What are factors and how do you find them in Mathematics?

Factors are numbers that divide another number exactly without leaving a remainder. To find the factors of a number, you can perform division tests starting from 1 up to the number itself. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12. This concept is fundamental in the Cambridge IGCSE Mathematics curriculum under the Number topic.

How do multiples differ from factors in Mathematics?

Multiples of a number are the products obtained by multiplying the number by integers. Unlike factors, which are limited, multiples are infinite. For example, the multiples of 4 include 4, 8, 12, 16, and so on. Understanding the difference between factors and multiples is crucial for mastering the Number subtopic in the Cambridge IGCSE Mathematics syllabus.

What is the greatest common factor (GCF) and how is it determined?

The greatest common factor (GCF), also known as the greatest common divisor (GCD), is the largest number that divides two or more numbers without a remainder. To find the GCF, list the factors of each number and identify the largest common factor. For instance, the GCF of 18 and 24 is 6. This concept is an essential part of the Number topic in Cambridge IGCSE Mathematics.

How can you find the least common multiple (LCM) of two numbers?

The least common multiple (LCM) of two numbers is the smallest multiple that is exactly divisible by both numbers. To find the LCM, list the multiples of each number and identify the smallest common multiple. For example, the LCM of 3 and 4 is 12. This method is frequently used in solving problems related to the Number subtopic in Cambridge IGCSE Mathematics.

Why are factors and multiples important in Mathematics?

Factors and multiples are foundational concepts in Mathematics because they are used in various applications such as simplifying fractions, finding common denominators, and solving problems involving ratios and proportions. Mastery of these concepts is vital for students studying the Cambridge IGCSE Mathematics curriculum, as they form the basis for more advanced mathematical topics.

Why is "Mixing up HCF and LCM rules in the prime factorisation method" a common mistake in Factors and Multiples?

Why it happens: Both rules use "common primes" but with different powers, and exam pressure leads to the wrong choice. How to avoid it: Memorise: **HCF = lower** powers, **LCM = larger** powers. The HCF must fit inside both numbers. Source: 0580/42 May/Jun 2024 — examiner report

Why is "Missing a prime when listing the LCM" a common mistake in Factors and Multiples?

Why it happens: Students take only common primes (HCF rule) instead of all primes that appear in either number. How to avoid it: For LCM, write every prime from either factorisation, including primes that appear in one number only.

Why is "Missing factors when listing all factors" a common mistake in Factors and Multiples?

Why it happens: Students stop early or skip factor pairs (e.g. forgetting 6 when listing factors of 48). How to avoid it: Pair factors from 1 outwards: 1 × n, 2 × (n/2), until pairs repeat.

Why is "Listing $1$ as a prime factor" a common mistake in Factors and Multiples?

Why it happens: Mixing up factors with prime factors. How to avoid it: 1 is **never** prime. Prime factorisations contain only primes ≥ 2.

Why is "Leaving prime factorisation as a long product instead of using indices" a common mistake in Factors and Multiples?

Why it happens: The factorisation looks complete with 2 × 2 × 3 × 3 × 5. How to avoid it: If the question says "in index form", group repeats: 2^2 × 3^2 × 5. Source: 0580/12 Oct/Nov 2022 — examiner report

Cambridge IGCSE Mathematics (0580)

Factors and Multiples

0% Complete

3 tasks remaining

Short Notes - Factors and Multiples

Detailed Study Notes

Detailed notes on Number for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Factors, Multiples, HCF and LCM — Cambridge IGCSE 0580 Maths Extended (2026)

Factor trees, the highest common factor and the lowest common multiple — three skills that appear on almost every paper, and almost always inside a word problem. This is the topic that rewards careful prime factorisation.

At a glance

Factor of $n$ : divides $n$ exactly. Multiple of $n$ : $n$ multiplied by an integer.
Every integer above $1$ has a UNIQUE prime factorisation (Fundamental Theorem of Arithmetic).
Two methods for prime factorisation: factor tree OR repeated division by primes (ladder).
HCF = product of the LOWEST power of each shared prime.
LCM = product of the HIGHEST power of each prime appearing in either number.
Useful identity: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ .
Divisibility shortcuts: 3 ↔ digit sum, 9 ↔ digit sum, 11 ↔ alternating sum.
Word problem keywords: 'greatest/largest' → HCF, 'smallest/least/again at the same time' → LCM.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E1.1 — Identify and use common factors and common multiples.
E1.8 — Use HCF and LCM to solve problems, including writing a number as a product of its prime factors.

Factors and multiples — the basics

Two definitions, two reliable lists. Skim past these only if they're already second nature.

A factor of a positive integer $n$ is any positive integer that divides $n$ with no remainder. The factors of $12$ are $1, 2, 3, 4, 6, 12$ .

Reading the pairs off the arcs gives every factor of 12: 1, 2, 3, 4, 6, 12.

A multiple of $n$ is any number obtained by multiplying $n$ by an integer. Multiples of $4$ are $4, 8, 12, 16, 20, \ldots$ (and technically $0$ , but Cambridge usually means positive multiples).

Common factor = a factor of every number in a set. Common factors of $12$ and $18$ : $1, 2, 3, 6$ . Common multiple = a multiple of every number in a set. Common multiples of $4$ and $6$ : $12, 24, 36, \ldots$

When a question asks for a factor or multiple, list them in increasing order — examiners spot missing entries.

Factor = exact divisor.
Multiple = product with a positive integer.
Common factor = divides EVERY number in the set.
Common multiple = is a multiple of EVERY number in the set.

Prime factorisation

Every integer above 1 has a unique prime fingerprint. Get the fingerprint and HCF/LCM fall out for free.

The Fundamental Theorem of Arithmetic says every integer above $1$ can be written as a product of primes in exactly one way (apart from the order of the factors). For example $60 = 2^{2} \times 3 \times 5 \quad \text{and} \quad 72 = 2^{3} \times 3^{2}.$

This unique fingerprint is what makes HCF and LCM straightforward.

Method 1: Factor tree. Start with $n$ at the top, branch into ANY pair of factors, and keep splitting any composite branch until every leaf is prime.

For $\mathbf{72}$ :

$72 = 8 \times 9$
$8 = 2 \times 2 \times 2$
$9 = 3 \times 3$
Combine: $72 = 2^{3} \times 3^{2}$ .

Keep splitting composite branches; the green prime leaves are the factorisation.

Method 2: Repeated-division (ladder). Divide by the smallest prime that goes in, write the quotient below, repeat until you reach $1$ . The primes you divided by are the factorisation.

Read down: $2 \times 2 \times 2 \times 3 \times 3 = 2^{3} \times 3^{2}$ . Same answer.

The ladder method is faster when the number has many small prime factors. The factor tree is friendlier when one factor is obvious.

Every integer above 1 has a UNIQUE prime factorisation.
Factor tree: split into any two factors, recurse on composites.
Ladder: divide by the smallest prime, repeat.
Always write the answer in INDEX form (e.g. $2^{3} \times 3^{2}$ ), not as a long chain.

HCF and LCM — three reliable methods

Listing for small numbers; prime factorisation for everything else. Choose by size of the input.

Highest Common Factor (HCF) of two integers is the largest integer that divides both. Lowest Common Multiple (LCM) is the smallest positive integer that BOTH numbers divide into.

Method 1: Listing. Best for small numbers ( $\le 30$ ). List all factors of each number; pick the largest common one for HCF. List multiples until both lists agree for LCM.

Method 2: Prime-factorisation tree (RECOMMENDED for $> 30$ ). Write each number as a product of primes in index form.

HCF = take the lowest power of every prime that appears in BOTH.
LCM = take the highest power of every prime that appears in EITHER.

For $\mathbf{72}$ and $\mathbf{108}$ :

$72 = 2^{3} \times 3^{2}$
$108 = 2^{2} \times 3^{3}$
HCF = $2^{2} \times 3^{2} = 4 \times 9 = 36$ (lowest power of each shared prime).
LCM = $2^{3} \times 3^{3} = 8 \times 27 = 216$ (highest power of each prime).

Method 3: Venn-diagram method. Draw two overlapping circles, write the prime factors into the appropriate region (shared primes in the overlap, unique primes outside). HCF = product of overlap; LCM = product of every prime in the diagram.

HCF is the product of the overlap; LCM is the product of every prime in the whole diagram.

Quick sanity checks. HCF should never be larger than the smaller of the two numbers. LCM should never be smaller than the larger.

Word problems. Real-paper questions disguise HCF/LCM as everyday situations. Spot the verb:

"Cut into the largest equal pieces" / "the greatest number of..." → HCF.
"Both events happen at the same time again" / "the smallest number that" → LCM.

The verb in the question tells you which to use: "greatest" → HCF, "smallest / again together" → LCM.

HCF: lowest power of each SHARED prime.
LCM: highest power of each prime in EITHER factorisation.
'Greatest/largest' word problem → HCF.
'Smallest/least/again together' word problem → LCM.
Sanity check: HCF ≤ smaller; LCM ≥ larger.

Divisibility shortcuts

Skip the calculator. These rules pick out a divisor from a glance at the digits.

Each rule lets you decide whether one number divides another by looking at digits — no division required.

Divisor	Test
$2$	Last digit is even ( $0, 2, 4, 6, 8$ ).
$3$	Digit sum is a multiple of $3$ .
$4$	Last two digits form a multiple of $4$ .
$5$	Last digit is $0$ or $5$ .
$6$	Divisible by both $2$ AND $3$ .
$8$	Last three digits form a multiple of $8$ .
$9$	Digit sum is a multiple of $9$ .
$10$	Last digit is $0$ .
$11$	Alternating digit sum is a multiple of $11$ (including $0$ ).

Worked: is $6{,}138$ divisible by $11$ ? Alternating sum from the left: $6 - 1 + 3 - 8 = 0$ . $0$ is a multiple of $11$ → yes. (Confirm: $6{,}138 \div 11 = 558$ .)

Worked: is $5{,}472$ divisible by $9$ ? Digit sum: $5 + 4 + 7 + 2 = 18$ . $18 = 9 \times 2$ → yes. ( $5{,}472 \div 9 = 608$ .)

$3$ and $9$ : use the digit sum.
$4$ and $8$ : look at the last two/three digits.
$11$ : alternating digit sum.
$6$ = divisible by $2$ AND $3$ — no separate test.

Quick recap

Factor = divisor; multiple = product with an integer.
Every integer $> 1$ has a UNIQUE prime factorisation. Use a factor tree or the ladder method.
HCF: lowest power of each shared prime. LCM: highest power of each prime in either.
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$ .
'Greatest' word problem → HCF; 'smallest/least' → LCM.
Memorise divisibility tests for 2, 3, 4, 5, 6, 8, 9, 11.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Factor — a positive integer that divides another exactly.
Multiple — the product of an integer with a non-zero integer.
Prime factorisation — writing a number as a product of prime numbers, usually in index form.
Highest Common Factor (HCF) — the largest integer that divides each of the given numbers.
Lowest Common Multiple (LCM) — the smallest positive integer that is a multiple of each given number.

How it’s examined

On Paper 2 Cambridge typically asks for the prime factorisation, the HCF, or the LCM directly — usually 1-2 marks each, with a third mark for showing the prime tree. On Paper 4 the same skills hide inside a word problem (cutting into equal pieces, lights flashing simultaneously, buses leaving together). Examiner reports flag the same two failure modes every series: confusing HCF with LCM, and forgetting to write the factorisation in index form.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E1.1, E1.8); 0580/22 May/Jun 2024 — Q4 (HCF / LCM word problem); 0580/42 Oct/Nov 2024 — Q3 (prime factorisation in index form); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Factors and Multiples

Step-by-step solutions to past-paper-style questions on factors and multiples, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Factors and Multiples

Almost every factors and multiples exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — list all factors, find the HCF/LCM, write … in index form — applied to one number or one given pair.

How to approach it

Choose the standard method: pair factors from $1$ outwards, or prime-factorise and apply the HCF rule (lower powers of common primes) or LCM rule (higher powers of all primes). Show every line of working.

Common trap

Examiner reports flag swapping the HCF and LCM power rules, and leaving a factorisation as a long product instead of index form when index form is requested.

Word problem

Recognise it by

A real-world context — flashing lights, ribbons cut into pieces, buses leaving together — with no formula stated, so you must decide between HCF and LCM yourself.

How to approach it

Strip the context to the maths: largest size that divides each / no remainder points to HCF; next time events coincide / smallest common amount points to LCM. Then prime-factorise and apply the rule.

Common trap

Examiner reports note candidates pick the method from the adjective — 'greatest' lures them to LCM, 'together' to HCF — usually the opposite of what the quantity must do.

Multi-step problem

Recognise it by

Several quantities are given — an HCF, an LCM and one of the numbers — and the missing value must be reached by chaining them.

How to approach it

Apply the identity $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$ , substitute the known values, solve, then verify by prime-factorising your answer.

Common trap

Examiner reports flag candidates resorting to trial-and-error instead of the identity, then running out of time — and skipping the verification check.

Show that / prove

Recognise it by

The words show that or prove with the result already stated — for example $\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}$ .

How to approach it

Argue the general case: state the relevant rule ('HCF takes the lower power'), apply it to the exponents, and finish with an explicit concluding line.

Common trap

Examiner reports stress that bare answers score no method marks — the HCF/LCM rule must be explicitly referenced in the working.

1List all factors of a number
CoreDirect calculation• factors
Question
List all the positive factors of $48$ .
Step-by-step solution
1. Step 1
  Pair factors systematically starting from $1$ : $1 \times 48$ , $2 \times 24$ , $3 \times 16$ , $4 \times 12$ , $6 \times 8$ .
2. Step 2
  Stop when the pair starts to repeat — the next would be $8 \times 6$ , already counted.
Answer
$1, 2, 3, 4, 6, 8, 12, 16, 24, 48$
Examiner tip
Pair factors carefully — missing one (often $6$ or $8$ ) costs the accuracy mark.
2Find the HCF by listing
CoreDirect calculation• Adapted from 0580/12 Oct/Nov 2023 Q3• HCF
Question
Find the highest common factor of $24$ and $36$ .
Step-by-step solution
1. Step 1
  Factors of $24$ : $1, 2, 3, 4, 6, 8, 12, 24$ .
2. Step 2
  Factors of $36$ : $1, 2, 3, 4, 6, 9, 12, 18, 36$ .
3. Step 3
  Common factors: $1, 2, 3, 4, 6, 12$ . The highest is $12$ .
Answer
$\text{HCF}(24, 36) = 12$
3Find LCM using prime factorisation
ExtendedDirect calculation• Adapted from 0580/22 May/Jun 2024 Q6• LCM, prime factorisation
Question
Find the LCM of $84$ and $90$ .
Step-by-step solution
1. Step 1
  Prime factorise each number.
  $84 = 2^{2} \times 3 \times 7,\quad 90 = 2 \times 3^{2} \times 5$
2. Step 2
  LCM uses the higher power of every prime that appears.
  $\text{LCM} = 2^{2} \times 3^{2} \times 5 \times 7$
3. Step 3
  Evaluate.
  $= 4 \times 9 \times 5 \times 7 = 1260$
Answer
$\text{LCM}(84, 90) = 1260$
Examiner tip
Show the prime factorisations explicitly. Mark schemes award method marks for them even if the final value is wrong.
4Apply LCM to a real-world timing problem
ExtendedWord problem• LCM, applied
Question
Two lighthouses flash every $24$ seconds and $36$ seconds. They flash together at $9{:}00$ . When do they next flash together?
Step-by-step solution
1. Step 1
  They flash together after the LCM of their cycle times.
2. Step 2
  Prime factorise: $24 = 2^{3} \times 3$ , $36 = 2^{2} \times 3^{2}$ .
3. Step 3
  $\text{LCM} = 2^{3} \times 3^{2} = 72$ seconds.
4. Step 4
  Add to start time: $9{:}00$ + $72$ s = $9{:}01{:}12$ .
Answer
At $9{:}01{:}12$ .
5Use the HCF·LCM identity
ExtendedDirect calculation• HCF, LCM, identity
Question
Two integers have product $720$ and HCF $4$ . Find their LCM.
Step-by-step solution
1. Step 1
  Use the identity $a \times b = \text{HCF} \times \text{LCM}$ .
  $720 = 4 \times \text{LCM}$
2. Step 2
  Divide.
  $\text{LCM} = \dfrac{720}{4} = 180$
Answer
$\text{LCM} = 180$
6HCF word problem — cutting ribbon into equal pieces
ExtendedWord problem• Adapted from 0580/42 Oct/Nov 2023 Q4• HCF, word problem, applied
Question
Three ribbons measure $120$ cm, $180$ cm and $300$ cm. They are cut into pieces of equal length with no ribbon left over. Find (a) the greatest possible length of each piece, (b) the total number of pieces.
Step-by-step solution
1. Step 1
  For each piece to be the same length with no waste, the piece length must divide all three numbers. The greatest such length is $\text{HCF}(120, 180, 300)$ .
2. Step 2
  Prime factorise each value.
  $120 = 2^{3} \times 3 \times 5,\ 180 = 2^{2} \times 3^{2} \times 5,\ 300 = 2^{2} \times 3 \times 5^{2}$
3. Step 3
  Take each common prime to its lowest power: $\text{HCF} = 2^{2} \times 3 \times 5 = 60$ cm.
  $\text{HCF} = 2^{2} \times 3 \times 5 = 60$
4. Step 4
  Number of pieces: $\dfrac{120}{60} + \dfrac{180}{60} + \dfrac{300}{60} = 2 + 3 + 5 = 10$ pieces.
Answer
(a) $60$ cm (b) $10$ pieces.
Examiner tip
The examiner report flags that candidates frequently apply LCM to this style of question, misled by the word 'greatest'. 'No ribbon left over' = piece must divide each length = HCF.
7LCM word problem — coinciding bus arrivals
ExtendedWord problem• Adapted from 0580/22 May/Jun 2024 Q12• LCM, word problem, applied
Question
Three buses leave a station together at $08{:}00$ . Bus A returns every $15$ minutes, bus B every $20$ minutes and bus C every $25$ minutes. At what time do all three next arrive at the station together?
Step-by-step solution
1. Step 1
  All three coincide every $\text{LCM}(15, 20, 25)$ minutes.
2. Step 2
  Prime factorise.
  $15 = 3 \times 5,\ 20 = 2^{2} \times 5,\ 25 = 5^{2}$
3. Step 3
  LCM takes the highest power of each prime: $\text{LCM} = 2^{2} \times 3 \times 5^{2} = 300$ minutes.
  $\text{LCM} = 2^{2} \times 3 \times 5^{2} = 300$
4. Step 4
  $300$ minutes $= 5$ hours. Add to $08{:}00 \Rightarrow 13{:}00$ .
Answer
$13{:}00$ (i.e. $1{:}00$ p.m.).
Examiner tip
Mark schemes award method marks for the prime factorisations and the explicit statement 'use LCM'. Listing multiples by hand is slow and error-prone — get used to the prime-factor method.
8Express a number as $2^a \times 3^b \times 5^c$
ExtendedDirect calculation• Adapted from 0580/22 Feb/Mar 2024 Q5• prime factorisation, index form
Question
Write $1\,800$ in the form $2^{a} \times 3^{b} \times 5^{c}$ , stating the values of $a$ , $b$ and $c$ .
Step-by-step solution
1. Step 1
  Divide by $2$ repeatedly until odd: $1800 \div 2 = 900$ , $900 \div 2 = 450$ , $450 \div 2 = 225$ . So $1800 = 2^{3} \times 225$ .
2. Step 2
  Factor $225$ : $225 \div 3 = 75$ , $75 \div 3 = 25$ . So $225 = 3^{2} \times 25$ .
3. Step 3
  Factor $25 = 5^{2}$ . Combine.
  $1800 = 2^{3} \times 3^{2} \times 5^{2}$
4. Step 4
  Compare with the given form: $a = 3$ , $b = 2$ , $c = 2$ .
Answer
$1800 = 2^{3} \times 3^{2} \times 5^{2}$ ; $a = 3$ , $b = 2$ , $c = 2$ .
Examiner tip
The 2024 mark scheme awards full marks only when the index form is fully simplified. A common slip is leaving one factor as $4$ or $9$ instead of $2^{2}$ or $3^{2}$ .
9Use the HCF · LCM identity to find a missing number
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q8• HCF, LCM, identity, stretch
Question
Two positive integers $a$ and $b$ have $\text{HCF}(a,b) = 15$ and $\text{LCM}(a,b) = 630$ . If $a = 45$ , find $b$ .
Step-by-step solution
1. Step 1
  Use the identity $a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$ .
  $a \times b = 15 \times 630$
2. Step 2
  Substitute $a = 45$ .
  $45 \times b = 15 \times 630 = 9450$
3. Step 3
  Solve for $b$ .
  $b = \dfrac{9450}{45} = 210$
4. Step 4
  Verify: $45 = 3^{2} \times 5$ , $210 = 2 \times 3 \times 5 \times 7$ . HCF $= 3 \times 5 = 15$ ✓, LCM $= 2 \times 3^{2} \times 5 \times 7 = 630$ ✓.
Answer
$b = 210$ .
Examiner tip
The 2024 examiner report notes candidates who memorise the HCF·LCM identity convert a 4-mark stretch question into a one-line calculation. Verify with prime factorisation to secure the final accuracy mark.
10Show that $\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}$
ChallengeShow that / prove• Adapted from 0580/42 Oct/Nov 2024 Q23• show that, proof, prime factorisation, algebra
Question
Let $n$ be a positive integer. Show that $\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}$ and find the corresponding LCM in terms of $n$ .
Step-by-step solution
1. Step 1
  Both numbers have the same prime, $2$ , with different exponents: $2^{n}$ has exponent $n$ , and $2^{n+1}$ has exponent $n+1$ .
2. Step 2
  HCF rule: take each common prime to the lower power. The lower power is $\min(n, n+1) = n$ .
  $\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}$
3. Step 3
  LCM rule: take each prime to the higher power. The higher power is $\max(n, n+1) = n+1$ .
  $\text{LCM}(2^{n}, 2^{n+1}) = 2^{n+1}$
4. Step 4
  Check using the HCF·LCM identity: $2^{n} \times 2^{n+1} = 2^{2n+1}$ and $\text{HCF} \times \text{LCM} = 2^{n} \times 2^{n+1} = 2^{2n+1}$ ✓.
Answer
$\text{HCF} = 2^{n}$ (lower power of the only common prime). $\text{LCM} = 2^{n+1}$ .
Examiner tip
On 'show that' questions the examiner report demands an explicit reference to the HCF rule ('lower power') as part of the working. State the rule, then apply it — bare answers score no method marks.

Key Formulae — Factors and Multiples

The formulae you need to memorise for factors and multiples on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

HCF · LCM identity
$a \times b = \text{HCF}(a,b) \times \text{LCM}(a,b)$
$a, b$
two positive integers
$\text{HCF}(a,b)$
their highest common factor
$\text{LCM}(a,b)$
their lowest common multiple
When to use
Quickest way to find one of HCF or LCM when the product and the other are known.
HCF rule (prime factorisation)
$\text{HCF} = \prod p_{i}^{\min(a_{i}, b_{i})}$
$p_{i}$
primes appearing in both factorisations
$a_{i}, b_{i}$
the powers of $p_{i}$ in each number
When to use
Standard method on Paper 4 extended HCF questions.
LCM rule (prime factorisation)
$\text{LCM} = \prod p_{i}^{\max(a_{i}, b_{i})}$
$p_{i}$
every prime in either factorisation
$a_{i}, b_{i}$
the powers of $p_{i}$ in each number
When to use
Standard method on Paper 4 extended LCM questions.

Key Definitions and Keywords — Factors and Multiples

Definitions to memorise and the exact keywords mark schemes credit for factors and multiples answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Factor (divisor)
Examiner keyword
An integer that divides another integer leaving no remainder.
Multiple
Examiner keyword
The result of multiplying an integer by another integer.
Common factor
A number that is a factor of two or more given numbers.
Common multiple
A number that is a multiple of two or more given numbers.
Highest common factor (HCF)
Examiner keyword
The largest positive integer that divides each of two or more given integers exactly.
Lowest common multiple (LCM)
Examiner keyword
The smallest positive integer that is a multiple of each of two or more given integers.
Prime factor
A factor that is itself a prime number.
Example
Prime factors of $30$ : $2, 3, 5$ .
Coprime (relatively prime)
Two numbers are coprime if their HCF is $1$ .
Example
$8$ and $15$ are coprime: $\gcd(8,15) = 1$ .

Common Mistakes and Misconceptions — Factors and Multiples

The traps other students keep falling into on factors and multiples questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Mixing up HCF and LCM rules in the prime factorisation method
0580/42 May/Jun 2024 — examiner report
Why it happens
Both rules use "common primes" but with different powers, and exam pressure leads to the wrong choice.
How to avoid it
Memorise: HCF = lower powers, LCM = larger powers. The HCF must fit inside both numbers.
✕Missing a prime when listing the LCM
Why it happens
Students take only common primes (HCF rule) instead of all primes that appear in either number.
How to avoid it
For LCM, write every prime from either factorisation, including primes that appear in one number only.
✕Missing factors when listing all factors
Why it happens
Students stop early or skip factor pairs (e.g. forgetting $6$ when listing factors of $48$ ).
How to avoid it
Pair factors from $1$ outwards: $1 \times n$ , $2 \times (n/2)$ , $\dots$ until pairs repeat.
✕Listing $1$ as a prime factor
Why it happens
Mixing up factors with prime factors.
How to avoid it
$1$ is never prime. Prime factorisations contain only primes $\geq 2$ .
✕Leaving prime factorisation as a long product instead of using indices
0580/12 Oct/Nov 2022 — examiner report
Why it happens
The factorisation looks complete with $2 \times 2 \times 3 \times 3 \times 5$ .
How to avoid it
If the question says "in index form", group repeats: $2^{2} \times 3^{2} \times 5$ .

Factors and Multiples — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Highest Common Factor (HCF) of two integers is the largest integer that divides both. Lowest Common Multiple (LCM) is the smallest positive integer that BOTH numbers divide into.

Method 1: Listing. Best for small numbers ( $\le 30$ ). List all factors of each number; pick the largest common one for HCF. List multiples until both lists agree for LCM.

Method 2: Prime-factorisation tree (RECOMMENDED for $> 30$ ). Write each number as a product of primes in index form.

HCF = take the lowest power of every prime that appears in BOTH.
LCM = take the highest power of every prime that appears in EITHER.

For $\mathbf{72}$ and $\mathbf{108}$ :

$72 = 2^{3} \times 3^{2}$
$108 = 2^{2} \times 3^{3}$
HCF = $2^{2} \times 3^{2} = 4 \times 9 = 36$ (lowest power of each shared prime).
LCM = $2^{3} \times 3^{3} = 8 \times 27 = 216$ (highest power of each prime).

HCF is the product of the overlap; LCM is the product of every prime in the whole diagram.

Quick sanity checks. HCF should never be larger than the smaller of the two numbers. LCM should never be smaller than the larger.

Word problems. Real-paper questions disguise HCF/LCM as everyday situations. Spot the verb:

"Cut into the largest equal pieces" / "the greatest number of..." → HCF.
"Both events happen at the same time again" / "the smallest number that" → LCM.

The verb in the question tells you which to use: "greatest" → HCF, "smallest / again together" → LCM.

Divisor

Test

2

Last digit is even (

0, 2, 4, 6, 8

3

Digit sum is a multiple of

3

4

Last two digits form a multiple of

4

5

Last digit is

0

5

6

Divisible by both

2

AND

3

8

Last three digits form a multiple of

8

9

Digit sum is a multiple of

9

10

Last digit is

0

11

Alternating digit sum is a multiple of

11

(including

0

Factors and Multiples

Short Notes - Factors and Multiples

Detailed Study Notes

At a glance

What you’ll learn

Factors and multiples — the basics

Prime factorisation

HCF and LCM — three reliable methods

Divisibility shortcuts

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Factors and Multiples

1List all factors of a number

2Find the HCF by listing

3Find LCM using prime factorisation

4Apply LCM to a real-world timing problem

5Use the HCF·LCM identity

6HCF word problem — cutting ribbon into equal pieces

7LCM word problem — coinciding bus arrivals

8Express a number as 2a×3b×5c2^a \times 3^b \times 5^c2a×3b×5c

9Use the HCF · LCM identity to find a missing number

10Show that HCF(2n,2n+1)=2n\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}HCF(2n,2n+1)=2n

HCF · LCM identity

HCF rule (prime factorisation)

LCM rule (prime factorisation)

Factor (divisor)

Multiple

Common factor

Common multiple

Highest common factor (HCF)

Lowest common multiple (LCM)

Prime factor

Coprime (relatively prime)

✕Mixing up HCF and LCM rules in the prime factorisation method

✕Missing a prime when listing the LCM

✕Missing factors when listing all factors

✕Listing 111 as a prime factor

✕Leaving prime factorisation as a long product instead of using indices

Factors and Multiples — frequently asked questions

Factors and Multiples

Short Notes - Factors and Multiples

Detailed Study Notes

At a glance

What you’ll learn

Factors and multiples — the basics

Prime factorisation

HCF and LCM — three reliable methods

Divisibility shortcuts

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Factors and Multiples

1List all factors of a number

2Find the HCF by listing

3Find LCM using prime factorisation

4Apply LCM to a real-world timing problem

5Use the HCF·LCM identity

6HCF word problem — cutting ribbon into equal pieces

7LCM word problem — coinciding bus arrivals

8Express a number as 2a×3b×5c2^a \times 3^b \times 5^c2a×3b×5c

9Use the HCF · LCM identity to find a missing number

10Show that HCF(2n,2n+1)=2n\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}HCF(2n,2n+1)=2n

HCF · LCM identity

HCF rule (prime factorisation)

LCM rule (prime factorisation)

Factor (divisor)

Multiple

Common factor

Common multiple

Highest common factor (HCF)

Lowest common multiple (LCM)

Prime factor

Coprime (relatively prime)

✕Mixing up HCF and LCM rules in the prime factorisation method

✕Missing a prime when listing the LCM

✕Missing factors when listing all factors

✕Listing 111 as a prime factor

8Express a number as $2^a \times 3^b \times 5^c$

10Show that $\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}$

✕Listing $1$ as a prime factor

8Express a number as $2^a \times 3^b \times 5^c$

10Show that $\text{HCF}(2^{n}, 2^{n+1}) = 2^{n}$

✕Listing $1$ as a prime factor