Worksheet - Prime Factors, HCF, and LCM of 3-digit numbers

Table of Contents
1. Instructions to the Learner
2. Section A: Multiple Choice
3. Section B: Short Answer and Structured Questions
4. Section C: Problem Solving and Word Problems
5. Bonus / Challenge Question
6. Answer Key
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Instructions to the Learner

1. Read all questions carefully before answering.
2. Show all working clearly. Marks are awarded for correct methods and processes.
3. Write your final answers clearly and circle or underline them where appropriate.
4. Calculators may be used unless otherwise stated.
5. All questions are based on prime factors, highest common factor (HCF), and lowest common multiple (LCM) of 3-digit numbers.
6. Answer all questions in the spaces provided or in your answer booklet.

Section A: Multiple Choice

Q1. What is the prime factorization of 180? [2 marks]

1. 2 × 3 × 5 × 6
2. 2² × 3² × 5
3. 2 × 3³ × 5
4. 2³ × 3 × 5

Q2. The HCF of 126 and 234 is: [2 marks]

1. 6
2. 9
3. 18
4. 36

Q3. The LCM of 120 and 150 is: [2 marks]

1. 300
2. 450
3. 600
4. 900

Q4. Which of the following is a prime factor of 315? [2 marks]

1. 6
2. 9
3. 7
4. 15

Q5. Two numbers have an HCF of 24 and an LCM of 360. If one of the numbers is 120, what is the other number? [2 marks]

1. 48
2. 72
3. 96
4. 144

Section B: Short Answer and Structured Questions

Q1. Consider the number 252.

1. Write 252 as a product of its prime factors using index notation. [3 marks]
2. How many distinct prime factors does 252 have? [1 mark]

Q2. Find the HCF and LCM of 144 and 180.

1. Express both 144 and 180 as products of their prime factors using index notation. [4 marks]
2. Use the prime factorizations to determine the HCF of 144 and 180. [2 marks]
3. Use the prime factorizations to determine the LCM of 144 and 180. [2 marks]

Q3. The prime factorization of a number is \(2^3 \times 3^2 \times 5\).

1. What is the number? [2 marks]
2. List all the prime factors of this number. [2 marks]

Q4. Find the HCF and LCM of 210, 315, and 420.

1. Write the prime factorization of each number using index notation. [6 marks]
2. Calculate the HCF of the three numbers. [2 marks]
3. Calculate the LCM of the three numbers. [2 marks]

Section C: Problem Solving and Word Problems

Q1. Thabo is organizing a community event in Soweto. He has 168 bottles of water and 252 packets of chips to distribute equally into identical gift packs with no items left over.

1. Find the prime factorization of 168 and 252. [4 marks]
2. What is the maximum number of identical gift packs Thabo can make? [2 marks]
3. How many bottles of water and packets of chips will be in each gift pack? [2 marks]

Q2. Nomsa and Sipho are training for a marathon. Nomsa runs a lap around the track every 135 seconds, while Sipho completes a lap every 180 seconds. They both start running from the same point at the same time.

1. Express 135 and 180 as products of their prime factors using index notation. [4 marks]
2. After how many seconds will they both be at the starting point together for the first time? [3 marks]
3. How many laps will each person have completed when they meet at the starting point for the first time? [2 marks]

Bonus / Challenge Question

This question is optional and for fast finishers.

Q1. Three bells ring at intervals of 108 minutes, 144 minutes, and 162 minutes respectively. If all three bells ring together at 6:00 AM:

1. Find the LCM of 108, 144, and 162 by first expressing each number as a product of its prime factors. [6 marks]
2. At what time will all three bells ring together again? [2 marks]

Answer Key

Section A: Multiple Choice

Q1. B - 2² × 3² × 5

Explanation: 180 = 2 × 90 = 2 × 2 × 45 = 2 × 2 × 9 × 5 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5. Option A is incorrect because 6 is not a prime number. Option C is incorrect because 180 ≠ 2 × 27 × 5 = 270. Option D is incorrect because 2³ × 3 × 5 = 8 × 3 × 5 = 120 ≠ 180.

Q2. C - 18

Explanation: 126 = 2 × 3² × 7 and 234 = 2 × 3² × 13. The common factors are 2 and 3², so HCF = 2 × 3² = 2 × 9 = 18. Option A (6) includes only 2 × 3. Option B (9) includes only 3². Option D (36) is too large and not a factor of 126.

Q3. C - 600

Explanation: 120 = 2³ × 3 × 5 and 150 = 2 × 3 × 5². The LCM takes the highest power of each prime: 2³ × 3 × 5² = 8 × 3 × 25 = 600. Option A (300) is missing a factor of 2. Option B (450) = 2 × 3² × 5². Option D (900) is too large.

Q4. C - 7

Explanation: 315 = 3² × 5 × 7. The prime factors are 3, 5, and 7. Option A (6) is not prime. Option B (9) is not prime. Option D (15) is not prime (15 = 3 × 5).

Q5. B - 72

Explanation: Using the relationship: HCF × LCM = Product of two numbers. Therefore, 24 × 360 = 120 × other number. Other number = (24 × 360) ÷ 120 = 8640 ÷ 120 = 72. Option A (48) gives an LCM of 240. Option C (96) gives an LCM of 480. Option D (144) gives an LCM of 720.

Section B: Short Answer and Structured Questions

Q1.

(a) 252 = 2 × 126
= 2 × 2 × 63
= 2 × 2 × 9 × 7
= 2 × 2 × 3 × 3 × 7
= 2² × 3² × 7
[1 mark for method, 1 mark for correct factorization, 1 mark for index notation]

(b) The distinct prime factors are 2, 3, and 7.
Number of distinct prime factors = 3
[1 mark]

Q2.

(a) 144 = 2 × 72
= 2 × 2 × 36
= 2 × 2 × 2 × 18
= 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
= 2⁴ × 3²
[2 marks]

180 = 2 × 90
= 2 × 2 × 45
= 2 × 2 × 9 × 5
= 2 × 2 × 3 × 3 × 5
= 2² × 3² × 5
[2 marks]

(b) 144 = 2⁴ × 3²
180 = 2² × 3² × 5
HCF = product of lowest powers of common prime factors
HCF = 2² × 3² = 4 × 9 = 36
[1 mark for method, 1 mark for correct answer]

(c) LCM = product of highest powers of all prime factors
LCM = 2⁴ × 3² × 5
= 16 × 9 × 5
= 720
[1 mark for method, 1 mark for correct answer]

Q3.

(a) The number = 2³ × 3² × 5
= 8 × 9 × 5
= 72 × 5
= 360
[1 mark for method, 1 mark for correct answer]

(b) The prime factors are 2, 3, and 5
[2 marks for all three correct]

Q4.

(a) 210 = 2 × 105
= 2 × 3 × 35
= 2 × 3 × 5 × 7
= 2 × 3 × 5 × 7
[2 marks]

315 = 3 × 105
= 3 × 3 × 35
= 3 × 3 × 5 × 7
= 3² × 5 × 7
[2 marks]

420 = 2 × 210
= 2 × 2 × 105
= 2 × 2 × 3 × 35
= 2 × 2 × 3 × 5 × 7
= 2² × 3 × 5 × 7
[2 marks]

(b) 210 = 2 × 3 × 5 × 7
315 = 3² × 5 × 7
420 = 2² × 3 × 5 × 7
HCF = product of lowest powers of common prime factors
Common factors: 3, 5, 7
HCF = 3 × 5 × 7 = 105
[1 mark for method, 1 mark for correct answer]

(c) LCM = product of highest powers of all prime factors
LCM = 2² × 3² × 5 × 7
= 4 × 9 × 5 × 7
= 36 × 35
= 1260
[1 mark for method, 1 mark for correct answer]

Section C: Problem Solving and Word Problems

Q1.

(a) 168 = 2 × 84
= 2 × 2 × 42
= 2 × 2 × 2 × 21
= 2 × 2 × 2 × 3 × 7
= 2³ × 3 × 7
[2 marks]

252 = 2 × 126
= 2 × 2 × 63
= 2 × 2 × 9 × 7
= 2 × 2 × 3 × 3 × 7
= 2² × 3² × 7
[2 marks]

(b) The maximum number of gift packs = HCF of 168 and 252
168 = 2³ × 3 × 7
252 = 2² × 3² × 7
HCF = 2² × 3 × 7 = 4 × 3 × 7 = 84 gift packs
[1 mark for method, 1 mark for correct answer]

(c) Bottles per pack = 168 ÷ 84 = 2 bottles
Packets per pack = 252 ÷ 84 = 3 packets
Thabo will put 2 bottles of water and 3 packets of chips in each gift pack.
[1 mark for each correct answer]

Q2.

(a) 135 = 5 × 27
= 5 × 3 × 9
= 5 × 3 × 3 × 3
= 3³ × 5
[2 marks]

180 = 2 × 90
= 2 × 2 × 45
= 2 × 2 × 9 × 5
= 2 × 2 × 3 × 3 × 5
= 2² × 3² × 5
[2 marks]

(b) They will meet at the starting point after a time equal to the LCM of 135 and 180 seconds.
135 = 3³ × 5
180 = 2² × 3² × 5
LCM = 2² × 3³ × 5
= 4 × 27 × 5
= 540 seconds
Nomsa and Sipho will meet at the starting point after 540 seconds.
[1 mark for identifying LCM method, 1 mark for calculation, 1 mark for correct answer]

(c) Nomsa's laps = 540 ÷ 135 = 4 laps
Sipho's laps = 540 ÷ 180 = 3 laps
When they meet, Nomsa will have completed 4 laps and Sipho will have completed 3 laps.
[1 mark for each correct answer]

Bonus / Challenge Question

Q1.

(a) 108 = 2 × 54
= 2 × 2 × 27
= 2 × 2 × 3 × 9
= 2 × 2 × 3 × 3 × 3
= 2² × 3³
[2 marks]

144 = 2 × 72
= 2 × 2 × 36
= 2 × 2 × 2 × 18
= 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
= 2⁴ × 3²
[2 marks]

162 = 2 × 81
= 2 × 3 × 27
= 2 × 3 × 3 × 9
= 2 × 3 × 3 × 3 × 3
= 2 × 3⁴
[2 marks]

LCM = highest power of each prime factor
LCM = 2⁴ × 3⁴
= 16 × 81
= 1296 minutes

(b) 1296 minutes = 1296 ÷ 60 hours
= 21.6 hours
= 21 hours 36 minutes
Starting time: 6:00 AM
6:00 AM + 21 hours 36 minutes = 3:36 AM (next day)
The three bells will ring together again at 3:36 AM the next day.
[1 mark for conversion, 1 mark for correct time]
This question tests multi-number LCM and real-world time application, requiring higher-order problem-solving skills.

Total Marks Summary