Primes, factors, multiples, HCF and LCM - Edexcel GCSE Maths (Calculator) Exam Questions

Practise 18 exam-style questions on Primes, factors, multiples, HCF and LCM for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

18 questions
1–5 marks each
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18 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Write 3960 as a product of its prime factors.
Give your answer in index form.

Model answer

\(3960 = 2 \times 1980\)
\(1980 = 2 \times 990\)
\(990 = 2 \times 495\)
\(495 = 3 \times 165\)
\(165 = 3 \times 55\)
\(55 = 5 \times 11\)
\(3960 = 2^3 \times 3^2 \times 5 \times 11\)

Explanation

Keep dividing by the smallest prime that goes in until you reach 1 (or use a factor tree). Every branch must end in a prime. In index form, repeated factors are written as powers: \(2 \times 2 \times 2 = 2^3\).

Mark scheme

1 mark
M1 for a correct method, e.g. a factor tree or repeated division with at least two correct steps
e.g. 3960 = 2 × 1980 = 2 × 2 × 990 ..; correct answer with no working scores full marks
1 mark
A1 for \(2^3 \times 3^2 \times 5 \times 11\)
cao; accept the factors in any order

Question 2

3 marks

Use prime factors to show that 1764 is a square number.

Model answer

\(1764 = 2 \times 882 = 2 \times 2 \times 441 = 2 \times 2 \times 3 \times 147 = 2 \times 2 \times 3 \times 3 \times 49\)
\(1764 = 2^2 \times 3^2 \times 7^2\)
Every prime factor has an even power, so the factors can be split into two equal groups:
\(1764 = (2 \times 3 \times 7) \times (2 \times 3 \times 7) = 42^2\)
So 1764 is a square number.

Explanation

A number is a square number exactly when every power in its prime factorisation is even, because the factors can then be shared equally into two identical groups. Halving each power gives the square root: \(2 \times 3 \times 7 = 42\). The question says use prime factors, so just finding \(\sqrt{1764}\) on a calculator is not enough.

Mark scheme

1 mark
M1 for a correct method to find the prime factors of 1764, with at least two correct steps
1 mark
A1 for \(1764 = 2^2 \times 3^2 \times 7^2\)
oe, e.g. 2 × 2 × 3 × 3 × 7 × 7
1 mark
C1 for a correct explanation, e.g. every power is even, so \(1764 = (2 \times 3 \times 7)^2 = 42^2\)
dependent on A1; answer only (e.g. \(\sqrt{1764} = 42\)) scores 0

Question 3

3 marks

Two numbers have a highest common factor of 18 and a lowest common multiple of 540

Both numbers are between 50 and 150

Find the two numbers.

Model answer

Both numbers are multiples of 18: \(18a\) and \(18b\), where \(a\) and \(b\) have no common factor.
LCM = \(18ab = 540\), so \(ab = 30\)
Pairs: 1 and 30, 2 and 15, 3 and 10, 5 and 6
Numbers: 18 and 540, 36 and 270, 54 and 180, 90 and 108
Only 90 and 108 are both between 50 and 150.
Check: \(90 = 2 \times 3^2 \times 5\), \(108 = 2^2 \times 3^3\), HCF = \(2 \times 3^2 = 18\), LCM = \(2^2 \times 3^3 \times 5 = 540\)

Explanation

Both numbers must be multiples of the HCF, and the parts left after taking out 18 must have no common factor (otherwise the HCF would be bigger). Their product with 18 gives the LCM. Testing multiples of 18 between 50 and 150 (54, 72, 90, 108, 126, 144) also works.

Mark scheme

1 mark
M1 for writing the numbers as \(18a\) and \(18b\) with \(a \times b = 30\) (\(540 \div 18\))
or listing multiples of 18 between 50 and 150 and testing pairs, or factors of 540; correct answer with no working scores full marks
1 mark
M1 for considering pairs with HCF 18, e.g. 54 and 180, 90 and 108
1 mark
A1 for 90 and 108
cao

Question 4

4 marks

\(A = 2^x \times 3^4 \times 5\)
\(B = 2^3 \times 3^y \times 7\)

where \(x\) and \(y\) are positive integers.

The highest common factor of \(A\) and \(B\) is 72
The lowest common multiple of \(A\) and \(B\) is 45 360

Find the value of \(x\) and the value of \(y\).

Model answer

HCF: \(72 = 2^3 \times 3^2\)
The HCF uses the lower power of 3, and A has \(3^4\), so \(3^y = 3^2\) and \(y = 2\).
LCM: \(45\,360 = 2^4 \times 3^4 \times 5 \times 7\)
The LCM uses the higher power of 2, and B has \(2^3\), so \(2^x = 2^4\) and \(x = 4\).
Check: HCF uses \(2^3\), the lower of \(2^4\) and \(2^3\).
\(x = 4\), \(y = 2\)

Explanation

Write the HCF and LCM as products of primes. The HCF takes the lower power of each shared prime and the LCM takes the higher power. Since A has \(3^4\) but the HCF has only \(3^2\), B must have \(3^2\); since B has only \(2^3\) but the LCM has \(2^4\), A must have \(2^4\).

Mark scheme

1 mark
M1 for \(72 = 2^3 \times 3^2\)
correct answer with no working scores full marks
1 mark
M1 for \(45\,360 = 2^4 \times 3^4 \times 5 \times 7\)
1 mark
A1 for \(x = 4\)
cao; dependent on the M1 for the LCM
1 mark
A1 for \(y = 2\)
cao; may follow from the HCF alone

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