Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Number
Practise Primes, factors, HCF and LCM. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Primes, factors and multiples, writing a number as a product of its prime factors, and finding the highest common factor (HCF) and lowest common multiple (LCM). Expect 2 to 3 mark questions on both papers, often a word problem about sharing into equal groups or things that happen together.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
1
Find factors and multiples of a numberFactors divide exactly into a number (factors of 12: 1, 2, 3, 4, 6, 12); multiples are in its times table (12, 24, 36, ...).
2
Recognise prime numbersA prime has exactly two factors, 1 and itself: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...
3
Find HCF and LCM by listingList factors for the HCF and multiples for the LCM: for 12 and 18, the HCF is 6 and the LCM is 36.
4
Write a number as a product of primesUse a factor tree or repeated division: \(360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5\).
5
Find HCF and LCM using prime factorsHCF: primes in both numbers with the lower power; LCM: every prime with the higher power (or use a Venn diagram).
5
Solve HCF and LCM word problemsSharing into equal groups or the largest size means HCF; things happening together again means LCM.
Notes
Factors, multiples and primes
A factor of a number divides into it exactly. Find factors in pairs: for 36, 1 × 36, 2 × 18, 3 × 12, 4 × 9 and 6 × 6.
A multiple of a number is in its times table: the multiples of 7 are 7, 14, 21, 28, ...
A prime number has exactly two factors: 1 and itself. 1 is not prime (it has only one factor). 2 is the only even prime.
Primes up to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
Product of prime factors
Every whole number greater than 1 can be written as a product of primes in exactly one way (apart from the order). This is the unique factorisation theorem.
Use a factor tree (split into factor pairs until every branch ends in a prime) or repeated division: 360 ÷ 2 = 180, ÷ 2 = 90, ÷ 2 = 45, ÷ 3 = 15, ÷ 3 = 5.
Write the answer as a product (with × signs), usually in index form: \(360 = 2^3 \times 3^2 \times 5\).
Check by multiplying back: 8 × 9 × 5 = 360.
HCF and LCM
HCF: the highest number that is a factor of both. LCM: the lowest number that is a multiple of both.
Sharing into equal groups (largest group size) → HCF; happening together again, or the smallest equal total from packs of different sizes → LCM
How to answer each type of question
Write a number as a product of its prime factors
2 marks4
Use a factor tree or divide repeatedly by primes (2, 3, 5, 7, ...).
Stop only when every branch ends in a prime.
Write the primes multiplied together, not as a list with commas. Index form is fine.
Check by multiplying back.
Example. Write 252 as a product of its prime factors.
Show the model answer
252 ÷ 2 = 126, 126 ÷ 2 = 63, 63 ÷ 3 = 21, 21 ÷ 3 = 7 (M1 for at least two correct steps in a factor tree or repeated division) \(252 = 2 \times 2 \times 3 \times 3 \times 7 = 2^2 \times 3^2 \times 7\) (A1)
Find the HCF and LCM of numbers in prime factor form
1 to 2 marks each5
Do not multiply the numbers out first; read the powers.
HCF: each prime that is in both, with the lower power.
LCM: every prime, with the higher power.
Give the answer in the form the question asks (index form or an ordinary number).
Example. \(A = 2^3 \times 3 \times 5^2\) and \(B = 2 \times 3^2 \times 5\). (a) Find the highest common factor (HCF) of A and B. (b) Find the lowest common multiple (LCM) of A and B.
Show the model answer
(a) Primes in both with the lower power: \(2 \times 3 \times 5 = 30\) (B1) (b) Every prime with the higher power: \(2^3 \times 3^2 \times 5^2 = 1800\) (B1)
HCF or LCM word problem
3 marks5
Decide: are things happening together again (LCM), or being shared into equal groups (HCF)?
List multiples (or factors), or use prime factors.
Answer the question that was asked, e.g. a time of day or a number of packs, not just the LCM.
Example. Buses to the airport leave a station every 18 minutes. Buses to the zoo leave the same station every 24 minutes. Both buses leave at 9:00 am. When will both buses next leave the station at the same time?
Show the model answer
Multiples of 18: 18, 36, 54, 72, ... Multiples of 24: 24, 48, 72, ... (P1 for listing multiples of both, or \(18 = 2 \times 3^2\) and \(24 = 2^3 \times 3\)) LCM = 72 minutes (P1) 9:00 am + 72 minutes = 10:12 am (A1)
Shortcuts and memory tricks
Divisibility tests: by 2 if even; by 3 if the digits add up to a multiple of 3; by 5 if it ends in 0 or 5; by 9 if the digits add up to a multiple of 9.
Sense check: the HCF can never be bigger than the smaller number, and the LCM can never be smaller than the bigger number.
Check a prime factorisation by multiplying it back out.
Many scientific calculators have a prime factor (FACT) function: use it to check on the calculator paper.
Where marks are lost
Including 1 as a prime number, or forgetting that 2 is prime.
Stopping a factor tree too early, e.g. leaving 9 or 15 at the end of a branch (they are not prime).
Writing the primes as a list (2, 2, 3, 3, 7) instead of a product (2 × 2 × 3 × 3 × 7).
Mixing up HCF and LCM: things happening together again needs the LCM.
Finding the LCM (72 minutes) but not answering the question (10:12 am).
Exam technique
'Write as a product of its prime factors' needs × signs between the primes; index form is accepted.
For word problems, decide HCF or LCM first, then finish with a sentence that answers the question, with units.
If the numbers are given in prime factor form, don't multiply them out; read the HCF and LCM straight from the powers.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
Write \(288\) as a product of its prime factors.[2]
M1 for a complete method with at most one arithmetic error, e.g. a factor tree or repeated division, or for all the prime factors found (2, 2, 2, 2, 2, 3, 3) but not written as a product
A1 for \(2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3\) or \(2^{5} \times 3^{2}\) (factors in any order)
\(A = 2^3 \times 3 \times 5\) and \(B = 2 \times 3^2 \times 7\). Jo says, 'The highest common factor (HCF) of \(A\) and \(B\) is \(2^3 \times 3^2 = 72\).'
(a) Explain why Jo must be wrong.[1]
(b) Find the HCF of \(A\) and \(B\).[1]
(c) Find the lowest common multiple (LCM) of \(A\) and \(B\). Give your answer as an ordinary number.[2]
Show the answer and mark scheme
(a)Answer: 72 is not a factor of \(A\) (\(A = 120\) and 120 ÷ 72 is not a whole number). For the HCF you take the lowest power of each common prime.
C1 for a correct reason, e.g. 72 is not a factor of \(A\) (or of \(B\)), or \(2^3\) is not a factor of \(B\), or \(3^2\) is not a factor of \(A\), or Jo has used the highest powers instead of the lowest
Worked solution: \(A = 120\) and \(B = 126\). The HCF must divide both numbers, but 72 divides neither. Jo used the highest power of each prime; the HCF uses the lowest power of each prime that is common to both.
(b)Answer: 6
B1 for 6 (or \(2 \times 3\))
Worked solution: The common primes are 2 and 3, each to the lowest power: \(2 \times 3 = 6\).
(c)Answer: 2520
M1 for \(2^3 \times 3^2 \times 5 \times 7\) oe
A1 for 2520
Worked solution: Take each prime to its highest power: \(2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 35 = 2520\).