How to Find HCF and LCM (Highest Common Factor and Lowest Common Multiple)
HCF and LCM are essential GCSE Maths skills, especially in number questions involving prime factors, simplifying, and problem solving. This page explains reliable methods for finding the Highest Common Factor and the Lowest Common Multiple using prime factorisation, with clear routines and worked examples for two and three numbers. You will learn how to identify the prime factors shared by every number to find the HCF, then build the LCM by taking the highest powers that appear.
Prior Knowledge
This page builds on
Prime Factorisation
(writing a number as a product of primes). You should also be confident with index notation such as \(2^3\) and \(3^2\).
HCF
Highest Common Factor. The largest integer that divides every given number exactly.
LCM
Lowest Common Multiple. The smallest positive integer that every given number divides into exactly.
How to find HCF and LCM using prime lists
- List each number as a product of primes (include powers). Keep work vertical with one equals per line.
- Bubble (circle or box) the primes common to every list. Multiply these to get the HCF.
- LCM (Option A): take every prime that appears in any list, using the largest power seen, then multiply them.
- LCM (Option B): start with the HCF and times all the leftover primes needed to cover each list (no double counting). In the worked solutions below, the HCF is shown in red.
Worked examples
💡 Example 1: HCF and LCM of 48 and 180
Step 1. Prime lists (bubble the common primes).
\[ \begin{aligned} 48 &= \boxed{2}\times \boxed{2}\times 2\times 2\times \boxed{3}\\ 180 &= \boxed{2}\times \boxed{2}\times \boxed{3}\times 3\times 5 \end{aligned} \]Step 2. Multiply the bubbled primes for the HCF.
\[ \begin{aligned} \text{HCF} &= 2\times 2\times 3\\ \text{HCF} &= 12 \end{aligned} \]Step 3 (Option A, highest powers).
\[ \begin{aligned} \text{LCM} &= 2^4 \times 3^2 \times 5\\ \text{LCM} &= 720 \end{aligned} \]Step 3 (Option B, HCF times the leftovers).
\[ \begin{aligned} \text{LCM} &= \textcolor{#d00000}{12} \times 2 \times 2 \times 3 \times 5\\ \text{LCM} &= 720 \end{aligned} \]What's happening?
- Both lists share two 2s and one 3. Those are the bubbled primes.
- The HCF is just the bubbled primes multiplied together.
- For Option A, the highest power of \(2\) anywhere is \(2^4\) (from 48); the highest power of \(3\) is \(3^2\) (from 180); \(5\) only appears once.
- For Option B, after using up the HCF, the leftover primes from 48 are \(2\times 2\), and the leftovers from 180 are \(3\times 5\). Both options give the same answer.
💡 Example 2: HCF and LCM of 45, 60 and 75
Step 1. Prime lists (bubble one copy of each prime that appears in all three lists).
\[ \begin{aligned} 45 &= \boxed{3}\times 3\times \boxed{5}\\ 60 &= \boxed{3}\times 2\times 2\times \boxed{5}\\ 75 &= \boxed{3}\times \boxed{5}\times 5 \end{aligned} \]Step 2. Multiply the bubbled primes for the HCF.
\[ \begin{aligned} \text{HCF} &= 3 \times 5\\ \text{HCF} &= 15 \end{aligned} \]Step 3 (Option A, highest powers).
\[ \begin{aligned} \text{LCM} &= 2^2 \times 3^2 \times 5^2\\ \text{LCM} &= 900 \end{aligned} \]Step 3 (Option B, HCF times the leftovers).
\[ \begin{aligned} \text{LCM} &= \textcolor{#d00000}{15} \times 3 \times 2 \times 2 \times 5\\ \text{LCM} &= 900 \end{aligned} \]What's happening?
- Only one \(3\) and one \(5\) appear in every list. Those are the only primes you bubble.
- The extra \(3\) in 45, the \(2\times 2\) in 60, and the extra \(5\) in 75 are not shared by all three, so they don't contribute to the HCF.
- For the LCM, you must still cover those leftover primes. Option B picks them up after the HCF.
💡 Example 3: 40, 60 and 90 (when a prime is shared by only two of the three)
Step 1. Prime lists. Bubble primes common to all three: one \(2\) and one \(5\). The prime \(3\) is shared by 60 and 90 only.
\[ \begin{aligned} 40 &= \boxed{2}\times 2\times 2\times \boxed{5}\\ 60 &= \boxed{2}\times 2\times \boxed{5}\times 3\\ 90 &= \boxed{2}\times \boxed{5}\times 3\times 3 \end{aligned} \]Step 2. HCF from the all-three bubbles.
\[ \begin{aligned} \text{HCF} &= 2 \times 5\\ \text{HCF} &= 10 \end{aligned} \]Step 3 (Option A, highest powers).
\[ \begin{aligned} \text{LCM} &= 2^3 \times 3^2 \times 5\\ \text{LCM} &= 360 \end{aligned} \]Step 3 (Option B, HCF times the leftovers).
\[ \begin{aligned} \text{LCM} &= \textcolor{#d00000}{10} \times 2 \times 2 \times 3 \times 3\\ \text{LCM} &= 360 \end{aligned} \]What's happening?
- The \(3\) appears in 60 and 90 but not in 40, so it does not get bubbled. Pairwise-only primes never contribute to the HCF.
- However, those pairwise-only primes do appear in the LCM, because the LCM must be divisible by every list.
- Highest power of \(2\) is \(2^3\) (from 40); highest power of \(3\) is \(3^2\) (from 90); \(5\) appears once everywhere.
🔑 Key points
- HCF: bubble the primes shared by every list, then multiply.
- LCM (Option A): take every prime, using the highest power seen.
- LCM (Option B): start with HCF, then multiply by the leftover primes once.
- One equals per line and clearly labelled prime lists earn method marks in IGCSE exams.
- Quick check: HCF must divide every number; LCM must be divisible by every number.
⚠️ Common pitfalls
- Bubbling a prime that appears in only some lists. For three numbers, only bubble what is in all three.
- Using the lowest power of a prime in the LCM. The LCM uses the highest power that appears.
- Stopping a factor tree before reaching primes (e.g. leaving \(15 = 3\times 5\) finished but not splitting \(15\) at all).
- Mixing up HCF and LCM: HCF is the small one (a factor), LCM is the big one (a multiple).
- Double counting in Option B by including primes that are already inside the HCF.