Prime Numbers, Factors and Multiples Revision Notes | GCSE AQA Higher Maths

1. The Building Blocks of Every Number

Every whole number you have ever worked with — 12, 60, 360 — can be broken apart into a unique set of prime pieces. That idea is not just a curiosity; it is the engine behind simplifying fractions, finding common denominators, and solving problems that appear throughout GCSE maths and beyond.

Before we can factorise anything, we need to be precise about the vocabulary. The terms prime, factor, and multiple each mean something specific, and examiners will use them exactly — so you need to as well.

A factor (also called a divisor) of a number $n$ is any integer that divides into $n$ with no remainder. A multiple of $n$ is any integer you get by multiplying $n$ by a positive integer. So 6 is a factor of 24, and 24 is a multiple of 6 — the relationship runs in both directions.

2. Prime Factorisation and the Unique Factorisation Theorem

Here is the central idea of this lesson: every integer greater than 1 can be written as a product of prime numbers, and that product is unique (apart from the order of the primes). This guarantee is called the unique factorisation theorem — sometimes called the fundamental theorem of arithmetic.

In practice, this means that no matter how you break a number apart, you will always arrive at the same set of prime factors. The standard method for finding them is a factor tree or repeated division by primes. We write the result using product notation, for example $360 = 2^3 \times 3^2 \times 5$.

Worked example: Prime Factorisation Using a Factor Tree

Express 360 as a product of its prime factors. Write your answer in product notation. (Without a Calculator)

1. $360 \div 2 = 180$ — 360 is even, so 2 divides it. Write 2 as one branch and 180 as the other. [M1 — correct first division]
2. $180 \div 2 = 90$ $90 \div 2 = 45$ — 180 and 90 are both even, so 2 divides again each time. We have found three factors of 2 so far.
3. $45 \div 3 = 15$ $15 \div 3 = 5$ — 45 ends in 5 but is odd, so 2 does not divide it. The digit sum of 45 is 4+5=9, which is divisible by 3, confirming 3 is a factor.
4. $5$ is prime — Once the remaining number is prime, the tree is complete. We have reached the end of the factorisation.
5. $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5$ — List every prime from the ends of the factor tree branches.
6. $360 = 2^3 \times 3^2 \times 5$ — Group repeated primes using powers. This is the required product notation. [A1 — correct final answer]

Answer: $360 = 2^3 \times 3^2 \times 5$

The unique factorisation theorem tells us that $2^3 \times 3^2 \times 5$ is the only way to write 360 as a product of primes (ignoring order). If you started your factor tree with $360 = 4 \times 90$ instead of $360 = 2 \times 180$, you would still end up with exactly $2^3 \times 3^2 \times 5$. That is the power of the theorem.

3. Common Factors, Common Multiples, HCF and LCM

Now that we can factorise numbers, we can use those factorisations to find what two numbers share. A common factor of two integers is any factor that divides into both of them. A common multiple is any multiple that both integers divide into.

The highest common factor (HCF) is the largest of all common factors. The lowest common multiple (LCM) is the smallest positive common multiple. Both can be found efficiently using prime factorisation and a Venn diagram — a method that is both reliable and straightforward to check in an exam.

Worked example: Finding HCF and LCM Using Prime Factorisation

Find the HCF and LCM of 360 and 252. (Without a Calculator)

1. $360 = 2^3 \times 3^2 \times 5$ — From our earlier factor tree. [M1 — correct factorisation of 360]
2. $252 \div 2 = 126$ $126 \div 2 = 63$ $63 \div 3 = 21$ $21 \div 3 = 7$ $7$ is prime $252 = 2^2 \times 3^2 \times 7$ — [M1 — correct factorisation of 252]
3. $360 = 2^3 \times 3^2 \times 5$ $252 = 2^2 \times 3^2 \times 7$ \text{Shared: } 2^2 \text{ and } 3^2 — For each prime that appears in BOTH factorisations, take the LOWER power. 2 appears as $2^3$ and $2^2$ — take $2^2$. 3 appears as $3^2$ in both — take $3^2$.
4. $\text{HCF} = 2^2 \times 3^2 = 4 \times 9 = 36$ — $4 \times 9 = 36$. [A1 — correct HCF]
5. \text{All primes: } 2^3, 3^2, 5, 7 — For the LCM, take EVERY prime that appears in EITHER factorisation, using the HIGHER power each time. 2 appears as $2^3$ and $2^2$ — take $2^3$.
6. $\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7$ $= 72 \times 5 \times 7$ $= 360 \times 7$ $= 2520$ — $8 \times 9 = 72$, then $72 \times 5 = 360$, then $360 \times 7 = 2520$. [A1 — correct LCM]

Answer: $\text{HCF} = 36$, $\text{LCM} = 2520$

4. Fluency Practice — Consolidating the Methods

The methods for prime factorisation, HCF, and LCM follow a clear pattern once you have seen them a few times. The best way to lock them in is to work through problems yourself before checking. Try each question below, then reveal the solution.

5. Reasoning and Problem-Solving with Factors and Multiples

GCSE Higher exam papers move beyond straightforward calculation. You will encounter problems where you need to decide whether a situation calls for HCF or LCM — and then carry out the method accurately. The context clues are usually about sharing (HCF) or synchronising / tiling (LCM).

Worked example: Problem-Solving: Choosing HCF or LCM

Priya has 96 red beads and 60 blue beads. She wants to make identical bracelets using all the beads, with no beads left over. What is the greatest number of bracelets she can make, and how many of each colour will each bracelet contain?

1. \text{We need the HCF of 96 and 60.} — Priya is sharing two quantities into equal groups with nothing left over — that is an HCF situation. The HCF gives the greatest number of equal groups possible. [M1 — correct method identification]
2. $96 \div 2 = 48$ $48 \div 2 = 24$ $24 \div 2 = 12$ $12 \div 2 = 6$ $6 \div 2 = 3$ $3$ is prime $96 = 2^5 \times 3$ — [M1 — correct factorisation of 96]
3. $60 \div 2 = 30$ $30 \div 2 = 15$ $15 \div 3 = 5$ $5$ is prime $60 = 2^2 \times 3 \times 5$ — [M1 — correct factorisation of 60]
4. \text{Shared: } 2^2 \text{ and } 3^1 \text{HCF} = 2^2 \times 3 = 4 \times 3 = 12 — 2 appears as $2^5$ and $2^2$ — take $2^2$. 3 appears as $3^1$ in both — take $3^1$. [A1 — HCF = 12]
5. \text{Red beads per bracelet: } 96 \div 12 = 8 \text{Blue beads per bracelet: } 60 \div 12 = 5 — Priya can make 12 bracelets. Each bracelet has 8 red beads and 5 blue beads. [A1 — correct interpretation]

Answer: Priya can make 12 bracelets, each containing 8 red beads and 5 blue beads.

6. Exam Technique, Common Mistakes, and Mark-Winning Habits

This topic appears on both questions without a calculator and calculator-allowed exam questions. In questions without a calculator, you must show all arithmetic working — a bare answer with no factor tree or division steps will lose method marks even if the answer is correct.

Examiners also reward correct use of vocabulary. If a question asks you to 'find the HCF', writing 'the biggest number that goes into both' without stating the value will not earn the answer mark. State the numerical result clearly and, in context questions, interpret what it means.