Prime Factorisation and Product Notation Revision Notes | GCSE AQA Higher Maths

1. Why Prime Factorisation Is the Backbone of Number Theory

Every whole number greater than 1 can be broken down into a unique set of prime building blocks. This idea — that primes are the atoms of arithmetic — is what makes prime factorisation so powerful. Once you can write any number as a product of primes, finding HCF, LCM, and solving divisibility problems becomes a systematic process rather than guesswork.

This topic appears on both questions without a calculator and calculator-allowed exam questions of GCSE Higher maths. The methods here are designed to work without a calculator, so every step of arithmetic is shown in full.

2. The Core Vocabulary You Must Know

Knowing your primes up to at least 50 is essential for questions without a calculator. The primes up to 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. Notice that 2 is the only even prime — every other even number has 2 as a factor, so it cannot be prime.

A multiple of a number is what you get when you multiply it by a positive integer. The multiples of 4 are 4, 8, 12, 16, 20, and so on — the four times table, extended indefinitely. Factors are finite; multiples are infinite.

3. Prime Factorisation: The Factor Tree Method

To write a number as a product of its prime factors, you repeatedly divide by the smallest prime that goes in, until you are left with only primes. The most reliable method at GCSE is the factor tree. You split the number into any two factors, then keep splitting each branch until every branch ends in a prime.

The result is written using product notation — the primes multiplied together, with repeated primes written using index (power) notation. For example, $2 \times 2 \times 2 \times 3$ is written as $2^3 \times 3$. This compact form is what examiners expect.

Worked example: Prime Factorisation of 360 Using a Factor Tree

Express 360 as a product of its prime factors. Write your answer in index notation. [3 marks]

1. $360 = 2 \times 180$ — 2 is the smallest prime. 360 is even, so 2 divides in. [M1 — starting the factor tree correctly]
2. $180 = 2 \times 90$ — 180 is still even, so divide by 2 again.
3. $90 = 2 \times 45$ — 90 is even, so divide by 2 once more.
4. $45 = 3 \times 15$ — 45 is odd, so 2 does not divide in. Try 3: $4 + 5 = 9$, which is divisible by 3, so 3 divides 45.
5. $15 = 3 \times 5$ — Both 3 and 5 are prime — both branches end here.
6. $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5$ — Read off every prime at the end of each branch. [M1 — all prime factors identified]
7. $360 = 2^3 \times 3^2 \times 5$ — Group repeated primes using powers. [A1 — correct index notation]

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

Notice that it does not matter which pair of factors you start with. You could have begun with $360 = 4 \times 90$ or $360 = 6 \times 60$ — the final set of prime factors will always be the same. This is guaranteed by the unique factorisation theorem.

The theorem tells you something reassuring: you cannot get two different prime factorisations for the same number. If your factor tree gives $2^3 \times 3^2 \times 5$ for 360, that is the only correct answer — there is no alternative set of primes that multiplies to 360.

4. Using Prime Factorisation to Find HCF and LCM

Once you have the prime factorisations of two numbers, finding the HCF and LCM becomes a matter of comparing the prime factors systematically. This is far more reliable than listing factors or multiples — especially for larger numbers.

The rule for HCF: take the lowest power of each prime that appears in both factorisations. The rule for LCM: take the highest power of each prime that appears in either factorisation. A Venn diagram of prime factors makes this visual and is a method examiners frequently reward.

Worked example: Finding HCF and LCM of 84 and 120

Find the HCF and LCM of 84 and 120. Show your working using prime factorisation. [4 marks]

1. $84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$ — Divide by 2 twice, then 3, then 7. All branches end in primes. [M1]
2. $120 = 2 \times 60 = 2 \times 2 \times 30 = 2 \times 2 \times 2 \times 15 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$ — Divide by 2 three times, then 3, then 5. [M1]
3. $84 = 2^2 \times 3 \times 7$ and $120 = 2^3 \times 3 \times 5$ Shared primes: $2$ and $3$ Lowest powers: $2^2$ and $3^1$ — HCF uses the lowest power of each prime common to both numbers.
4. $\text{HCF} = 2^2 \times 3 = 4 \times 3 = 12$ — $4 \times 3 = 12$. [A1 — correct HCF]
5. Primes appearing in either: $2, 3, 5, 7$ Highest powers: $2^3, 3^1, 5^1, 7^1$ — LCM uses the highest power of every prime that appears in either factorisation.
6. $\text{LCM} = 2^3 \times 3 \times 5 \times 7 = 8 \times 3 \times 5 \times 7$ $= 8 \times 3 = 24$ $24 \times 5 = 120$ $120 \times 7 = 840$ — Multiply step by step to avoid errors. [A1 — correct LCM]

Answer: $\text{HCF}(84, 120) = 12$ and $\text{LCM}(84, 120) = 840$

That verification formula is worth memorising. For 84 and 120: $\text{HCF} \times \text{LCM} = 12 \times 840 = 10080$, and $84 \times 120 = 10080$. They match — so the answers are confirmed.

5. Common Factors, Common Multiples, and How They Connect

A common factor of two numbers is any factor that divides exactly into both. The highest common factor is the largest of these. For example, the common factors of 12 and 18 are 1, 2, 3, and 6 — and the HCF is 6.

A common multiple of two numbers is any number that both divide into exactly. The lowest common multiple is the smallest positive one. For 4 and 6, the common multiples are 12, 24, 36,... and the LCM is 12. In practice, listing multiples works for small numbers, but prime factorisation is the reliable method for larger ones.

6. Deliberate Practice

These problems cover the full range of the full range of higher-tier questions on this topic. Work through each one before checking the solution — the method marks come from showing clear prime factorisation, not from writing down the answer alone.

7. Exam Technique and Common Mistakes

This topic is reliable marks territory — but only if you avoid the traps that cost students method marks. The mistakes below appear repeatedly in mark schemes, and each one has a straightforward fix.

One final point on exam language: if a question says 'express as a product of prime factors', the word 'product' means multiplication. Your answer must be written as primes multiplied together using $\times$ signs and index notation — a list of primes separated by commas is not acceptable.

Worked example: Exam-Style: HCF Applied to a Real Context

Aisha has two pieces of ribbon. One is 84 cm long and the other is 120 cm long. She wants to cut both pieces into equal lengths with no ribbon left over, using the longest possible pieces. How long should each piece be? [3 marks]

1. We need the longest length that divides exactly into both 84 and 120. — 'Longest possible equal pieces with no remainder' is the definition of HCF. [M1 — correct identification]
2. $84 = 2^2 \times 3 \times 7$ — $84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7$
3. $120 = 2^3 \times 3 \times 5$ — $120 = 2 \times 60 = 2 \times 2 \times 30 = 2 \times 2 \times 2 \times 15 = 2 \times 2 \times 2 \times 3 \times 5$
4. Shared primes: $2$ and $3$. Lowest powers: $2^2$ and $3^1$. $\text{HCF} = 2^2 \times 3 = 4 \times 3 = 12$ — [M1 — correct method for HCF]
5. Each piece should be $12$ cm long. — [A1 — correct answer with units and context]

Answer: Each piece should be 12 cm long.