1.6 Prime factorisation, HCF and LCM
9 exam questions worth 23 marks. Write your answer, open the mark scheme, then give yourself the marks you earned. Your answers are saved on this device.
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Which of these is the prime factorisation of 90?
Choose one answer.
Correct Not quite: the answer is 2 × 3² × 5
Mark scheme
B: 2 × 3² × 5
Model answer
2 × 3² × 5 (B). 90 = 2 × 45 = 2 × 3 × 15 = 2 × 3 × 3 × 5 = 2 × 3² × 5.
Write 84 as a product of its prime factors.
Mark scheme
M1: a factor tree or division method started; M1: continuing until all factors are prime; A1: 2 × 2 × 3 × 7 or 2² × 3 × 7
Model answer
84 = 2 × 42 = 2 × 2 × 21 = 2 × 2 × 3 × 7 = 2² × 3 × 7.
Find the highest common factor (HCF) of 48 and 72.
Mark scheme
M1: 48 = 2⁴ × 3 or 72 = 2³ × 3²; M1: common primes to the lowest powers (2³ × 3); A1: 24
Model answer
48 = 2⁴ × 3 and 72 = 2³ × 3². The HCF is 2³ × 3 = 24.
Find the lowest common multiple (LCM) of 15 and 20.
Mark scheme
M1: 15 = 3 × 5, 20 = 2² × 5 or lists of multiples; A1: 60
Model answer
15 = 3 × 5 and 20 = 2² × 5. The LCM is 2² × 3 × 5 = 60.
Write 360 as a product of its prime factors in index form.
Mark scheme
M1: a correct method; M1: 2, 2, 2, 3, 3, 5 identified; A1: 2³ × 3² × 5
Model answer
360 = 2 × 180 = 2 × 2 × 90 = 2 × 2 × 2 × 45 = 2³ × 3² × 5.
A = 2² × 3 × 5² and B = 2 × 3³ × 5. Find (a) the HCF of A and B (b) the LCM of A and B. Give each answer as a number.
Mark scheme
B1: HCF = 2 × 3 × 5; M1: LCM uses highest powers 2² × 3³ × 5²; A1: HCF = 30 and LCM = 2700
Model answer
(a) HCF = 2 × 3 × 5 = 30. (b) LCM = 2² × 3³ × 5² = 4 × 27 × 25 = 2700.
Two lighthouses flash every 12 seconds and every 20 seconds. They flash together at midnight. After how many seconds do they next flash together?
Mark scheme
M1: attempt at LCM of 12 and 20; A1: 60
Model answer
12 = 2² × 3 and 20 = 2² × 5, so the LCM is 2² × 3 × 5 = 60 seconds.
72 = 2³ × 3². Find the smallest whole number that 72 should be multiplied by to give a square number.
Mark scheme
M1: indices must be even / 2⁴ × 3²; A1: 2 (giving 144 = 12²)
Model answer
72 × 2 = 144 = 2⁴ × 3² = 12². All the powers are now even, so the smallest number is 2.
A floor is 84 cm long and 126 cm wide. It is covered with identical square tiles with no gaps and no cutting. Work out the largest possible side length of a tile and the number of tiles needed.
Mark scheme
M1: prime factorisations of 84 and 126; M1: HCF = 2 × 3 × 7; A1: 42 cm; A1: 6 tiles (2 × 3)
Model answer
84 = 2² × 3 × 7 and 126 = 2 × 3² × 7. The HCF is 2 × 3 × 7 = 42 cm. The floor is 84 ÷ 42 = 2 tiles by 126 ÷ 42 = 3 tiles, so 2 × 3 = 6 tiles.
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