Factors, multiples, primes, HCF & LCM – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator3 marks

Here is a list of numbers.

$12\quad 17\quad 21\quad 25\quad 27\quad 36\quad 39$

From the list, write down

(a) a prime number [1]

(b) a factor of 75 [1]

(a) $17$B1
(b) $25$B1
(c) $39$B1
Answer: (a) 17 (b) 25 (c) 39
Examiner tip: 21, 27 and 39 are not prime: check divisibility by 3 (add the digits) before choosing.

2Grade 4–5Non-calculator2 marks

Write 360 as a product of its prime factors. Give your answer in index form. [2]

for a correct method to find the prime factors, e.g. a complete factor tree or repeated division with at most one error, or for the factors 2, 2, 2, 3, 3, 5 identifiedM1
$2^3 \times 3^2 \times 5$ (accept in any order; accept $5^1$; do not accept $2 \times 2 \times 2 \times 3 \times 3 \times 5$ for this mark)A1
Answer: $2^3 \times 3^2 \times 5$
Examiner tip: Index form means powers: 2 × 2 × 2 × 3 × 3 × 5 only scores the method mark.

3Grade 4–5Non-calculator4 marks

(a) Find the highest common factor (HCF) of 84 and 126. [2]

(b) Find the lowest common multiple (LCM) of 84 and 126. [2]

(a) for $84 = 2^2 \times 3 \times 7$ and $126 = 2 \times 3^2 \times 7$, or lists of factors of both numbers (at least 3 correct each)M1
(a) $42$A1
(b) for $2^2 \times 3^2 \times 7$ or lists of multiples of both numbers (at least 3 correct each) or $84 \times 126 \div 42$M1
(b) $252$A1
Answer: (a) 42 (b) 252
Examiner tip: Do not mix them up: the HCF can be no bigger than the smaller number; the LCM is at least as big as the larger number.

4Grade 4–5Non-calculator3 marks

Bus A leaves the bus station every 12 minutes. Bus B leaves the bus station every 18 minutes.

Both buses leave the bus station at 7:00 am.

At what time will both buses next leave the bus station at the same time? [3]

for listing multiples of 12 and 18 (at least 3 of each) or prime factorisations $2^2 \times 3$ and $2 \times 3^2$P1
for identifying 36 (the LCM) as the number of minutesP1
7:36 am (accept 07:36)A1
Answer: 7:36 am
Examiner tip: The answer is a time, not '36' – add the LCM on to 7:00 am to get the final mark.

5Grade 6–7HigherNon-calculator4 marks

$A = 2^3 \times 3^2 \times 5$ and $B = 2^2 \times 3^4 \times 7$

(a) Find the HCF of $A$ and $B$. [1]

(b) Find the LCM of $A$ and $B$. Give your answer as a product of powers of its prime factors. [1]

(a) $36$ or $2^2 \times 3^2$B1
(b) $2^3 \times 3^4 \times 5 \times 7$B1
(c) for identifying that the powers of 2 and 5 are odd, or that each prime needs an even power, e.g. $2^4 \times 3^2 \times 5^2$M1
(c) $10$A1
Answer: (a) 36 (b) $2^3 \times 3^4 \times 5 \times 7$ (c) 10
Examiner tip: For a square number every prime power must be even – multiply by just enough to make each odd power even.

6Grade 4–5Non-calculator4 marks

Ella has 72 red counters and 108 blue counters.

She puts all the counters into bags so that

every bag contains the same number of red counters
every bag contains the same number of blue counters
there are no counters left over.

(a) Work out the greatest number of bags she can use. [3]

(b) How many counters are in each bag? [1]

for starting a correct process, e.g. listing at least 3 correct factors of each of 72 and 108 (no wrong ones), or a correct prime factorisation of one of the numbers, e.g. $72 = 2^3 \times 3^2$P1
for a complete process to find the HCF, e.g. full factor lists of both with 36 identified as the highest common factor, or $72 = 2^3 \times 3^2$ and $108 = 2^2 \times 3^3$ with $2^2 \times 3^2$ selected, or a correct Venn diagram of prime factorsP1
(a) $36$ (bags)A1
(b) $5$ ft their answer to (a) if it is a common factor of 72 and 108, i.e. $\frac{72}{\text{their (a)}} + \frac{108}{\text{their (a)}}$; accept '2 red and 3 blue' (or 2 + 3)B1
Answer: (a) 36 bags (b) 5 counters (2 red, 3 blue)
Examiner tip: 'Greatest number of equal groups' means HCF; 'when will they next coincide' means LCM.

7Grade 8–9HigherNon-calculator3 marks

Two numbers are both between 50 and 100.

The highest common factor of the two numbers is 12.

The lowest common multiple of the two numbers is 360.

Find the two numbers. [3]

for writing the numbers as $12a$ and $12b$ with $ab = 360 \div 12 = 30$, or for finding multiples of 12 between 50 and 100 (60, 72, 84, 96)P1
for identifying that $a$ and $b$ must share no common factor and selecting $5$ and $6$, or testing pairs of multiples of 12 against LCM 360P1
$60$ and $72$A1
Answer: 60 and 72
Examiner tip: Check both conditions at the end: HCF(60, 72) = 12 and LCM(60, 72) = 360 – a pair like 24 and 180 fails the 50–100 condition.

8Grade 1–3FoundationNon-calculator4 marks

(a) List all the factors of 28. [2]

(b) Write down a multiple of 6 that is between 40 and 50. [1]

(a) at least 4 correct factors of 28 with no more than one incorrect valueB1
(a) 1, 2, 4, 7, 14, 28 — all six and no extras (any order; ignore repeats; accept factor pairs such as 1×28, 2×14, 4×7)B1
(b) 42 or 48 (accept either; do not accept 36 or 54)B1
(c) 14 (do not accept 7, 2 or 1 as they are not greater than 7)B1
Answer: (a) 1, 2, 4, 7, 14, 28 (b) 42 or 48 (c) 14
Examiner tip: Work through factor pairs (1×28, 2×14, 4×7) so you don't miss 1 and the number itself — leaving out 28 costs the second mark.

9Grade 1–3FoundationNon-calculator4 marks

(a) Write down all the prime numbers between 30 and 50. [2]

(b) Kai says,

“When you add two prime numbers together, the answer is always even.”

Give an example to show that Kai is wrong. [1]

(a) at least 3 of 31, 37, 41, 43, 47 with no more than one incorrect valueB1
(a) 31, 37, 41, 43, 47 — all five and no extrasB1
(b) a correct counter-example using 2 and an odd prime, e.g. $2 + 3 = 5$ (must show an odd total)C1
(c) No, with reason $91 = 7 \times 13$ (accept “91 ÷ 7 = 13” or “7 is a factor of 91”); “No” alone scores 0C1
Answer: (a) 31, 37, 41, 43, 47 (b) e.g. $2+3=5$, which is odd (c) No, $91 = 7\times13$
Examiner tip: To show a number is not prime you must give an actual factor — saying “No” without $7\times13$ scores nothing.

10Grade 4–5Non-calculator4 marks

Here is a factor tree for 588.

(a) Write the three missing numbers in the boxes. [2]

(b) Write 588 as a product of its prime factors. Give your answer in index form. [2]

(a) 4 in the box below 588B1
(a) 49 in the box below 147 and 7 in the box below 49B1
(b) $2 \times 2 \times 3 \times 7 \times 7$ (any order) or all prime factors 2, 2, 3, 7, 7 identifiedM1
(b) $2^2 \times 3 \times 7^2$ (accept $2^2 \times 3^1 \times 7^2$, any order; do not accept with “+” or a 1 included as a factor)A1
Answer: (a) 4, 49, 7 (b) $2^2 \times 3 \times 7^2$
Examiner tip: Index form means repeated primes written as powers — $2\times2\times3\times7\times7$ only gets the method mark.

11Grade 4–5Non-calculator4 marks

The Venn diagram shows the prime factors of two numbers, $P$ and $Q$.

(a) Work out the value of $P$. [1]

(b) Write down the highest common factor (HCF) of $P$ and $Q$. [1]

(a) 60B1
(b) 10 (accept $2\times5$)B1
(c) $2\times3\times2\times5\times5\times7$ (product of every number in the diagram) or $2^2\times3\times5^2\times7$, or $Q = 350$ found and lists multiples of 60 and 350M1
(c) 2100A1
Answer: (a) 60 (b) 10 (c) 2100
Examiner tip: P is every number inside circle P, including the overlap; the LCM uses every number in the diagram exactly once.

12Grade 4–5Non-calculator4 marks

Sam is buying sausages and bread rolls for a barbecue.

Sausages are sold in packs of 12. Each pack costs £2.40
Bread rolls are sold in packs of 20. Each pack costs £1.15

Sam wants to buy exactly the same number of sausages as bread rolls.

He buys the smallest possible number of packs.

Work out the total amount Sam pays. [4]

lists at least 3 multiples of 12 and at least 3 multiples of 20, or finds prime factors $2^2\times3$ and $2^2\times5$P1
identifies 60 as the LCM and works out 5 packs of sausages and 3 packs of rollsP1
$5 \times 2.40\ (=12)$ and $3 \times 1.15\ (=3.45)$, or their numbers of packs × prices addedP1
£15.45 (accept 15.45; do not accept £15.4)A1
Answer: £15.45
Examiner tip: The LCM (60) tells you the number of items, not packs — divide by 12 and 20 before using the prices.

13Grade 6–7HigherNon-calculator5 marks

$N = 2^3 \times 3^2 \times 7$

(a) Explain why $N$ is a multiple of 12. [1]

(b) Write $N^2$ as a product of powers of its prime factors. [1]

(d) Find the highest common factor (HCF) of $N$ and 1050. [2]

(a) $12 = 2^2 \times 3$ and $N$ contains $2^2$ and $3$ (e.g. $N = 12 \times 2 \times 3 \times 7$ or $N = 504 = 12 \times 42$)C1
(b) $2^6 \times 3^4 \times 7^2$B1
(c) 63 (accept $3^2 \times 7$)B1
(d) $1050 = 2 \times 3 \times 5^2 \times 7$, or lists factors of both 504 and 1050 including the HCFM1
(d) 42 (accept $2 \times 3 \times 7$)A1
Answer: (a) $12 = 2^2\times3$, which is contained in $N$ (b) $2^6 \times 3^4 \times 7^2$ (c) 63 (d) 42
Examiner tip: For the largest odd factor just remove every factor of 2 — there is no need to work out $N$.

14Grade 6–7HigherCalculator4 marks

Two gear wheels, A and B, turn together. Gear A has 36 teeth and gear B has 48 teeth.

At the start, the two marked teeth are touching, as shown in the diagram.

(a) Work out the smallest number of complete turns gear A must make before the two marked teeth are touching again. [3]

(b) How many complete turns does gear B make in this time? [1]

(a) lists multiples of 36 and 48 (at least 3 of each) or writes $36 = 2^2\times3^2$ and $48 = 2^4\times3$M1
(a) 144 (teeth) identified as the LCMA1
(a) 4 (turns), from $144 \div 36$A1
(b) 3 (ft their LCM ÷ 48 if a whole number)B1ft
Answer: (a) 4 turns (b) 3 turns
Examiner tip: The LCM counts teeth, not turns — 144 alone is not the answer to (a); divide by the number of teeth on each gear.

15Grade 8–9HigherNon-calculator3 marks

$n$ is a positive integer.

The lowest common multiple (LCM) of 12 and $n$ is 60.

Find all the possible values of $n$. [3]

$12 = 2^2 \times 3$ and $60 = 2^2 \times 3 \times 5$, or states/uses that $n$ must be a factor of 60M1
recognises $n$ must be a multiple of 5 (contain the factor 5), or finds at least 4 correct values with no more than one incorrectM1
5, 10, 15, 20, 30, 60 — all six and no extrasA1
Answer: $n = 5, 10, 15, 20, 30, 60$
Examiner tip: Every value must divide 60 and include the factor 5 — students often stop at 60 and 5 and miss 10, 15, 20 and 30.

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