Practise 28 exam-style questions on Primes, factors, multiples, HCF and LCM for Edexcel GCSE Maths (Non-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.
28 questions
1–5 marks each
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20 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 180 as a product of its prime factors. Give your answer in index form.
Split the number into factors repeatedly (a factor tree) until every branch ends in a prime. Then collect equal primes using powers. Whatever factors you start with, you always get the same primes - this is the unique factorisation theorem.
Mark scheme
1 mark
M1 for a correct method, e.g. a factor tree or repeated division with at least two correct steps
or 2, 2, 3, 3, 5 listed; correct answer with no working scores full marks
1 mark
A1 for \(2^2 \times 3^2 \times 5\)
cao; accept factors in any order; 2 × 2 × 3 × 3 × 5 scores M1 A0
Question 2
2 marks
Explain why the sum of two different prime numbers that are both greater than 2 can never be a prime number.
Model answer
Every prime number greater than 2 is odd (an even number greater than 2 has 2 as a factor). Odd + odd = even, so the sum is even. The sum is even and greater than 2, so 2 is a factor of it, as well as 1 and the sum itself. So the sum has at least three factors and is not prime.
Explanation
Use the facts that 2 is the only even prime and that odd + odd is always even. An even number bigger than 2 cannot be prime. Giving examples such as 3 + 5 = 8 does not explain why it is true for every pair.
Mark scheme
1 mark
C1 for stating that all primes greater than 2 are odd, so the sum of two of them is even
1 mark
C1 for a complete explanation: the sum is even and greater than 2, so it has 2 as a factor (and is not 2 itself), so it is not prime
examples alone score 0
Question 3
3 marks
Three lights flash at regular intervals. Light A flashes every 18 seconds. Light B flashes every 24 seconds. Light C flashes every 30 seconds. At 7 pm all three lights flash at the same time.
At what time will all three lights next flash at the same time?
Model answer
\(18 = 2 \times 3^2\), \(24 = 2^3 \times 3\), \(30 = 2 \times 3 \times 5\) LCM \(= 2^3 \times 3^2 \times 5 = 360\) seconds 360 seconds = 6 minutes They next flash together at 7:06 pm.
Explanation
The lights flash together again after a time that is a multiple of all three intervals; the first such time is the LCM. Use the highest power of each prime. Remember to convert seconds to minutes and give a time, not just a number of seconds.
Mark scheme
1 mark
P1 for a correct method to find the LCM, e.g. \(18 = 2 \times 3^2\), \(24 = 2^3 \times 3\), \(30 = 2 \times 3 \times 5\), or lists of multiples of at least two of the numbers
correct answer with no working scores full marks
1 mark
P1 for 360 (seconds)
1 mark
A1 for 7:06 pm
oe, e.g. 19:06, 6 minutes past 7
Question 4
4 marks
\(A\) is a positive integer. The highest common factor (HCF) of \(A\) and 540 is 36. The lowest common multiple (LCM) of \(A\) and 540 is 7560.
Find the value of \(A\).
Model answer
\(540 = 2^2 \times 3^3 \times 5\) \(36 = 2^2 \times 3^2\) \(7560 = 2^3 \times 3^3 \times 5 \times 7\) Prime 2: HCF has \(2^2\), LCM has \(2^3\); 540 has \(2^2\), so \(A\) has \(2^3\). Prime 3: HCF has \(3^2\), LCM has \(3^3\); 540 has \(3^3\), so \(A\) has \(3^2\). Prime 5: HCF has no 5, so \(A\) has no 5. Prime 7: LCM has 7 but 540 does not, so \(A\) has \(7^1\). \(A = 2^3 \times 3^2 \times 7 = 504\)
Explanation
For each prime, the HCF uses the lower of the two powers and the LCM uses the higher. Compare 540's power with the HCF and LCM powers to deduce \(A\)'s power for each prime. Check: HCF(504, 540) = 36 and LCM(504, 540) = 7560.
Mark scheme
1 mark
P1 for prime factorising at least two of 540, 36, 7560: \(540 = 2^2 \times 3^3 \times 5\), \(36 = 2^2 \times 3^2\), \(7560 = 2^3 \times 3^3 \times 5 \times 7\)
correct answer with no working scores full marks
1 mark
P1 for deducing the power of 2 in \(A\) is 3, or the power of 3 in \(A\) is 2
1 mark
P1 for deducing that \(A\) has no factor 5 and has a factor 7, e.g. \(A = 2^3 \times 3^2 \times 7\)
1 mark
A1 for 504
cao; accept \(2^3 \times 3^2 \times 7\)
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