🌳 HCF, LCM & Prime Factorisation

Cambridge Lower Secondary Stage 7 · Number & Calculation

Prime Factorisation
60 = 2² × 3 × 5
HCF via Venn Diagram
HCF(12, 18) = 2 × 3 = 6
LCM via Venn Diagram
LCM(4, 6) = 2 × 2 × 3 = 12

What you'll learn:

  • Write any number as a product of prime factors using a factor tree
  • Use index notation: e.g. 2³ × 3² × 5
  • Find the HCF using a Venn diagram (intersection)
  • Find the LCM using a Venn diagram (union)
  • Use HCF to simplify fractions and solve tiling problems
  • Use LCM to solve timetable / scheduling problems

Watch 60 grow into its prime factors:


Created by Mr Josh

📖 Learn: HCF, LCM & Prime Factorisation

Part 1: Prime Numbers & the Sieve of Eratosthenes

A prime number has exactly two factors: 1 and itself. Examples: 2, 3, 5, 7, 11, 13 …

Composite numbers have more than two factors. 1 is neither prime nor composite.

Interactive Sieve of Eratosthenes (1–50)

Click a prime to cross out all its multiples. Remaining unhighlighted numbers are prime!

Part 2: Factor Trees & Prime Factorisation

A factor tree breaks a number into two factors, then keeps splitting composite numbers until every branch ends in a prime.

Rule: Keep splitting until ALL leaves are prime (circle them or colour them gold).
Example: 60 → 6 × 10 → (2 × 3) × (2 × 5) → 60 = 2² × 3 × 5
Example: 84 → 4 × 21 → (2 × 2) × (3 × 7) → 84 = 2² × 3 × 7

Index notation: Write repeated prime factors using powers: 2 × 2 × 2 = 2³.

Number Prime Factorisation Index Form
122 × 2 × 32² × 3
302 × 3 × 52 × 3 × 5
722 × 2 × 2 × 3 × 32³ × 3²
1002 × 2 × 5 × 52² × 5²

Part 3: HCF using a Venn Diagram

The Highest Common Factor (HCF) is the largest number that divides into both numbers exactly.

Method: Find prime factorisations → draw a Venn diagram → shared primes go in the middle.
HCF = product of the primes in the overlapping region.
Example: HCF(36, 60)
36 = 2² × 3²  |  60 = 2² × 3 × 5
Shared: 2² × 3 → HCF = 4 × 3 = 12
💡 HCF is useful for simplifying fractions: divide top and bottom by the HCF.

Part 4: LCM using a Venn Diagram

The Lowest Common Multiple (LCM) is the smallest number that is a multiple of both numbers.

Method: Venn diagram as above → LCM = product of ALL primes in the diagram (each prime counted once).
Example: LCM(36, 60)
Venn left-only: 3 (one extra 3 from 36)  |  Middle: 2², 3  |  Right-only: 5
LCM = 3 × 2² × 3 × 5 = 3 × 4 × 3 × 5 = 180
💡 LCM is useful for timetable problems: when do two buses next meet?
💡 Also used to find a common denominator when adding fractions.

Part 5: Applications

Simplifying fractions: Divide numerator and denominator by their HCF.

Simplify 36/60: HCF(36,60) = 12. → 36÷12 = 3, 60÷12 = 5. Answer: 3/5

Timetable / scheduling: Two events repeat at intervals a and b. They next coincide after LCM(a, b) units of time.

Bus A every 8 min, Bus B every 12 min. LCM(8,12)=24. They next meet after 24 minutes.

Tiling problems: Largest square tile that fits a rectangle with no cutting → HCF(length, width).

Floor 48 cm × 36 cm. Largest square tile = HCF(48,36) = 12 cm
Created by Mr Josh

💡 Worked Examples

Example 1: Factor Tree for 72

Write 72 as a product of its prime factors in index notation.

Step 1: Split 72 into any two factors: 72 = 8 × 9
Step 2: Split further: 8 = 2 × 4 = 2 × 2 × 2    9 = 3 × 3
Step 3: All leaves are prime: 2, 2, 2, 3, 3
Answer: 72 = 2³ × 3²
💡 Different starting splits always give the same answer — prime factorisation is unique!

Example 2: HCF using a Venn Diagram — HCF(24, 36)

Find the HCF of 24 and 36.

Step 1: 24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3 = 2³ × 3
Step 2: 36 = 4 × 9 = 2 × 2 × 3 × 3 = 2² × 3²
Step 3: Venn diagram overlap — take the LOWER power of each shared prime:
  Both have 2: take 2² (the lower of 2³ and 2²)
  Both have 3: take 3¹ (the lower of 3¹ and 3²)
HCF = 2² × 3 = 4 × 3 = 12

Example 3: LCM using a Venn Diagram — LCM(24, 36)

Find the LCM of 24 and 36.

Step 1: As above: 24 = 2³ × 3    36 = 2² × 3²
Step 2: Venn diagram — take the HIGHER power of each prime:
  2: take 2³    3: take 3²
LCM = 2³ × 3² = 8 × 9 = 72
Quick check: 72 ÷ 24 = 3 ✅   72 ÷ 36 = 2 ✅

Example 4: Application — Traffic Light Problem

Traffic light A changes every 45 seconds. Traffic light B changes every 30 seconds. They both just changed together. After how many seconds will they next change together?

Step 1: 45 = 3² × 5    30 = 2 × 3 × 5
Step 2: LCM: take highest power of each prime: 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90
Answer: They next change together after 90 seconds (1 minute 30 seconds).
Created by Mr Josh

🌳 Factor Tree Builder

Enter any whole number from 2 to 200. Click a composite node to split it — keep splitting until all leaves glow gold (prime)!



Try these numbers:

Created by Mr Josh

✏️ Exercise 1: Prime Factorisation

Write each number as a product of prime factors in index notation.
Use ^ for powers, × or * for multiply. E.g. 2^3×3 or 2^2*5

Created by Mr Josh

✏️ Exercise 2: Highest Common Factor (HCF)

Find the HCF of each pair of numbers. Type just the number.

Created by Mr Josh

✏️ Exercise 3: Lowest Common Multiple (LCM)

Find the LCM of each pair of numbers. Type just the number.

Created by Mr Josh

🔵 Exercise 4: Venn Diagram — Sort Prime Factors

For each question, drag the prime factor bubbles into the correct Venn region.
Left only = factors of A only  ·  Middle = factors of both  ·  Right only = factors of B only

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🚌 Exercise 5: Real-World Problems

Use HCF or LCM to solve each problem. Type your numeric answer.

🚌 Bus Timetable Visualiser

Bus A leaves every 8 minutes. Bus B leaves every 12 minutes.
They both just left together at time 0. Watch them on the timeline!

Created by Mr Josh

📝 Practice Questions

  1. Is 37 a prime number? Explain.
  2. Write 18 as a product of prime factors in index notation.
  3. Write 45 as a product of prime factors.
  4. Write 56 as a product of prime factors in index notation.
  5. Write 90 as a product of prime factors in index notation.
  6. Write 144 as a product of prime factors in index notation.
  7. Find HCF(12, 20).
  8. Find HCF(18, 24).
  9. Find HCF(30, 42).
  10. Find HCF(48, 72).
  11. Find LCM(4, 6).
  12. Find LCM(9, 15).
  13. Find LCM(8, 12).
  14. Find LCM(14, 21).
  15. Simplify the fraction 24/36 using HCF.
  16. Simplify the fraction 45/60 using HCF.
  17. Two alarm clocks ring together. Clock A rings every 20 minutes, Clock B every 30 minutes. After how many minutes will they next ring together?
  18. A rectangular floor is 48 cm by 36 cm. What is the side length of the largest square tile that fits exactly?
  19. Find the HCF and LCM of 60 and 84.
  20. Explain why HCF(a, b) × LCM(a, b) = a × b for any two positive integers a and b. Verify with a = 12, b = 18.
  1. Yes. 37 has no factors other than 1 and 37 (not divisible by 2, 3, 5, or 7).
  2. 18 = 2 × 3²
  3. 45 = 3² × 5
  4. 56 = 2³ × 7
  5. 90 = 2 × 3² × 5
  6. 144 = 2⁴ × 3²
  7. HCF(12, 20) = 4 (12 = 2²×3, 20 = 2²×5, shared: 2²)
  8. HCF(18, 24) = 6 (18 = 2×3², 24 = 2³×3, shared: 2×3)
  9. HCF(30, 42) = 6 (30 = 2×3×5, 42 = 2×3×7, shared: 2×3)
  10. HCF(48, 72) = 24 (48 = 2⁴×3, 72 = 2³×3², shared: 2³×3)
  11. LCM(4, 6) = 12
  12. LCM(9, 15) = 45 (9 = 3², 15 = 3×5, LCM = 3²×5)
  13. LCM(8, 12) = 24
  14. LCM(14, 21) = 42 (14 = 2×7, 21 = 3×7, LCM = 2×3×7)
  15. HCF(24,36) = 12. 24÷12 = 2, 36÷12 = 3. Answer: 2/3
  16. HCF(45,60) = 15. 45÷15 = 3, 60÷15 = 4. Answer: 3/4
  17. LCM(20, 30) = 60. They next ring together after 60 minutes.
  18. HCF(48, 36) = 12. Largest square tile is 12 cm × 12 cm.
  19. 60 = 2²×3×5  |  84 = 2²×3×7. HCF = 2²×3 = 12. LCM = 2²×3×5×7 = 420.
  20. 12 = 2²×3, 18 = 2×3². HCF = 6, LCM = 36. 6 × 36 = 216 = 12 × 18 ✅
Created by Mr Josh

🏆 Challenge: Multi-Step Problems

  1. Three bells ring at intervals of 12 minutes, 15 minutes, and 20 minutes. They all ring together at 9:00 am. At what time will they next all ring together?
  2. A rectangular garden measures 84 m by 126 m. A gardener wants to divide it into identical square plots using the largest possible square size. How many square plots will there be in total?
  3. Find two numbers whose HCF is 6 and whose LCM is 180. Find all possible pairs.
  4. Given that 360 = 2³ × 3² × 5, find: (a) the number of factors of 360, (b) HCF(360, 504), given 504 = 2³ × 3² × 7.
  5. A shop sells ribbons in lengths of 36 cm and 48 cm. A customer wants to cut both ribbons into equal pieces with no waste, using the longest possible piece length. How long should each piece be, and how many pieces of each ribbon does the customer get?
  6. Without using a calculator, show that HCF(a, b) × LCM(a, b) = a × b when a = 15 and b = 35. Verify your answer.
  7. The LCM of two numbers is 120 and one of the numbers is 24. List all possible values for the second number.
  8. A swimming pool is used by two clubs. Club A practises every 4 days and Club B every 6 days. They both practised on 1st January. On which dates in January will they both practise together?
  1. LCM(12, 15, 20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60 min. They next ring together at 10:00 am.
  2. HCF(84, 126): 84=2²×3×7, 126=2×3²×7. HCF = 2×3×7 = 42 m. Number of plots = (84×126)/(42²) = 10584/1764 = 6 plots. (84/42=2 columns, 126/42=3 rows → 6 total)
  3. HCF=6, LCM=180. LCM/HCF = 30 = a' × b' where gcd(a',b')=1. Pairs for (a',b'): (1,30), (2,15), (3,10), (5,6). Multiply each by 6: (6,180), (12,90), (18,60), (30,30).
  4. (a) Number of factors of 360 = (3+1)(2+1)(1+1) = 4×3×2 = 24 factors. (b) 504=2³×3²×7. HCF(360,504) = 2³×3² = 8×9 = 72.
  5. HCF(36, 48): 36=2²×3², 48=2⁴×3. HCF=2²×3=12 cm. 36÷12=3 pieces of 36 cm ribbon. 48÷12=4 pieces of 48 cm ribbon.
  6. 15=3×5, 35=5×7. HCF=5, LCM=3×5×7=105. 5×105=525 = 15×35 ✅
  7. 120 = 2³×3×5. Second number must be a divisor of 120 and LCM with 24 must equal 120. 24=2³×3. Candidates: must include factor 5 (else LCM ≤ 72): 5, 10, 15, 20, 40, 60, 120. Check each: LCM(24,5)=120✅ LCM(24,10)=120✅ LCM(24,15)=120✅ LCM(24,20)=120✅ LCM(24,40)=120✅ LCM(24,60)=120✅ LCM(24,120)=120✅
  8. LCM(4,6)=12. They meet on day 1, 13, 25. In January: 1st, 13th, and 25th January.
Created by Mr Josh