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4.3H Prime Factors, HCF and LCM: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Express 2310 as a product of prime factors.

(Total for Question 1 is 2 marks)

Work out the highest common factor (HCF) of 378 and 630.

(Total for Question 2 is 2 marks)

Find the least common multiple (LCM) of 40 and 96 using prime factors.

(Total for Question 3 is 3 marks)

The number A = 2⁴ x 3² x 5 and the number B = 2² x 3³ x 7.
(a) Find HCF(A, B).
(b) Find LCM(A, B).

(a)Find HCF(A, B).(2)

(b)Find LCM(A, B).(2)

(Total for Question 4 is 4 marks)

Show that for any two positive integers a and b with prime factorisations a = product p_i^{α_i} and b = product p_i^{β_i} (over the same primes p_i), the product a x b equals HCF(a, b) x LCM(a, b). You must give a reason referring to prime exponents. This is a 'show that' question.

(Total for Question 5 is 4 marks)

Two buses leave the same stop at 08:00. One bus completes a lap every 18 minutes, the other every 30 minutes. At what time after 08:00 will both buses next arrive at the stop together? Give your answer in hours and minutes.

(Total for Question 6 is 3 marks)

Find the smallest positive integer n such that 360 x n is a perfect cube.

(Total for Question 7 is 3 marks)

Work out the largest three-digit number that is divisible by both 18 and 28.

(Total for Question 8 is 2 marks)

Let x = 2⁴ x 3¹ x 5².
(a) Find the number of positive divisors of x.
(b) How many of these divisors are multiples of 5?

(a)Find the number of positive divisors of x.(2)

(b)How many of these divisors are multiples of 5?(2)

(Total for Question 9 is 4 marks)

Find all positive integers n such that HCF(n, 18) = 6 and LCM(n, 18) = 90. Show your working.

(a)Find all positive integers n satisfying the conditions.(2)

(Total for Question 10 is 2 marks)

Mark scheme · 4.3H Prime Factors, HCF and LCM: Higher Tier Practice

Question 1

M1 correct division tree or primes shown, e.g. 2310 = 2 x 3 x 5 x 7 x 11 (method)
A1 2310 = 2 x 3 x 5 x 7 x 11 cao
Answer: 2 x 3 x 5 x 7 x 11 (2310 = 2 x 3 x 5 x 7 x 11)

Question 2

M1 correct prime factorisation or identification of common prime powers, e.g. 378 = 2 x 3³ x 7, 630 = 2 x 3² x 5 x 7
A1 HCF = 2 x 3² x 7 = 126 cao
Answer: 126

Question 3

M1 correct prime factorisations, e.g. 40 = 2³ x 5, 96 = 2⁵ x 3
M1 take highest powers: 2⁵ x 3 x 5
A1 LCM = 32 x 3 x 5 = 480 cao
Answer: 480

Question 4

(a) M1 choose minimum powers: 2² x 3² oe
(a) A1 HCF = 4 x 9 = 36 cao
(a) Answer: 36
(b) M1 take highest powers: 2⁴ x 3³ x 5 x 7 oe
(b) A1 LCM = 16 x 27 x 5 x 7 = 15120 cao
(b) Answer: 15120

Question 5

M1 express HCF exponents as min(α_i, β_i) and LCM exponents as max(α_i, β_i)
M1 note that for each i, α_i + β_i = min(α_i,β_i) + max(α_i,β_i)
M1 multiply over all primes to get a x b = product p_i^{α_i+β_i} = product p_i^{min+max} = HCF x LCM (method)
A1 clear conclusion: therefore a x b = HCF(a,b) x LCM(a,b) cso
Answer: For each prime p_i, exponent in HCF = min(α_i,β_i), exponent in LCM = max(α_i,β_i). α_i + β_i = min + max, so multiplying p_i^{α_i} p_i^{β_i} = p_i^{min} p_i^{max}. Product over all primes gives a x b = HCF x LCM. Hence shown.

Question 6

M1 find LCM of 18 and 30, e.g. 18 = 2 x 3², 30 = 2 x 3 x 5 so LCM = 2 x 3² x 5 = 90
M1 convert 90 minutes to hours and minutes
A1 90 minutes = 1 hour 30 minutes, so both buses meet at 09:30 cao
Answer: 09:30 (1 hour 30 minutes after 08:00)

Question 7

M1 prime factorise 360 = 2³ x 3² x 5¹
M1 choose multipliers to make exponents multiples of 3: need 3¹ and 5²
A1 n = 3 x 25 = 75 cao
Answer: 75

Question 8

M1 find LCM(18,28) = 2² x 3² x 7 = 252
A1 largest three-digit multiple is 252 x 3 = 756 cao
Answer: 756

Question 9

(a) M1 use formula (exponents + 1) product: (4+1)(1+1)(2+1) method
(a) A1 number of divisors = 5 x 2 x 3 = 30 cao
(a) Answer: 30
(b) M1 count divisors that include at least one factor 5: treat exponent of 5 as 1 or 2 -> (4+1)(1+1) x (choices for 5 exponent 2) or more direct: (4+1)(1+1) where exponent of 5≥1 gives 5 x 2 x 2? (method shown)
(b) A1 divisors multiple of 5 = 5 x 2 x 2 = 20 cao
(b) Answer: 20

Question 10

(a) M1 use relation n x 18 = HCF x LCM = 6 x 90 = 540 so n = 540 / 18
(a) A1 n = 30 cao
(a) Answer: 30

Prime Factors, HCF and LCM: Higher Tier Practice - Worksheets, Questions and Revision