**Real-Life Problem:** Sameeksha is building her new house. The main room is 12 ft by 16 ft. She wants to cover the floor with square tiles of the same size. She wants to:
**Why We Need HCF Here:** For square tiles to fit perfectly:
**Finding Factors:**
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 16: 1, 2, 4, 8, 16
Common factors (numbers in both lists): 1, 2, 4
The largest common factor is 4 ft.
Therefore, Sameeksha should buy 4 ft × 4 ft tiles.
Number of tiles needed: (12 ÷ 4) × (16 ÷ 4) = 3 × 4 = 12 tiles
**HCF (Highest Common Factor)** = The highest (or greatest) of the common factors of two or more numbers. It is also called **GCD (Greatest Common Divisor)**.
**Problem:** Lekhana bought 84 kg of rice from one farm and 108 kg from another farm. She wants to pack them in bags where:
**Solution Method:**
Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
Factors of 108: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108
Common factors: 1, 2, 3, 4, 6, 12
To minimize bags, we choose the **largest** common factor = 12 kg per bag
Number of bags: 84/12 + 108/12 = 7 + 9 = 16 bags
The longest jump size that lands on given treasure numbers is the HCF. For example:
This works because a jump size that lands exactly on both numbers must divide both numbers, and the longest such jump is the HCF.
---
**Prime Number:** A number greater than 1 that has only 1 and itself as factors.
Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47...
Note: 1 is NOT prime. 2 is the only even prime number.
**Prime Factorisation:** Expressing a number as a product of only prime numbers.
**Key Property:** No matter how you break down a number into factors initially, the final prime factors will always be the same (though the order may differ).
**Example:** For 90:
Both give the same prime factors: 2, 3, 3, 5
**Steps:**
1. Divide the number by the smallest prime that divides it
2. Write the prime on the left, quotient below
3. Repeat until you reach 1
4. Write all primes on the left and the final 1 at bottom (or the last prime factor)
**Example 1: Prime Factorisation of 105**
```
3 | 105
5 | 35
| 7
```
Read the primes from left column and bottom: 105 = 3 × 5 × 7
**Example 2: Prime Factorisation of 30**
```
2 | 30
3 | 15
| 5
```
Therefore: 30 = 2 × 3 × 5
**Example 3: Prime Factorisation of 1200**
```
2 | 1200
2 | 600
2 | 300
2 | 150
3 | 75
5 | 25
| 5
```
Therefore: 1200 = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 2⁴ × 3 × 5²
**Why This Method Is Better:** Much faster and more reliable than random factorisation, especially for large numbers.
**Method:** A factor of a number consists of any combination of its prime factors.
**Example: Find all factors of 225**
First, find prime factorisation: 225 = 3 × 3 × 5 × 5
Now list all possible combinations:
**All factors of 225:** 1, 3, 5, 9, 15, 25, 45, 75, 225
**Verification:** 225 has (2+1) × (2+1) = 9 factors (if written as 3² × 5², the exponents are 2 and 2)
**Conjecture:** A statement or claim made without proof.
**Example of a False Conjecture:** "The larger a number is, the longer its prime factorisation will be"
**Counterexample (disproof):**
121 > 96, but 121 has fewer prime factors. This shows the conjecture is false.
---
**HCF Rule:** The HCF contains only the prime factors that appear in **both** numbers, and each prime appears the **minimum** number of times it appears in either factorisation.
1. Find the prime factorisation of both numbers
2. Identify which primes are **common** to both
3. For each common prime, count how many times it appears in each number
4. Choose the **minimum** count for that prime
5. Multiply all these minimum-count primes together
**Example 1: Find HCF of 45 and 75**
Prime factorisations:
Common primes: 3 and 5
HCF = 3¹ × 5¹ = 15
**Verification:** Factors of 45: 1, 3, 5, 9, 15, 45
Factors of 75: 1, 3, 5, 15, 25, 75
Common factors: 1, 3, 5, 15
HCF = 15 ✓
---
**Example 2: Find HCF of 112 and 84**
Prime factorisations:
Common primes: 2 and 7
HCF = 2² × 7 = 4 × 7 = 28
**Verification:** 112 ÷ 28 = 4, 84 ÷ 28 = 3 (both whole numbers, and 3 and 4 share no common factors) ✓
---
**Example 3: Find HCF of 96 and 275**
Prime factorisations:
Common primes: **None!**
HCF = 1
These are called **coprime numbers** (numbers with HCF = 1)
---
**Example 4: Find HCF of 30 and 72**
Prime factorisations:
Common primes: 2 and 3
HCF = 2 × 3 = 6
---
**Example 5: Find HCF of 225 and 750**
Prime factorisations:
Common primes: 3 and 5
HCF = 3¹ × 5² = 3 × 25 = 75
---
**Method:** Same principle—find minimum count for each common prime across **all** numbers.
**Example: Find HCF of 30, 60, and 90**
Prime factorisations:
Common primes: 2, 3, and 5
HCF = 2 × 3 × 5 = 30
---
**Real-Life Problem:** Anshu and Guna make torans (decorative cloth hangings).
**Solution:**
Any toran by Anshu is a multiple of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54...
Any toran by Guna is a multiple of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72...
Common multiples: 24, 48, 72... (numbers appearing in both lists)
The **smallest** common multiple is 24 cm.
So they can both make 24 cm long torans. This is the shortest toran both can make.
**Problem:** A sweet shop gives free gajak every Monday. Kabamai visits once every 10 days (today is Monday). When will she get free gajak again?
**Solution:**
Free gajak given on days: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70... (multiples of 7)
Kabamai visits on days: 10, 20, 30, 40, 50, 60, 70... (multiples of 10)
Common multiples: 70, 140, 210...
The **smallest** is 70 days.
Kabamai will get free gajak again after 70 days.
**LCM (Lowest Common Multiple)** = The lowest (smallest or least) of the common multiples of two or more numbers. Also called **Least Common Multiple**.
In the Idli-Vada game, two numbers are chosen:
The first number where "Idli-Vada" is called out is the **LCM**.
**Example:** For 4 and 6:
---
**LCM Rule:** The LCM contains **all** prime factors that appear in **either** number, and each prime appears the **maximum** number of times it appears in either factorisation.
1. Find the prime factorisation of both numbers
2. List **all** primes that appear in either factorisation
3. For each prime, count how many times it appears in each number
4. Choose the **maximum** count for that prime
5. Multiply all these maximum-count primes together
**Example 1: Find LCM of 14 and 35**
Prime factorisations:
All primes present: 2, 5, 7
LCM = 2 × 5 × 7 = 70
**Verification:** 70 ÷ 14 = 5, 70 ÷ 35 = 2 (both are whole numbers) ✓
---
**Example 2: Find LCM of 96 and 360**
Prime factorisations:
All primes present: 2, 3, 5
LCM = 2⁵ × 3² × 5 = 32 × 9 × 5 = 1440
**Verification:**
---
**Conceptual Reason:** The LCM must contain ALL the prime factors of both numbers so it's divisible by both. To keep it "least," we use each prime the minimum times needed to satisfy this condition.
For 96 = 2⁵ × 3:
For 360 = 2³ × 3² × 5:
Therefore: LCM = 2⁵ × 3² × 5
---
**Method:** Same principle—find maximum count for each prime across **all** numbers.
**Example: Find LCM of 30, 60, and 90**
Prime factorisations:
All primes: 2, 3, 5
LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180
---
| Property | HCF | LCM |
|----------|-----|-----|
| Full form | Highest Common Factor | Lowest Common Multiple |
| Meaning | Largest number that divides both | Smallest number divisible by both |
| Using primes | Use **minimum** count of each **common** prime | Use **maximum** count of each **any** prime |
| Result | Smaller or equal to original numbers | Larger or equal to original numbers |
| Example: 12, 18 | HCF = 6 | LCM = 36 |
---
**Observation:** HCF(6, 18) = 6
Notice that 6 is one of the two numbers. When does this happen?
**Answer:** When one number is a **factor** of the other (or equivalently, when one number is a **multiple** of the other).
**General Statement:** If a number m divides another number n completely, then:
**Algebraic Expression:** For any number n and positive integer k:
**Examples:**
---
**Example:**
**General Statement:** The HCF of any two consecutive even numbers is always **2**.
**Why?** Any two consecutive even numbers can be written as 2k and 2(k+1), where k is a positive integer.
Both have 2 as a factor. Since k and (k+1) are consecutive integers, they have no common factor other than 1.
Therefore, HCF(2k, 2(k+1)) = 2
---
**Example:**
**General Statement:** The HCF of any two consecutive odd numbers is always **1**.
**Why?** Any two consecutive odd numbers differ by 2. If they had a common factor d > 1, then d would divide both numbers, so d would also divide their difference (which is 2). The only divisors of 2 are 1 and 2. Since both numbers are odd, neither is divisible by 2.
Therefore, HCF(odd, odd) = 1
---
**Example:**
**General Statement:** The HCF of any two consecutive numbers is always **1**.
**Why?** Any two consecutive numbers are n and (n+1). If they shared a common factor d > 1, then:
The only divisor of 1 is 1 itself.
Therefore, HCF(n, n+1) = 1 for any positive integer n
---
**Example:**
**General Statement:** The HCF of any two even numbers is always **even** (at least 2).
**Why?** Both numbers have 2 as a factor. Therefore, their HCF must be at least 2, which is even.
Mathematical form: If HCF(m, n) = d where both m and n are even, then d is even.
---
**Definition:** Two numbers are **coprime** (or **relatively prime**) if their HCF is 1.
**Examples:**
**Important Note:** Coprime numbers don't need to be prime themselves. For example:
**General Statement:** If HCF(m, n) = 1, then m and n are coprime.
---
**Observation:** LCM(3, 24) = 24
Notice that 24 is one of the two numbers. When does this happen?
**Answer:** When one number is a **multiple** of the other.
**General Statement:** If a number n divides another number m completely, then:
**Algebraic Expression:** For any number n and positive integer k:
**Examples:**
---
**Key Property:** For any two numbers m and n:
**HCF(m, n) × LCM(m, n) = m × n**
**Proof Example:** For 12 and 18:
**Why This Works:**
**Practical Use:** If you know HCF and want LCM (or vice versa):
---
**Definition:** A **general statement** describes a pattern or property that holds true in all possible cases.
**Process of Generalisation:**
1. Observe specific examples
2. Find patterns
3. Express the pattern in words (general statement)
4. Verify with more examples
5. Optionally, express using algebra
**Example:** From specific cases:
The general statement: "If one number is a multiple of another, the HCF equals the smaller number."
---
**Rule:** Use the **minimum power** of each **common prime factor**.
**Formula:** For numbers m and n with prime factorisation:
HCF(m,n) = p₁^min(a₁,b₁) × p₂^min(a₂,b₂) × p₃^min(a₃,b₃) × ...
(Only include primes common to both)
---
**Rule:** Use the **maximum power** of each **prime factor** appearing in either number.
**Formula:** For numbers m and n with prime factorisation:
LCM(m,n) = p₁^max(a₁,b₁) × p₂^max(a₂,b₂) × p₃^max(a₃,b₃) × ...
(Include all primes from both factorizations)
---
**Steps:**
1. Divide by smallest prime that divides the number
2. Write prime divisor on left
3. Write quotient below
4. Repeat until quotient is 1
5. Product of all left-side primes = prime factorisation
**Quick Check:** For n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ, the number of factors = (a₁+1)(a₂+1)...(aₖ+1)
---
Every positive integer greater than 1 has a **unique prime factorisation** (the order of primes may vary, but the primes themselves and their powers are unique).
**Importance:** This allows us to reliably use prime factorisation for finding HCF and LCM.
For any two positive integers m and n:
**HCF(m, n) × LCM(m, n) = m × n**
This is one of the most useful properties and can be used to find one if the other is known.
If m is a factor of n, then:
The HCF of any two consecutive positive integers is always 1.
---
1. **Mistake:** Confusing HCF with LCM
2. **Mistake:** Using only common primes for LCM
3. **Mistake:** Using minimum for LCM or maximum for HCF
4. **Mistake:** Forgetting that 1 is always a common factor
5. **Mistake:** Prime factorisation stops too early
6. **Mistake:** Not verifying prime factorisation
7. **Mistake:** Assuming larger number always has longer prime factorisation
---
**Problem:** A room is 18 ft × 24 ft. Find the largest square tile size that fits exactly.
**Solution:**
Step 1: Understand — tile side must divide both dimensions
Step 2: Find prime factorisation:
Step 3: Find HCF using minimum powers:
Step 4: Calculate tiles needed:
**Answer:** 6 ft × 6 ft tiles; 12 tiles needed
---
**Problem:** Two traffic lights blink at intervals of 20
Q1. What is the HCF of 20 and 30?
Answer: B — Factors of 20 are 1, 2, 4, 5, 10, 20 and factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30; the highest common factor is 10.
Q2. Which of the following is the prime factorisation of 36?
Answer: C — 36 = 2 × 2 × 3 × 3 = 2² × 3²; the other options contain composite numbers, not only primes.
Q3. What is the HCF of 84 and 108?
Answer: B — 84 = 2² × 3 × 7 and 108 = 2² × 3³; common primes are 2² and 3, so HCF = 4 × 3 = 12.
Q4. Is 15 a common factor of 45 and 75?
Answer: A — 15 divides both 45 and 75 exactly, so it is a common factor; it is also the HCF.
Q5. Sameeksha's room is 12 ft by 16 ft. What is the largest square tile size (whole number) she can use?
Answer: C — The tile size must divide both 12 and 16 exactly; the HCF of 12 and 16 is 4 ft.
Q6. Lekhana has 84 kg and 108 kg of rice. What is the maximum weight of each equal bag?
Answer: B — The bag weight must divide both 84 and 108 equally; the HCF is 12 kg, using fewest bags.
Q7. A market vendor has 30 apples and 45 oranges. She wants to pack them into identical fruit baskets with the same number of each fruit in each basket. What is the maximum number of baskets she can make?
Answer: C — The HCF of 30 and 45 is 15, so she can make 15 baskets with 2 apples and 3 oranges each.
Q8. Two ropes are 56 m and 84 m long. They need to be cut into equal-length pieces with no waste. What is the longest possible length of each piece?
Answer: B — 56 = 2³ × 7 and 84 = 2² × 3 × 7; common primes are 2² and 7, so HCF = 4 × 7 = 28 m.
Q9. Which pair of numbers has HCF equal to 1?
Answer: C — 15 = 3 × 5 and 28 = 2² × 7 share no common prime factors, so their HCF is 1.
Q10. A school has 60 pencils and 90 erasers. These must be packed into identical gift boxes with equal numbers of pencils and erasers in each box. What is the maximum number of gift boxes that can be made?
Answer: C — HCF of 60 and 90 is 30; so 30 boxes can be made with 2 pencils and 3 erasers each.
What is the HCF of 12 and 16?
The HCF of 12 and 16 is 4, because 4 is the largest number that divides both 12 and 16 exactly.
Write the prime factorisation of 90.
The prime factorisation of 90 is 2 × 3 × 3 × 5.
What are the common factors of 45 and 75?
The common factors of 45 and 75 are 1, 3, 5, and 15.
Define a prime number.
A prime number is a number greater than 1 that has only 1 and itself as its factors.
What is the HCF of 96 and 275?
The HCF of 96 and 275 is 1, because they share no common prime factors.
How do you find HCF using prime factorisation?
Find common primes in both numbers, then multiply the minimum count of each common prime together.
What is a counterexample in mathematics?
A counterexample is a specific example that proves a conjecture false.
What is the HCF of 30 and 72 using their prime factorisations?
Since 30 = 2 × 3 × 5 and 72 = 2³ × 3², the HCF is 2 × 3 = 6.
Are the prime factors of a number always the same?
Yes, the prime factorisation of any number is unique, though the order of primes may change.
In Sameeksha's room problem, why is 4 ft the best tile size?
Because 4 ft is the HCF of 12 ft and 16 ft, making it the largest size that fits both dimensions exactly with fewest tiles.
What is the HCF of 12 and 16? [1 mark]
List factors of both numbers or use prime factorisation: 12 = 2² × 3 and 16 = 2⁴; common prime is 2², so HCF = 4.
Find all common factors of 45 and 75. [2 marks]
Prime factorise both: 45 = 3² × 5 and 75 = 3 × 5²; common subparts are 1, 3, 5, and 15. List them in order.
Sameeksha wants to tile her floor measuring 12 ft by 16 ft with square tiles of the largest possible size. How many tiles will she need? Show your working. [3 marks]
Step 1: Find HCF of 12 and 16 (= 4 ft). Step 2: Calculate number of tiles along length = 16 ÷ 4 = 4 tiles. Step 3: Calculate number of tiles along breadth = 12 ÷ 4 = 3 tiles. Step 4: Total = 4 × 3 = 12 tiles.
A shopkeeper has 84 kg of rice from one farm and 108 kg from another. She wants to pack them in equal-weight bags (whole kg) such that each bag contains rice from only one farm. (a) What should be the weight of each bag to use the fewest bags? (b) How many bags of each type will she need? Show all steps. [5 marks]
Step 1: Find HCF of 84 and 108 using prime factorisation: 84 = 2² × 3 × 7 and 108 = 2² × 3³; HCF = 2² × 3 = 12 kg. Step 2: Number of bags from farm 1 = 84 ÷ 12 = 7 bags. Step 3: Number of bags from farm 2 = 108 ÷ 12 = 9 bags. Step 4: Total bags = 7 + 9 = 16 bags. State both answers clearly.
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