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GCSE · Number

4.3 Prime Factors, HCF and LCM

EDEXCEL 1MA1 · Calculator allowed · about 70 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Here is a list of numbers.
8 11 15 19 21 29
From the list, write down all the prime numbers.

(Total for Question 1 is 2 marks)

Write each number as a product of its prime factors. Give each answer in index form.

(a)12(2)

(b)18(2)

(Total for Question 2 is 4 marks)

A factor tree for 60 has been started.
60 splits into 6 and 10. Complete the factor tree and use it to write 60 as a product of prime factors, giving your answer in index form.

(Total for Question 3 is 3 marks)

Which of these numbers is a prime number?

A) 51
B) 57
C) 61
D) 63

(Total for Question 4 is 1 mark)

Express 84 as a product of its prime factors. Give your answer in index form.

(Total for Question 5 is 3 marks)

Find the highest common factor (HCF) of 24 and 36. You must show your method.

(Total for Question 6 is 3 marks)

Find the lowest common multiple (LCM) of 8 and 12.

(Total for Question 7 is 2 marks)

60 = 2² x 3 x 5 and 84 = 2² x 3 x 7.

(a)Draw a Venn diagram to show the prime factors of 60 and 84.(2)

(b)Use your Venn diagram to find the HCF of 60 and 84.(1)

(c)Use your Venn diagram to find the LCM of 60 and 84.(2)

(Total for Question 8 is 5 marks)

Show that 91 is not a prime number.

(Total for Question 9 is 2 marks)

Two buses leave a bus station at the same time, 08:00. Bus A leaves every 15 minutes. Bus B leaves every 20 minutes. Work out the next time both buses will leave the bus station together.

(Total for Question 10 is 3 marks)

A florist has 42 roses and 56 lilies. She wants to make identical bouquets, each with the same number of roses and the same number of lilies, using all the flowers with none left over. Work out the greatest number of bouquets she can make.

(Total for Question 11 is 3 marks)

180 = 2² x 3² x 5. Find the smallest positive integer k such that 180 x k is a square number.

(Total for Question 12 is 3 marks)

2³ x 3² and 2² x 3 x 5 are the prime factorisations of two numbers. What is their HCF?

A) 2² x 3
B) 2³ x 3² x 5
C) 2² x 3 x 5
D) 2 x 3

(Total for Question 13 is 1 mark)

a = 2² x 3 x 5 and b = 2 x 3^3.

(a)Find the HCF of a and b.(2)

(b)Find the LCM of a and b.(2)

(Total for Question 14 is 4 marks)

Elena has three lengths of ribbon: 90 cm, 126 cm and 168 cm. She wants to cut all three ribbons into equal-length pieces, as long as possible, with no ribbon left over. Work out the length of each piece.

(Total for Question 15 is 3 marks)

Three lighthouses flash at regular intervals. Lighthouse P flashes every 12 seconds, Lighthouse Q every 18 seconds and Lighthouse R every 24 seconds. All three lighthouses flash together at exactly 21:00:00. Work out the next time all three will flash together.

(Total for Question 16 is 4 marks)

The HCF of two positive integers is 6. The LCM of the two integers is 90. One of the integers is 18. Work out the other integer.

(Total for Question 17 is 3 marks)

A = 2³ x 3² x 5 and B = 2 x 3^a x 5², where a is a positive integer. The HCF of A and B is 30.

(a)Find the value of a.(2)

(b)Find the LCM of A and B.(2)

(Total for Question 18 is 4 marks)

N = 2² x 3 x 5^3. Find the smallest positive integer m such that N x m is a cube number.

(Total for Question 19 is 3 marks)

p = 2^a x 3² and q = 2³ x 3^b, where a and b are positive integers. The HCF of p and q is 36 and the LCM of p and q is 648.

(a)Find the value of a.(2)

(b)Find the value of b.(2)

(Total for Question 20 is 4 marks)

Mark scheme · 4.3 Prime Factors, HCF and LCM

Question 1

B1 any two of 11, 19, 29 correctly identified
B1 all three of 11, 19, 29 identified with no additional incorrect numbers
Answer: 11, 19, 29

Question 2

(a) M1 correct method shown, e.g. factor tree or repeated division, 12 = 2 x 6 = 2 x 2 x 3
(a) A1 2² x 3 oe cao
(a) Answer: 2² x 3
(b) M1 correct method shown, e.g. factor tree or repeated division, 18 = 2 x 9 = 2 x 3 x 3
(b) A1 2 x 3² oe cao
(b) Answer: 2 x 3²

Question 3

M1 6 split as 2 x 3 and 10 split as 2 x 5 (or an equivalent complete split of 60 down to primes)
M1 all four prime factors 2, 2, 3, 5 identified from the completed tree
A1 2² x 3 x 5 oe cao
Answer: 2² x 3 x 5

Question 4

B1 C cao
Answer: C) 61

Question 5

M1 attempt to divide 84 by a prime number, e.g. 84 = 2 x 42
M1 84 = 2 x 2 x 21 = 2 x 2 x 3 x 7, correct sequence of prime divisions shown
A1 2² x 3 x 7 oe cao
Answer: 2² x 3 x 7

Question 6

M1 factors of 24 listed: 1, 2, 3, 4, 6, 8, 12, 24 (or use of prime factorisation)
M1 factors of 36 listed: 1, 2, 3, 4, 6, 9, 12, 18, 36 and common factors identified
A1 12 cao
Answer: 12

Question 7

M1 multiples of 8 and/or 12 listed, e.g. 8, 16, 24, 32, ... and 12, 24, 36, ...
A1 24 cao
Answer: 24

Question 8

(a) B1 2, 2, 3 placed correctly in the intersection of the two circles
(a) B1 5 placed only in the 60 circle and 7 placed only in the 84 circle
(a) Answer: Intersection: 2, 2, 3. In 60 only: 5. In 84 only: 7.
(b) B1 12 cao, ft from their Venn diagram
(b) Answer: 12
(c) M1 all prime factors from the Venn diagram identified (2, 2, 3, 5, 7)
(c) A1 420 cao, ft from their Venn diagram
(c) Answer: 420

Question 9

M1 attempt to divide 91 by a prime number other than 2, 3 or 5, e.g. divides by 7
A1 cso, 91 = 7 x 13 stated, so 91 has factors other than 1 and itself, therefore 91 is not prime
Answer: 91 = 7 x 13, so 91 is not prime

Question 10

M1 attempt to find the LCM of 15 and 20, e.g. multiples listed or prime factorisation used
A1 60 (minutes) found
A1 ft, 09:00 stated as the final answer
Answer: 09:00

Question 11

M1 42 = 2 x 3 x 7 and 56 = 2³ x 7 (or equivalent factorisation or listing of factors)
M1 HCF identified as 2 x 7, dependent on first method mark
A1 14 cao
Answer: 14 bouquets

Question 12

M1 recognises that for a square number, every prime factor must have an even power
M1 identifies that the power of 5 (currently 1, an odd power) must be increased to 2
A1 5 cao
Answer: 5

Question 13

B1 A cao
Answer: A) 2² x 3 (= 12)

Question 14

(a) M1 lowest powers of common prime factors identified: 2¹ and 3¹
(a) A1 6 cao
(a) Answer: 6
(b) M1 highest powers of all prime factors identified: 2², 3³, 5¹
(b) A1 540 cao
(b) Answer: 540

Question 15

M1 prime factorisation (or listing of factors) of at least two of 90, 126, 168 attempted
M1 90 = 2 x 3² x 5, 126 = 2 x 3² x 7, 168 = 2³ x 3 x 7 (or equivalent), with HCF identified as 2 x 3
A1 6 (cm) cao
Answer: 6 cm

Question 16

M1 12 = 2² x 3, 18 = 2 x 3², 24 = 2³ x 3 (or equivalent method)
M1 LCM identified using highest powers: 2³ x 3²
A1 72 (seconds) found
A1 ft, 21:01:12 stated as the final answer
Answer: 21:01:12

Question 17

M1 uses HCF x LCM = product of the two numbers (oe uses prime factorisation of 18 with the required HCF and LCM)
M1 6 x 90 (= 540) calculated, dependent on first method mark
A1 30 cao (540 / 18)
Answer: 30

Question 18

(a) M1 30 = 2 x 3 x 5 identified and compared to the exponents in A and B, e.g. min(2, a) = 1
(a) A1 a = 1 cao
(a) Answer: a = 1
(b) M1 highest powers identified using a = 1: 2³, 3², 5²
(b) A1 1800 cao
(b) Answer: 1800

Question 19

M1 recognises that for a cube number, every prime factor must have a power that is a multiple of 3
M1 identifies 2 needs one more factor of 2 (power 2 to 3) and 3 needs two more factors of 3 (power 1 to 3); 5³ already a multiple of 3
A1 18 cao (2 x 3²)
Answer: 18

Question 20

(a) M1 36 = 2² x 3² identified; compares exponent of 2, min(a, 3) = 2
(a) A1 a = 2 cao
(a) Answer: a = 2
(b) M1 648 = 2³ x 3⁴ identified; compares exponent of 3, max(2, b) = 4
(b) A1 b = 4 cao
(b) Answer: b = 4

Prime Factors, HCF and LCM - Worksheets, Questions and Revision