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# Factorisation and prime numbers

## Factors and multiples

A number is a multiple of another number if it is the product of this number with an integer.

An integer is a divisor of a number if the ratio of the number with the integer is an integer. It is also called a factor.

$\Tblue{6}$ is a multiple of $\Tred{2}$, $\Tred{2}$ is a divisor of $\Tblue{6}$, since $\displaystyle\frac{\Tblue{6}}{\Tred{2}} = 3$ is an integer.

To rephrase, take two integers $\Tblue{m}$ and $\Tred{d}$. $\Tblue{m}$ is a multiple of $\Tred{d}$ and $\Tred{d}$ is a divisor of $\Tblue{m}$ if the ratio $\displaystyle \frac{\Tblue{m}}{\Tred{d}}$ is an integer.

$\Tred{4}$ is a factor of $\Tblue{12}$, because $\displaystyle\frac{12}{\Tred{4}} = 3$ is an integer

$\Tred{8}$ is not a factor of $\Tblue{12}$ because $\displaystyle\frac{12}{\Tred{8}} = 1.5$ is not an integer.

60 has many factors. An hour can be divided into 12 times five minutes.

## Properties of factors and multiples

An integer can have many factors. Every number has itself and $1$ as a factor. No integer can have infinitely many factors.

The factors of $\Tblue{12}$ are $1$, $2$, $3$, $4$, $6$ and $12$.

A number has an infinite number of multiples.

The multiples of $\Tred{6}$ are $6$, $12$, $18$, $24$, $30$, etc.

## Highest common factor

A common factor is a factor that two numbers share.

$4$ is a common factor of $8$ and $12$.

$6$ is not a common factor of $18$ and $15$.

Two integers have a unique highest common factor (HCF). It is also known as their greatest common divisor (GCD).

The factors of $35$ are $1$, $\Tred{5}$, $7$ and $35$. The factors of $15$ are $1$, $3$ , $\Tred{5}$ and $15$. So the highest common factor of $15$ and $35$ is $\Tred{5}$.

The common factors of $100$ and $70$ are $1$, $2$, $5$ and $\Tred{10}$. So $\HCF(100,70) = \Tred{10}$.

It is possible for the highest common factor of two numbers to be $1$ or one of the numbers.

$$\HCF(20, 9) = \Tred{1},\quad \HCF(6,9) = \Tred{3},\quad\HCF(6,18) = \Tred{6}.$$

## Lowest common multiple

A common multiple of two integers is a multiple of both numbers.

Common multiples of $\Tblue{3}$ and $\Tgreen{7}$ are $\Tred{21}$, $42$, $63$, etc.

$15$ is not a common multiple of $\Tblue{5}$ and $\Tgreen{9}$, because it is not a multiple of $9$.

Two integers have a unique lowest common multiple (LCM).

The LCM of $\Tblue{12}$ and $\Tgreen{15}$ is $\Tred{60}$ because $\Tred{60} = \Tblue{12} \times 5 = \Tgreen{15} \times 4$

The LCM of $\Tblue{8}$ and $\Tgreen{120}$ is $\Tred{120}$ because $\Tred{120} = \Tblue{8}\times 15$

The highest common factor of two numbers can be one of the numbers or the product of the numbers.

$$\LCM(\Tblue{6},\Tgreen{18}) = \Tred{18},\quad \LCM(\Tblue{6},\Tgreen{9}) = \Tred{18},\quad \LCM(\Tblue{20}, \Tgreen{9}) = \Tred{180}.$$

The product of the HCF and LCM is the product of the numbers.

$$\Tblue{m}\Tgreen{n} = \HCF(\Tblue{m},\Tgreen{n})\LCM(\Tblue{m},\Tgreen{n})$$

We have $\HCF(\Tblue{15}, \Tgreen{10}) = 5$, $\LCM(\Tblue{15}, \Tgreen{10}) = 30$ and $$150 = \Tblue{15}\times \Tgreen{10} = \HCF(\Tblue{15}, \Tgreen{10})\LCM(\Tblue{15},\Tgreen{10}).$$

## Prime number

A prime number is a positive integer that has no factors apart from one and itself. $1$ is not considered to be a prime number.

$2$, $3$ and $5$ are prime numbers.

$6$ is not a prime number because it has factors $2$ and $3$.

There are infinitely many prime numbers. The first ones are

$$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,\dots$$
These cicadas hatch in a 17 year lifecycle - a prime number. This means that predator lifecycles are less likely to overlap, and so more cicadas survive.

## Prime factorisation

Every positive integer is the product of prime numbers. The list of prime factors of a number is called its prime factorisation. The prime factorisation of every number is unique.

$$6 = \Tblue{2}\times \Tgreen{3},\qquad 12 = \Tblue{2}\times\Tblue{2}\times \Tgreen{3},\qquad 1105 = \Tblue{5}\times \Tgreen{13}\times \Tviolet{17}.$$

The number of times that a prime factor appears is its multiplicity.

$\Tblue{2}$ has multiplicity $\Tred{3}$ in the factorisation of $24 = \Tblue{2}^\Tred{3} \times \Tgreen{3}$.

If the prime factorisation of a number is known, it is easy to find all the factors of that number. The factors are all possible products of the prime factors.

We have the prime factorisation $$24 = \Tblue{2}^\Tred{3} \times \Tgreen{3}.$$ So all the factors of $24$ are : $1$, $\Tblue{2}$, $\Tblue{2}^\Tred{2} = 4$, $\Tblue{2}^\Tred{3} = 8$, $\Tgreen{3}$, $\Tblue{2}\times\Tgreen{3} = 6$, $\Tblue{2}^\Tred{2} \times\Tgreen{3} = 12$ and $\Tblue{2}^\Tred{3} \times\Tgreen{3} = 24$.

The prime factorisation of a number can be found using a tree method. Find the highest

## Finding the HCF and LCM using prime factorisation

We can use prime factorisation to find the highest common factor (HCF) and the lowest common multiple (LCM) of two numbers.

• Compute the prime factorisation of each number. $$90 = \Tblue{2} \times \Tred{3}\times \Tblue{3}\times \Tred{5}, \qquad 105 = \Tred{3} \times \Tred{5} \times \Tgreen{7}$$
• The HCF is the product of all the prime factors that the two numbers have in common. $$\LCM(90,105) = \Tred{3}\times \Tred{5} = 15$$
• The LCM is the product of all the common prime numbers once and all the remaining prime numbers $$\HCF(90,105) = \Tblue{2}\times \Tred{3}\times \Tblue{3}\times \Tred{5}\times \Tgreen{7} = 630$$ The LCM can be computed as well by the formula $$\LCM(m,n) = \frac{mn}{\HCF(m,n)}.$$
\begin{align*} 45 = \Tred{3}\times\Tblue{3}\times\Tblue{5},\qquad 42 = \Tgreen{2}\times \Tred{3} \times \Tgreen{7},\qquad\HCF = \Tred{3},\\ \LCM = \Tgreen{2}\times \Tred{3}\times \Tblue{3}\times \Tblue{5}\times \Tgreen{7} = 630 = \frac{45\times 42}{3} \end{align*}