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Factorisation and prime numbers

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

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.

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.

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

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 $$$150 = \Tblue{15}\times \Tgreen{10} = \HCF(\Tblue{15}, \Tgreen{10})\LCM(\Tblue{15},\Tgreen{10}).$$$

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.
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.

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

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*}