remainder. Therefore, 2 is a common factor of 10 and 12.

common factor A common factor is a factor of two or more numbers.

Worked example 2.4: Common factors

Find the common factors of 12 and 18.

Step 1: List all the factors for both numbers, from the smallest factor to the largest factor.

The factors of 12 are: 1; 2; 3; 4; 6; 12
The factors of 18 are: 1; 2; 3; 6; 9; 18
Step 2: Identify the numbers that appear in both lists.

The numbers 1, 2 and 3 are the common factors of 12 and 18.

The number 1 is a common factor of all whole numbers.

Exercise 2.4: Find common factors

Find the common factors of the following sets of numbers:

15 and 27

The factors of 15 are: 1; 3; 5; 15
The factors of 27 are: 1; 3; 9; 27

The common factors are 1 and 3.
30 and 45

The factors of 30 are: 1; 2; 3; 5; 6; 10; 15; 30
The factors of 45 are: 1; 3; 5; 9; 15; 45

The common factors are 1, 3, 5 and 15.
10, 20 and 30

The factors of 10 are: 1; 2; 5; 10
The factors of 20 are: 1; 2; 4; 5; 10; 20
The factors of 30 are: 1; 2; 3; 5; 6; 10; 15; 30

The common factors are 1, 2, 5 and 10.

The highest common factor (HCF) of two numbers is the largest number that is a factor of both. For example, the common factors of 12 and 16 are 1, 2 and 4. The number 4 is the largest number on the list, so it is the highest common factor.

highest common factor (HCF) The highest common factor is the largest whole number that divides into two or more numbers without a remainder.

Exercise 2.5: Find the HCF

In Exercise 2.4, you determined the common factors of the following sets of numbers. Now identify the HCF in each case.

15 and 27

The common factors are 1 and 3.
$\therefore \$ HCF = 3
30 and 45

The common factors are 1, 3, 5 and 15.
$\therefore$ HCF = 15
10, 20 and 30

The common factors are 1, 2, 5 and 10.
$\therefore$ HCF = 10

Worked example 2.5: HCF using prime factors

Find the HCF of 60 and 80.

Step 1: Express each number as a product of prime factors.
\begin{array}{r | r} 2 & 60 \newline \hline 2 & 30 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $60=2 \times 2 \times 3 \times 5$ \begin{array}{r | r} 2 & 80 \newline \hline 2 & 40 \newline \hline 2 & 20 \newline \hline 2 & 10 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $80=2 \times 2 \times 2 \times 2 \times 5$
Step 2: Identify all the prime factors that are contained in both numbers.

The number 2 appears twice for 80 and 60: $2 \times 2$
The number 3 does not appear for 80
The number 5 appears once for 80 and 60: $5$
Step 3: Multiply the prime factors you identified to get the HCF.
$\text{HCF} = 2 \times 2 \times 5 = 20$

Worked example 2.6: HCF using prime factors in index form

Find the HCF of 90 and 108.

Step 1: Express each number as a product of prime factors in index form.
\begin{array}{r | r} 2 & 90 \newline \hline 3 & 45 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $90=2 \times 3^2 \times 5$ \begin{array}{r | r} 2 & 108 \newline \hline 2 & 54 \newline \hline 3 & 27 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $108=2^2 \times 3^3$
Step 2: Identify the prime factors that are contained in both numbers, to the lowest index.

The number 2 divided into both 90 and 108. The lowest index both numbers have as a factor is one: $2$
The number 3 divided into both 90 and 108. The lowest index both numbers have as a factor is two: $3^2$
Step 3: Multiply the prime factors you identified to get the HCF.
$\text{HCF} = 2 \times 3^2=2 \times 3 \times 3 = 18$

Exercise 2.6: Use different methods to find the HCF

For the following sets of numbers, find all the factors of each number and then find the HCF.
1. 15 and 80
$\begin{array}{1} 1 \times 15 = 15 \newline 3 \times 5 = 15 \end{array}$
Factors of 15: 1; 3; 5; 15
$\begin{array}{1} 1 \times 80 = 80 \newline 2 \times 40 = 80 \newline 4 \times 20 = 80 \newline 5 \times 16 = 80 \newline 8 \times 10 = 80 \end{array}$
Factors of 80: 1; 2; 4; 5; 8; 10; 16; 20; 40; 80

$\therefore$ HCF = 5
1. 40 and 64
$\begin{array}{1} 1 \times 40 = 40 \newline 2 \times 20 = 40 \newline 4 \times 10 = 40 \newline 5 \times 8 = 40 \end{array}$
Factors of 40: 1; 2; 4; 5; 8; 10; 20; 40
$\begin{array}{1} 1 \times 64 = 64 \newline 2 \times 32 = 64 \newline 4 \times 16 = 64 \newline 8 \times 8 = 64 \newline \end{array}$
Factors of 64: 1; 2; 4; 8; 16; 32; 64

$\therefore$ HCF = 8
1. 50 and 70
$\begin{array}{1} 1 \times 50 = 50 \newline 2 \times 25 = 50 \newline 5 \times 10 = 50 \newline \end{array}$
Factors of 50: 1; 2; 5; 10; 25; 50
$\begin{array}{1} 1 \times 70 = 70 \newline 2 \times 35 = 70 \newline 5 \times 14 = 70 \newline 7 \times 10 = 70 \newline \end{array}$
Factors of 70: 1; 2; 5; 7; 10; 14; 35; 70

$\therefore$ HCF = 10
1. 18, 24 and 32
$\begin{array}{1} 1 \times 18 = 18 \newline 2 \times 9 = 18 \newline 3 \times 6 = 18 \newline \end{array}$
Factors of 18: 1; 2; 3; 6; 9; 18
$\begin{array}{1} 1 \times 24 = 24 \newline 2 \times 12 = 24 \newline 3 \times 8 = 24 \newline 4 \times 6 = 24 \newline \end{array}$
Factors of 24: 1; 2; 3; 4; 6; 8; 12; 24
$\begin{array}{1} 1 \times 32 = 32 \newline 2 \times 16 = 32 \newline 4 \times 8 = 32 \newline \end{array}$
Factors of 32: 1; 2; 4; 8; 16; 32

$\therefore$ HCF = 2
1. 30, 45 and 75
$\begin{array}{1} 1 \times 30 = 30 \newline 2 \times 15 = 30 \newline 3 \times 10 = 30 \newline 5 \times 6 = 30 \newline \end{array}$
Factors of 30: 1; 2; 3; 5; 6; 10; 15; 30
$\begin{array}{1} 1 \times 45 = 45 \newline 3 \times 15 = 45 \newline 5 \times 9 = 45 \newline \end{array}$
Factors of 45: 1; 3; 5; 9; 15; 45
$\begin{array}{1} 1 \times 75 = 75 \newline 3 \times 25 = 75 \newline 5 \times 15 = 75 \newline \end{array}$
Factors of 75: 1; 3; 5; 15; 25; 75

$\therefore$ HCF = 15
For the following sets of numbers, express each number as a product of prime factors and then find the HCF.
1. 39 and 169
\begin{array}{r | r} 3 & 39 \newline \hline 13 & 13 \newline \hline & 1 \end{array} $39=3 \times 13$ \begin{array}{r | r} 13 & 169 \newline \hline 13 & 13 \newline \hline & 1 \end{array} $169=13 \times 13$
$\therefore$ HCF = 13
1. 45 and 63
\begin{array}{r | r} 3 & 45 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $45=3 \times 3\times 5$ \begin{array}{r | r} 3 & 63 \newline \hline 3 & 21 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $63=3 \times 3 \times 7$ $\therefore \text{HCF} = 3\times 3 = 9$
1. 144 and 216
\begin{array}{r | r} 2 & 144 \newline \hline 2 & 72 \newline \hline 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $144=2 \times 2\times 2\times 2\times 3 \times 3$ \begin{array}{r | r} 2 & 216 \newline \hline 2 & 108 \newline \hline 2 & 54 \newline \hline 3 & 27 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $216=2 \times 2\times 2\times 3\times 3 \times 3$ $\therefore \text{HCF} = 2\times 2\times 2\times 3 \times 3 = 72$
1. 28, 42 and 70
\begin{array}{r | r} 2 & 28 \newline \hline 2 & 14 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $28=2 \times 2 \times 7$ \begin{array}{r | r} 2 & 42 \newline \hline 3 & 21 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $42=2 \times 3 \times 7$ \begin{array}{r | r} 2 & 70 \newline \hline 5 & 35 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $70=2 \times 5 \times 7$ $\therefore \text{HCF} = 2 \times 7 = 14$
1. 36, 63 and 72
\begin{array}{r | r} 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $36=2 \times 2 \times 3\times 3$ \begin{array}{r | r} 3 & 63 \newline \hline 3 & 21 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $63=3 \times 3 \times 7$ \begin{array}{r | r} 2 & 72 \newline \hline 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $72=2\times 2\times 2\times 3 \times 3$ $\therefore \text{HCF} = 3 \times 3 = 9$
For the following sets of numbers, express each number as a product of prime factors in index form and then find the HCF.
1. 36 and 120
\begin{array}{r | r} 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $36=2^2 \times 3^2$ \begin{array}{r | r} 2 & 120 \newline \hline 2 & 60 \newline \hline 2 & 30 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $120=2^3 \times 3 \times 5$ $\therefore \text{HCF} = 2^2 \times 3 = 2 \times 2 \times 3 = 12$
1. 125 and 300
\begin{array}{r | r} 5 & 125 \newline \hline 5 & 25 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $125=5^3$ \begin{array}{r | r} 2 & 300 \newline \hline 2 & 150 \newline \hline 3 & 75 \newline \hline 5 & 25 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $300=2^2 \times 3 \times 5^2$ $\therefore \text{HCF} = 5^2 = 5 \times 5 = 25$
1. 288 and 360
\begin{array}{r | r} 2 & 288 \newline \hline 2 & 144 \newline \hline 2 & 72 \newline \hline 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $288=2^5 \times 3^2$ \begin{array}{r | r} 2 & 360 \newline \hline 2 & 180 \newline \hline 2 & 90 \newline \hline 3 & 45 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $360=2^3 \times 3^2 \times 5$ $\therefore \text{HCF} = 2^3 \times 3^2 = 2\times 2\times 2\times 3\times 3 = 72$
1. 32, 56 and 140
\begin{array}{r | r} 2 & 32 \newline \hline 2 & 16 \newline \hline 2 & 8 \newline \hline 2 & 4 \newline \hline 2 & 2 \newline \hline & 1 \end{array} $32=2^5$ \begin{array}{r | r} 2 & 56 \newline \hline 2 & 28 \newline \hline 2 & 14 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $56=2^3 \times 7$ \begin{array}{r | r} 2 & 140 \newline \hline 2 & 70 \newline \hline 5 & 35 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $140=2^2 \times 5 \times 7$ $\therefore \text{HCF} = 2^2 = 2 \times 2 = 4$
1. 96, 160 and 224
\begin{array}{r | r} 2 & 96 \newline \hline 2 & 48 \newline \hline 2 & 24 \newline \hline 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $96=2^5\times 3$ \begin{array}{r | r} 2 & 160 \newline \hline 2 & 80 \newline \hline 2 & 40 \newline \hline 2 & 20 \newline \hline 2 & 10 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $160=2^5 \times 5$ \begin{array}{r | r} 2 & 224 \newline \hline 2 & 112 \newline \hline 2 & 56 \newline \hline 2 & 28 \newline \hline 2 & 14 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $224=2^5 \times 7$ $\therefore \text{HCF} = 2^5 = 2 \times 2\times 2\times 2\times 2 = 32$

2.2 Lowest common multiple

Multiples

Natural numbers are the numbers we use for counting. They include all the positive whole numbers from 1 onwards: $1;\space2;\space3;\space4;\space5;\space\ldots$ . The multiples of a given number are the numbers you get when you multiply the given number with the natural numbers. For example, the multiples of the number 5 are: $5;\space10;\space15;\space20;\space\ldots$

A multiple is also described as a whole number into which a smaller number divides without a remainder. For example: $\frac{12}{6} = 2$ . The number 6 divides perfectly into 12. Therefore, the number 12 is a multiple of 6.

multiple A multiple is a whole number into which a smaller number divides without a remainder.

Exercise 2.7: Find multiples

Write down the first five multiples of the following numbers.

8

$\begin{array}{1} 8\times 1 = 8 \newline 8\times 2 = 16 \newline 8\times 3 = 24 \newline 8\times 4 = 32 \newline 8\times 5 = 40 \end{array}$
The first five multiples are: $8;\space16;\space24;\space32;\space40$
13

$\begin{array}{1} 13\times 1 = 13 \newline 13\times 2 = 26 \newline 13\times 3 = 39 \newline 13\times 4 = 52 \newline 13\times 5 = 65 \end{array}$
The first five multiples are: $13;\space26;\space39;\space52;\space65$
21

$\begin{array}{1} 21\times 1 = 21 \newline 21\times 2 = 42 \newline 21\times 3 = 63 \newline 21\times 4 = 84 \newline 21\times 5 = 105 \end{array}$
The first five multiples are: $21;\space42;\space63;\space84;\space105$

Common multiples

A multiple is a number that can be divided by another number without a remainder. Two numbers have a common multiple if the same number is a multiple of both those numbers. For example, the number 12 is a common multiple of the numbers 3 and 4. We know this because both 3 and 4 divide into 12 without a remainder.

common multiple A common multiple is a multiple of two or more numbers.

Worked example 2.7: Common multiples

Find three common multiples of 5 and 6.

Step 1: List the first ten multiples of each number.

The first ten multiples of 5 are: $5;\space10;\space 15;\space 20;\space 25;\space 30;\space 35;\space 40;\space 45;\space 50$
The first ten multiples of 6 are: $6;\space 12;\space 18;\space 24;\space 30;\space 36;\space 42;\space 48;\space 54;\space 60$
Step 2: Identify the numbers that appear in both lists.

The number 30 is the only number that appears in both lists.
Step 3: Expand the lists until you find another number that appear in both lists.

Multiples of 5: $5;\space 10;\space 15;\space 20;\space 25;\space 30;\space 35;\space 40;\space 45;\space 50;\space 55;\space 60$
Multiples of 6: $6;\space 12;\space 18;\space 24;\space 30;\space 36;\space 42;\space 48;\space 54;\space 60$

The number 60 also appears in both lists.
Step 4: Repeat Step 3 until you find the required number of common multiples.

Multiples of 5: $5;\space 10;\space 15;\space 20;\space 25;\space 30;\space 35;\space 40;\space 45;\space 50;\space 55;\space 60;\space 65;\space 70;\space 75;\space 80;\space 85;\space 90$
Multiples of 6: $6;\space 12;\space 18;\space 24;\space 30;\space 36;\space 42;\space 48;\space 54;\space 60;\space 66;\space 72;\space 78;\space 84;\space 90$

The number 90 also appears in both lists.
Step 5: Write down the common multiples.

Three common multiples of 5 and 6 are: $30;\space 60;\space 90$

An easy way to find a common multiple of two or more numbers is to multiply the numbers. For example, $5 \times 6 = 30$ . This tells us that 30 is a common multiple of 5 and 6. If you multiply 30 two more times, you get 60 and then 90, which gives you the three common multiples required.

Exercise 2.8: Find common multiples

Find two common multiples of the following sets of numbers:

12 and 15

Multiples of 12: $12;\space24;\space36;\space48;\space60$
Multiples of 15: $15;\space 30;\space 45;\space 60$
Common multiple: 60

$12 \times 15 = 180$
Common multiple: 180

Two common multiples of 12 and 15 are: $60;\space 180$
5, 10 and 15

Multiples of 5: $5;\space 10;\space 15;\space 20;\space 25;\space 30$
Multiples of 10: $10;\space 20;\space 30$
Multiples of 15: $15;\space 30;\space 45$
Common multiple: 30

$5 \times 10 \times 15 = 750$
Common multiple: 750

Two common multiples of 5, 10 and 15 are: $30;\space 750$
2, 4 and 6

Multiples of 2: $2;\space 4;\space 6;\space 8;\space 10;\space 12$
Multiples of 4: $4;\space 8;\space 12$
Multiples of 6: $6;\space 12;\space 18$
Common multiple: 12

$2 \times 4 \times 6 = 48$
Common multiple: 48

Two common multiples of 2, 4 and 6 are: $12;\space 48$

The lowest common multiple (LCM) of two numbers is the smallest number that is a multiple of both. For example, the first three common multiples of the numbers 4 and 6 are 12, 24 and 36. The number 12 is the smallest of these multiples.

lowest common multiple (LCM) The lowest common multiple is the smallest whole number into which two or more numbers divide without a remainder.

Exercise 2.9: Find the LCM

In Exercise 2.8, you determined two common multiples of the following sets of numbers. Now identify the LCM in each case.

12 and 15

Multiples of 12: $12;\space24;\space36;\space48;\space60$
Multiples of 15: $15;\space 30;\space 45;\space 60$
$\therefore \text{LCM}=60$
5, 10 and 15

Multiples of 5: $5;\space 10;\space 15;\space 20;\space 25;\space 30$
Multiples of 10: $10;\space 20;\space 30$
Multiples of 15: $15;\space 30$
$\therefore \text{LCM}=30$
2, 4 and 6

Multiples of 2: $2;\space 4;\space 6;\space 8;\space 10;\space 12$
Multiples of 4: $4;\space 8;\space 12$
Multiples of 6: $6;\space 12$
$\therefore \text{LCM}=12$

Worked example 2.8: LCM using prime factors

Find the LCM of 12 and 18.

Step 1: Express each number as a product of prime factors.
\begin{array}{r | r} 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $12=2 \times 2 \times 3$ \begin{array}{r | r} 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $18=2 \times 3 \times 3$
Step 2: Identify all the prime factors that must be included in a common multiple.

Any multiple of 12 must contain: $2 \times 2$ (The factor $3$ is included in $3 \times3$ ; see next line)
Any multiple of 18 must contain: $3 \times 3$ (The factor $2$ is included in $2 \times 2$ ; see previous line)

A multiple that contains $2\times 2$ automatically contains $2$ . Similarly, a multiple that contains $3\times 3$ automatically contains $3$ .
Step 3: Multiply the prime factors that have to be included in both multiples to get the LCM.
$\text{LCM} = 2 \times 2 \times 3 \times 3 = 36$

Worked example 2.9: LCM using prime factors in index form

Find the LCM of 60 and 45.

Step 1: Express each number as a product of prime factors in index form.
\begin{array}{r | r} 2 & 60 \newline \hline 2 & 30 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $60=2^2 \times 3 \times 5$ \begin{array}{r | r} 3 & 45 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $45=3^2 \times 5$
Step 2: List all the prime factors that were used, to the highest index.

The factor 2 was used and its highest index is two: $2^2$
The factor 3 was used and its highest index is two: $3^2$
The factor 5 was used: $5$
Step 3: Multiply the prime factors you listed to get the LCM.
$\text{LCM} = 2^2 \times 3^2 \times 5 =2 \times 2 \times 3 \times 3 \times 5 = 180$

Exercise 2.10: Use different methods to find the LCM

For the following sets of numbers, use lists of multiples to find the LCM.
1. 6 and 21
Multiples of 6: $6;\space 12;\space 18;\space 24;\space30;\space36;\space42$
Multiples of 21: $21;\space 42$
$\therefore \text{LCM} = 42$
1. 4 and 11
Multiples of 4: $4;\space 8;\space 12;\space 16;\space20;\space24;\space28;\space 32; \space 36;\space 40;\space 44$
Multiples of 11: $11;\space 22;\space 33;\space 44$
$\therefore \text{LCM} = 44$
1. 7 and 9
Multiples of 7: $7;\space 14;\space 21;\space 28;\space35;\space42;\space49;\space 56; \space 63$
Multiples of 9: $9;\space 18;\space 27;\space 36;\space 45; \space 54; \space63$
$\therefore \text{LCM} = 63$
1. 2, 3 and 5
Multiples of 2: $2;\space 4;\space 6;\space 8;\space10;\space12;\space14;\space 16; \space 18;\space 20;\space 22;\space 24;\space 26;\space 28;\space 30$
Multiples of 3: $3;\space 6;\space 9;\space 12;\space 15;\space 18;\space 21;\space 24;\space 27;\space30$
Multiples of 5: $5;\space 10;\space 15;\space 20;\space 25;\space 30$
$\therefore \text{LCM} = 30$
1. 6, 10 and 12
Multiples of 6: $6;\space 12;\space 18;\space 24;\space30;\space36;\space42;\space 48; \space 54;\space 60$
Multiples of 10: $10;\space 20;\space 30;\space 40;\space 50;\space 60$
Multiples of 12: $12;\space 24;\space 36;\space 48;\space 60$
$\therefore \text{LCM} = 60$
For the following sets of numbers, express each number as a product of prime factors and then find the LCM.
1. 9 and 12
\begin{array}{r | r} 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $9=3 \times 3$ \begin{array}{r | r} 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $12=2 \times 2 \times 3$
Any multiple of 9 must contain: $3\times3$
Any multiple of 12 must contain: $2\times2$ ( $3$ is included in $3\times3$ )
$\therefore \text{LCM} = 3 \times 3 \times 2 \times 2 = 36$
1. 20 and 50
\begin{array}{r | r} 2 & 20 \newline \hline 2 & 10 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $20=2 \times 2 \times 5$ \begin{array}{r | r} 2 & 50 \newline \hline 5 & 25 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $50=2 \times 5 \times 5$
Any multiple of 20 must contain: $2\times2$ ( $5$ is included in $5\times5$ below)
Any multiple of 50 must contain: $5\times5$ ( $2$ is included in $2\times2$ above)
$\therefore \text{LCM} = 2 \times 2 \times 5 \times 5 = 100$
1. 24 and 40
\begin{array}{r | r} 2 & 24 \newline \hline 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $24=2 \times 2 \times 2 \times 3$ \begin{array}{r | r} 2 & 40 \newline \hline 2 & 20 \newline \hline 2 & 10 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $40=2 \times 2 \times 2 \times 5$
Any multiple of both 24 and 40 must contain: $2 \times 2 \times 2$
Any multiple of 24 must contain: $3$
Any multiple of 40 must contain: $5$
$\text{LCM} = 2 \times 2 \times 2 \times 3 \times 5 = 120$
1. 6, 8 and 12
\begin{array}{r | r} 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $6=2 \times 3$ \begin{array}{r | r} 2 & 8 \newline \hline 2 & 4 \newline \hline 2 & 2 \newline \hline & 1 \end{array} $8=2 \times 2 \times 2$ \begin{array}{r | r} 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $12=2 \times 2 \times 3$
Any multiple of 6 must contain: $3$ ( $2$ is included in $2\times2\times2$ below)
Any multiple of 8 must contain: $2 \times 2\times 2$
Requirements for multiples of 12 are already listed ( $2\times2$ is included in $2\times2\times2$ ; $3$ listed for the number $6$ )
$\therefore \text{LCM} = 2 \times 2 \times 2 \times 3 = 24$
1. 10, 16 and 18
\begin{array}{r | r} 2 & 10 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $10=2 \times 5$ \begin{array}{r | r} 2 & 16 \newline \hline 2 & 8 \newline \hline 2 & 4 \newline \hline 2 & 2 \newline \hline & 1 \end{array} $16=2 \times 2 \times 2 \times 2$ \begin{array}{r | r} 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $18=2 \times 3 \times 3$
Any multiple of 10 must contain: $5$ ( $2$ is included in $2 \times 2 \times 2 \times 2$ below)
Any multiple of 16 must contain: $2 \times 2 \times 2 \times 2$
Any multiple of 18 must contain: $3 \times 3$ ( $2$ is included in $2 \times 2 \times 2 \times 2$ above)
$\text{LCM} = 5 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 720$
For the following sets of numbers, express each number as a product of prime factors in index form and then find the LCM.
1. 12 and 27
\begin{array}{r | r} 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $12=2^2 \times 3$ \begin{array}{r | r} 3 & 27 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $27=3^3$ $\therefore \text{LCM} = 2^2 \times 3^3 = 2\times 2\times 3\times 3\times 3=108$
1. 20 and 25
\begin{array}{r | r} 2 & 20 \newline \hline 2 & 10 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $20=2^2 \times 5$ \begin{array}{r | r} 5 & 25 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $25=5^2$ $\therefore \text{LCM} = 2^2 \times 5^2 = 2 \times 2 \times 5 \times 5 = 100$
1. 24 and 28
\begin{array}{r | r} 2 & 24 \newline \hline 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $24=2^3 \times 3$ \begin{array}{r | r} 2 & 28 \newline \hline 2 & 14 \newline \hline 7 & 7 \newline \hline & 1 \end{array} $28=2^2 \times 7$ $\therefore \text{LCM} = 2^3 \times 3 \times 7 = 2 \times 2 \times 2 \times 3 \times 7 = 168$
1. 9, 12 and 15
\begin{array}{r | r} 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $9=3^2$ \begin{array}{r | r} 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $12=2^2 \times 3$ \begin{array}{r | r} 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $15=3 \times 5$ $\therefore \text{LCM} = 3^2 \times 2^2 \times 5 = 3 \times 3 \times 2 \times 2 \times 5 = 180$
1. 48, 72 and 120
\begin{array}{r | r} 2 & 48 \newline \hline 2 & 24 \newline \hline 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $48=2^4\times3$ \begin{array}{r | r} 2 & 72 \newline \hline 2 & 36 \newline \hline 2 & 18 \newline \hline 3 & 9 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $72=2^3\times3^2$ \begin{array}{r | r} 2 & 120 \newline \hline 2 & 60 \newline \hline 2 & 30 \newline \hline 3 & 15 \newline \hline 5 & 5 \newline \hline & 1 \end{array} $120=2^3\times3\times5$ $\therefore \text{LCM} = 2^4 \times 3^2 \times 5 = 2 \times 2 \times 2 \times 2 \times 3\times 3\times5 = 720$

2.3 Compare HCF and LCM

Be careful to not confuse HCF and LCM. The differences are listed in the table.

HCF (highest common factor)	LCM (lowest common multiple)
A factor of two or more numbers	A multiple of two of more numbers
HCF must divide into the given numbers	Given numbers must divide into LCM
Only common prime factors to lowest index	All prime factors to highest index
Smaller than or equal to given numbers	Larger than or equal to given numbers

Worked example 2.10: HCF and LCM

Find the HCF and the LCM of 24 and 32.

Step 1: Express each number as a product of prime factors in index form.
\begin{array}{r | r} 2 & 24 \newline \hline 2 & 12 \newline \hline 2 & 6 \newline \hline 3 & 3 \newline \hline & 1 \end{array} $24=2^3 \times 3$ \begin{array}{r | r} 2 & 32 \newline \hline 2 & 16 \newline \hline 2 & 8 \newline \hline 2 & 4 \newline \hline 2 & 2 \newline \hline & 1 \end{array} $32=2^5$
Step 2: For the HCF, identify the common prime factors to the lowest index. For the LCM, list all the prime factors to the highest index.

HCF	LCM
$2^3$	$2^5\times3$
$2\times2\times2=8$	$2\times2\times2\times2\times2\times3=96$
Largest number that can divide into 24 and 32	Smallest number into which 24 and 32 can divide

Exercise 2.10: Determine HCF and LCM

The prime factorisation of three numbers are as follows:
$2^2 \times 3 \times 5$
$2^3 \times 3 \times 5$
$2^2 \times 3^2 \times 5$

Determine the LCM of the three numbers.

$\text{LCM}=2^3 \times 3^2 \times 5=2 \times 2 \times 2 \times 3 \times 3 \times 5=360$
Determine the HCF of the three numbers.

$\text{HCF}=2^2 \times 3 \times 5=60$
Apart from the HCF, identify another common factor for the three numbers.

$2 \times 3 \times 5 = 30$
Calculate the three original numbers.

$2^2 \times 3 \times 5 = 60$
$2^3 \times 3 \times 5 = 120$
$2^2 \times 3^2 \times 5= 180$

2.4 Practical applications

You can use your knowledge of factors and multiples to solve problems. Here are some hints:

Decide whether the problem is about factors or multiples. Factors divide into a given number. Multiples are the products of a given number and the natural numbers.
If you have to divide two or more amounts into smaller units, finding the HCF might solve the problem.
If you have to find a larger amount from smaller units, finding the LCM might solve the problem.
When you get stuck, list all the factors, or list the first ten multiples of the given numbers. Try to connect the lists to the question.

Exercise 2.11: Use factors and multiples to solve problems

Chinasa is a student. This year, her age is a multiple of 6. Next year, her age will be a multiple of 5. If Chinasa is younger than 40, how old will she be next year?

Multiples of 6 smaller than 40: $6;\space 12;\space 18;\space 24;\space 30;\space 36$
Multiples of 5 smaller than 40: $5;\space 10;\space 15;\space 20;\space 25;\space 30;\space 35$

Her age next year must be one more than her age this year. The number 25 (a multiple of 5) is one more than the number 24 (a multiple of 6).

Therefore, she will be 25 next year.
Zahrah decides to sell sweets for extra pocket money. She buys a bag of chocolate bars and a bag of toffees. At home, she opens the bags and discovers that she has 24 chocolate bars and 80 toffees. She decides to make up smaller packets to sell. She divides all the sweets into packets that have equal amounts of chocolate bars and toffees. What is the maximum number of packets she can make up?

Find the largest number that can divide into 24 and 80, that is, the HCF.

$24=2^3 \times 3$
$80=2^4 \times 5$
$\text{HCF}=2^3=2 \times 2 \times 2=8$
Therefore, the maximum number of packets Zahrah can make up is 8.
It is school holidays. Suleiman, Umar and Faruq want to earn some pocket money. They decide to wash cars on the days that they do not have chores to do at home. Suleiman washes 4 cars per day and Uman washes 5 cars per day. Faruq is the fastest and washes 6 cars per day. At the end of the holidays, they find to their surprise that each of them washed the same total number of cars over the holidays.
1. Determine the minimum number of cars each person could have washed over the holidays.
Find the smallest number that is a multiple of 4, 5 and 6, that is, the LCM.

$4=2 \times 2$
$5=5$
$6=2\times3$
$\text{LCM}=2\times 2 \times 5 \times3 = 60$
1. Calculate how many days of the school holidays each person washed cars.
Suleiman: $\frac{60}{4}=15$ days

Umar: $\frac{60}{5}=12$ days

Faruq: $\frac{60}{6}=10$ days
Chika's family has a few dogs and a few chickens. They have a total of 8 animals. The animals have 22 legs altogether. How many dogs and how many chickens does the family have?

\begin{array}{l r r r r r r r r} \text{number of chickens}\space\space&1&2&3&4&5&6&7 \newline \text{number of legs}\space\space&2&4&6&8&10&12&14 \newline \space \newline \text{number of dogs}\space\space&1&2&3&4&5&6&7 \newline \text{number of legs}\space\space&4&8&12&16&20&24&28 \newline \end{array}
1 chicken and 7 dogs have $2+28=30$ legs
2 chickens and 6 dogs have $4+24=28$ legs
3 chickens and 5 dogs have $6+20=26$ legs
4 chickens and 4 dogs have $8+16=24$ legs
5 chickens and 3 dogs have $10+12=22$ legs
A class of 26 boys and 39 girls must be divided into small groups for an activity.
1. If the teacher wants equal numbers of boys and girls in each group, what is the maximum number of groups she can form?
Find the largest number that can divide into 26 and 39, that is, the HCF.

$26=2 \times 13$
$39=3 \times 13$
$\text{HCF}=13$
The teacher can form 13 groups.
1. How many boys and how many girls are in each group?
$\frac{26}{13}=2$ boys

$\frac{39}{13}=3$ girls
1. After the first activity, the teacher realises that there should be only 2 boys and 2 girls in each group. She decides to form a few groups with only girls. How many additional groups with 4 girls in each can she form?
There are $39-26=13$ girls to divide into groups.
$\frac{13}{4}=3\text{ remainder 1}$
She can from 3 additional groups.
1. Will there be any of the additional groups with more than 4 members?
Yes. One girl is left over, so one group will have 5 members.

2.4 Summary

A factor is a whole number that divides into a larger number without a remainder. For example, 4 is a factor of 24.
A prime factor is a factor that is also a prime number. For example, 3 is a prime factor of 24, but 8 is not a prime factor.
We can express any non-prime number as a product of its prime factors. For example, 24 can be expressed as $2\times 2\times 2\times 3$ .
Index form is a way of writing as a superscript how many times a number is multiplied by itself. For example, $3^4$ means $3\times 3\times 3\times 3$ .
A common factor of two or more numbers is a factor of all the numbers. For example, 4 is a common factor of 8, 24 and 60.
The highest common factor (HCF) of two or more numbers is the largest factor of all the numbers. For example, 2 is a common factor of 12, 24 and 60, but 4 is the highest common factor of the three numbers.
A multiple is a whole number into which a smaller number divides without a remainder. For example, 24 is a multiple of 4.
A common multiple of two or more numbers is a multiple of all the numbers. For example, 24 is a common multiple of the numbers 2, 3, 4 and 6.
The lowest common multiple (LCM) of two or more numbers is the smallest whole number that is a multiple of all the numbers. For example, 24 is a common multiple of 3, 4 and 6, but 12 is the lowest common multiple of the three numbers.
When we express two or more numbers as prime factors in index form:
- the HCF is the product of the common prime factors to the lowest index
- the LCM is the product of all the prime factors to the highest index.

Chapter 1: Numbers big and small

Table of Contents

Chapter 3: Binary numbers