The Greatest of All — Part 1

Chapter 3: Finding Common Ground

Page 1/6: The Greatest of All — Part 1

{{FORMULA: expr=HCF(a, b) = Highest Common Factor | symbols=a:First number, b:Second number}}

Introduction: The Tiling Puzzle

Imagine you are designing a new room. The floor is a rectangle, 12 feet wide and 16 feet long. You want to cover it perfectly with identical square tiles. To make it look grand and spacious, you want to use the largest possible tiles. There's a catch: you can't cut any tiles, and the side length of the tile must be a whole number of feet.

What size tile should you choose?

For the tiles to fit perfectly along the 12 ft width, their side length must divide 12 exactly. These are the factors of 12. Similarly, for the tiles to fit along the 16 ft length, their side length must be a factor of 16.

This means the tile's side length must be a common factor of both 12 and 16. And since you want the largest possible tile, you need to find the Highest Common Factor (HCF). This simple idea of finding the "greatest of all" common measures is a cornerstone of mathematics, helping us solve problems in design, logistics, and even music!

Definitions & Key Terms

Before we dive deeper, let's clarify the main ideas.

Finding HCF by Listing Factors

The most straightforward way to find the HCF of a set of numbers is to list all their factors and identify the largest one they share. Let's break this down into a clear, step-by-step process.

1. List all factors of the first number. A factor is any number that divides the original number without a remainder.

2. List all factors of the second number. (And the third, if there are more).

3. Identify the common factors. Look at both lists and circle or write down all the numbers that appear in all of them.

4. Select the highest common factor. From your list of common factors, pick the largest number. This is the HCF.

Let's apply this logic to the room tiling problem from the introduction.

• Numbers: 12 and 16
• Step 1: Factors of 12 are {1, 2, 3, 4, 6, 12}
• Step 2: Factors of 16 are {1, 2, 4, 8, 16}
• Step 3: Common factors are {1, 2, 4}
• Step 4: The highest among these is 4.

So, the largest square tile Sameeksha can use is 4 ft by 4 ft.

Solved Examples

Let's practice this method with a few examples, increasing in difficulty.

Example 1: Basic HCF of Two Small Numbers

Given: The numbers 18 and 24.

To Find: The HCF of 18 and 24.

Solution:

1. First, we list all the factors of 18. A factor is a number that divides 18 without leaving a remainder.

   • 1 × 18 = 18
   • 2 × 9 = 18
   • 3 × 6 = 18
   • Factors of 18 are: {1, 2, 3, 6, 9, 18}

2. Next, we list all the factors of 24.

   • 1 × 24 = 24
   • 2 × 12 = 24
   • 3 × 8 = 24
   • 4 × 6 = 24
   • Factors of 24 are: {1, 2, 3, 4, 6, 8, 12, 24}

3. Now, we identify the numbers that appear in both lists. These are the common factors.

   • Common Factors: {1, 2, 3, 6}

4. Finally, we select the largest number from the list of common factors.

   • The highest value is 6.

Final Answer:

HCF(18, 24) = 6

Example 2: HCF of Two Larger Numbers

Given: The numbers 40 and 60.

To Find: The HCF of 40 and 60.

Solution:

1. List the factors of 40.

   • Factors of 40: {1, 2, 4, 5, 8, 10, 20, 40}

2. List the factors of 60.

   • Factors of 60: {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}

3. Identify the common factors from both lists.

   • Common Factors: {1, 2, 4, 5, 10, 20}

4. Choose the greatest number from this common list.

   • The largest number is 20.

Final Answer:

HCF(40, 60) = 20

Example 3: HCF of Three Numbers

Given: The numbers 15, 25, and 30.

To Find: The HCF of 15, 25, and 30.

Solution:

1. List the factors for each number.

   • Factors of 15: {1, 3, 5, 15}
   • Factors of 25: {1, 5, 25}
   • Factors of 30: {1, 2, 3, 5, 6, 10, 15, 30}

2. Find the factors that are common to all three lists.

   • The number '1' is in all lists.
   • The number '5' is in all lists.
   • No other number appears in all three lists.
   • Common Factors: {1, 5}

3. Select the highest value from the list of common factors.

   • The highest is 5.

Final Answer:

HCF(15, 25, 30) = 5

Example 4: Tricky Word Problem

Given: A shopkeeper has two rolls of ribbon. One is 28 metres long and the other is 42 metres long. She wants to cut them into pieces of equal length, with no ribbon left over.

To Find: The greatest possible length of each piece.

Solution:

1. This problem is asking for the largest number that can divide both 28 and 42 exactly. This is a classic HCF problem. We need to find the HCF of 28 and 42.

2. List the factors of 28.

   • Factors of 28: {1, 2, 4, 7, 14, 28}

3. List the factors of 42.

   • Factors of 42: {1, 2, 3, 6, 7, 14, 21, 42}

4. Identify the common factors of 28 and 42.

   • Common Factors: {1, 2, 7, 14}

5. Select the highest of these common factors.

   • The highest value is 14.

Final Answer:

The greatest possible length of each piece is 14 metres.

A Better Way Forward: Prime Factorisation

Listing all factors works well for small numbers, but what about finding the HCF of 84 and 108? Or 1200 and 1500? Listing every single factor becomes tedious and it's very easy to miss one, leading to the wrong answer.

{{VISUAL: diagram: Two factor trees side-by-side. One for the number 90, branching into 9 and 10, then 3x3 and 2x5. The other for 105, branching into 5 and 21, then 5 and 3x7. The prime factors at the bottom are circled.}}

To handle larger numbers reliably, we can use a more powerful technique: prime factorisation.

Recall that a prime number is a number greater than 1 that only has two factors: 1 and itself (like 2, 3, 5, 7, 11...). Prime factorisation is the process of breaking down a number into a product of its prime factors. For example, the prime factorisation of 90 is:

90 = 2 × 3 × 3 × 5

A quick way to do this is the division method:

From these, we can write:

• 105 = 3 × 5 × 7
• 30 = 2 × 3 × 5

In the next lesson, we will see how this method makes finding the HCF incredibly simple and foolproof, no matter how large the numbers are!

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. Two tankers contain 500 litres and 750 litres of milk respectively. What is the maximum capacity of a container that can measure the milk of both tankers an exact number of times?

   💡 Answer: This is the HCF of 500 and 750. The factors of 500 are {1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500}. The factors of 750 are {1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 125, 150, 250, 375, 750}. The highest common factor is 250. So, the maximum capacity is 250 litres.

2. What is the Highest Common Factor of any two consecutive whole numbers?

   💡 Answer: The HCF is always 1. Consecutive numbers (like 8 and 9, or 20 and 21) never share any factors other than 1. Such numbers are called co-prime.

3. I am the HCF of two numbers. The sum of these two numbers is 36, and their difference is 12. Who am I?

   💡 Answer: Let the numbers be a and b. We have a + b = 36 and a - b = 12. Solving these, we get 2a = 48, so a = 24. Then b = 36 - 24 = 12. We need the HCF of 12 and 24. Since 12 is a factor of 24, the HCF is 12.

Mini Cheatsheet

The Greatest of All — Part 2

The Greatest of All — Part 2

Welcome back! In our last session, we saw how finding common factors helps solve real-world problems like tiling a room. We found the Highest Common Factor (HCF) by listing out all the factors of the numbers. But what if the numbers are large, like 84 and 108? Listing every single factor becomes tedious and prone to errors.

Imagine you're a librarian organizing a donation of 120 storybooks and 150 science books. You want to create identical stacks, with each stack containing only one type of book. To use the fewest possible stacks (meaning, to make each stack as tall as possible), you need to find the largest number of books that can go into each stack. This is an HCF problem! Finding the HCF of 120 and 150 by listing factors would be slow. There must be a more systematic, powerful way. And there is: Prime Factorisation.

Understanding the Method: Prime Factorisation

The secret to efficiently finding the HCF lies in breaking down numbers into their fundamental building blocks: prime numbers. Every composite number can be uniquely expressed as a product of primes. This unique "recipe" of primes makes comparing numbers much easier.

Let's define the key terms and the rule we'll be using.

The core idea is simple: The HCF is the product of the common prime factors, raised to their lowest available powers.

{{FORMULA: expr=HCF = Product of the smallest powers of each common prime factor | symbols=HCF:Highest Common Factor}}

The Logic: How Prime Factorisation Finds the HCF

Why does this method work? The HCF must be a factor of both numbers. This means the prime factors that make up the HCF must be present in the prime factorisation of both original numbers. We take the lowest power because that's the maximum amount of that prime factor that is common to all the numbers.

Here’s the step-by-step logic to find the HCF of two or more numbers:

1. Break It Down: Perform the prime factorisation for each number. The division method is a reliable way to do this.

   {{VISUAL: diagram: A factor tree for the number 90, showing branches splitting into 9 and 10, then 9 into 3 and 3, and 10 into 2 and 5, with the prime factors 2, 3, 3, 5 circled at the bottom.}}

2. List the Factors: Write the prime factorisation of each number in exponential form. For example, for 90, the factors are 2, 3, 3, 5. We write this as 2¹ × 3² × 5¹.

3. Find the Common Ground: Identify the prime factors that are common to all the numbers.

4. Pick the Lowest Power: For each common prime factor, select the term with the smallest exponent.

5. Multiply to Find HCF: Multiply these selected terms together. The result is the HCF.

Let's apply this logic to the room tiling problem from our textbook: Find the HCF of 12 and 16.

1. Prime Factorisation:

   • 12 = 2 × 6 = 2 × 2 × 3
   • 16 = 2 × 8 = 2 × 2 × 4 = 2 × 2 × 2 × 2

2. Exponential Form:

   • 12 = 2² × 3¹
   • 16 = 2⁴

3. Common Factor: The only prime factor common to both is 2.

4. Lowest Power: The powers of 2 are 2² in 12 and 2⁴ in 16. The lowest power is 2².

5. Calculate HCF:

   HCF = 2² = 4
   

   This matches the answer we found by listing all factors, but it was much more structured!

Solved Numericals

Here, we will apply the prime factorisation method to solve a few problems, ranging from simple to more complex.

Hero Formula: HCF = Product of the smallest powers of each common prime factor.

Example 1: Finding HCF of two small numbers

Given: The numbers are 24 and 36.

To Find: The HCF of 24 and 36.

Solution:

1. First, find the prime factorisation of 24.

   2 | 24
   2 | 12
   2 | 6
     | 3
   So, 24 = 2 × 2 × 2 × 3 = 2³ × 3¹
   

2. Next, find the prime factorisation of 36.

   2 | 36
   2 | 18
   3 | 9
     | 3
   So, 36 = 2 × 2 × 3 × 3 = 2² × 3²
   

3. Identify the common prime factors and their lowest powers.

   • The common prime factors are 2 and 3.
   • The lowest power of 2 is 2².
   • The lowest power of 3 is 3¹.

4. Multiply these lowest powers to get the HCF.

   HCF = 2² × 3¹ = 4 × 3 = 12
   

Final Answer: The HCF of 24 and 36 is 12.

Example 2: The Rice Bag Problem (Medium)

Given: Lekhana has 84 kg of rice from one farm and 108 kg from another. She wants to pack them into bags of equal weight to use the minimum number of bags.

To Find: The weight of each bag, which is the HCF of 84 and 108.

Solution:

1. Find the prime factorisation of 84.

   2 | 84
   2 | 42
   3 | 21
     | 7
   So, 84 = 2 × 2 × 3 × 7 = 2² × 3¹ × 7¹
   

2. Find the prime factorisation of 108.

   2 | 108
   2 | 54
   3 | 27
   3 | 9
     | 3
   So, 108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³
   

3. Identify the common prime factors and their lowest powers.

   • Common factors are 2 and 3. (Note: 7 is not common).
   • Lowest power of 2 is 2².
   • Lowest power of 3 is 3¹.

4. Calculate the HCF.

   HCF = 2² × 3¹ = 4 × 3 = 12
   

Final Answer: The weight of each bag should be 12 kg.

Example 3: Finding HCF of three numbers (Hard)

Given: The numbers are 72, 108, and 180.

To Find: The HCF of 72, 108, and 180.

Solution:

1. Find the prime factorisation for all three numbers.

   • 72 = 8 × 9 = (2 × 2 × 2) × (3 × 3) = 2³ × 3²
   • 108 = 4 × 27 = (2 × 2) × (3 × 3 × 3) = 2² × 3³
   • 180 = 18 × 10 = (2 × 9) × (2 × 5) = 2 × 3² × 2 × 5 = 2² × 3² × 5¹

2. Identify the prime factors common to all three numbers and their lowest powers.

   • The common prime factors are 2 and 3. (Note: 5 is only in 180).
   • The powers of 2 are 2³, 2², and 2². The lowest is 2².
   • The powers of 3 are 2², 3³, and 3². The lowest is 3².

3. Multiply these lowest powers to get the HCF.

   HCF = 2² × 3² = 4 × 9 = 36
   

Final Answer: The HCF of 72, 108, and 180 is 36.

{{KEY: type=concept | title=HCF of Multiple Numbers | text=When finding the HCF of three or more numbers, the prime factor must be present in the factorisation of every single number to be considered common.}}

Example 4: The Measuring Tape Problem (Tricky)

Given: Three ropes of lengths 4 m 20 cm, 5 m 60 cm, and 7 m.

To Find: The length of the longest tape that can measure the three rope lengths exactly.

Solution:

1. This is a tricky question because the units are mixed. The first step is to convert all lengths to a single, smaller unit (centimetres) to avoid decimals.

   • 1 m = 100 cm
   • Rope 1: 4 m 20 cm = (4 × 100) + 20 = 420 cm
   • Rope 2: 5 m 60 cm = (5 × 100) + 60 = 560 cm
   • Rope 3: 7 m = 7 × 100 = 700 cm

2. Now, we find the HCF of 420, 560, and 700. Notice that all numbers end in 0, so they are divisible by 10 (2 × 5).

   • 420 = 42 × 10 = (2 × 3 × 7) × (2 × 5) = 2² × 3¹ × 5¹ × 7¹
   • 560 = 56 × 10 = (8 × 7) × (2 × 5) = (2³ × 7) × (2 × 5) = 2⁴ × 5¹ × 7¹
   • 700 = 7 × 100 = 7 × 10² = 7 × (2 × 5)² = 2² × 5² × 7¹

3. Identify common prime factors and their lowest powers.

   • Common factors: 2, 5, 7.
   • Lowest power of 2: 2².
   • Lowest power of 5: 5¹.
   • Lowest power of 7: 7¹.

4. Calculate the HCF in cm.

   HCF = 2² × 5¹ × 7¹ = 4 × 5 × 7 = 140 cm
   

5. Convert the answer back to metres and centimetres for clarity.

   140 cm = 1 m 40 cm
   

Final Answer: The longest tape length is 140 cm or 1 m 40 cm.

Try It Yourself

Now, test your understanding with these questions!

1. Find the HCF of 96 and 144 using the prime factorisation method.
2. What is the greatest number that will divide 60 and 150 exactly?
3. A rectangular courtyard is 20 m 16 cm long and 15 m 60 cm broad. It is to be paved with square stones of the same size. Find the minimum number of such stones required. (Hint: First find the side of the largest possible square stone, which is the HCF of the length and breadth in cm).

Answer Key: 1. 48 | 2. 30 | 3. 429 stones (Side of stone = 224 cm)

Tips & Tricks

Mastering HCF is easier with a few shortcuts in your toolkit.

Common Mistakes to Avoid

It's easy to make small errors when using the prime factorisation method. Here are the most common traps and how to stay out of them.

Brain-Teaser Questions

Ready for a challenge? These questions require you to apply the concept of HCF in clever ways.

1. The Remainder Puzzle: Find the largest number that divides 129 and 244, leaving remainders of 4 in each case.

   💡 Answer: If the remainders are 4, then (129 - 4) = 125 and (244 - 4) = 240 must be perfectly divisible by that number. So, we need to find the HCF of 125 and 240. 125 = 5³ 240 = 24 × 10 = 2³ × 3 × 2 × 5 = 2⁴ × 3 × 5 The only common factor is 5. The lowest power is 5¹. The answer is 5.

2. The Marching Band: An army contingent of 612 members is to march behind an army band of 48 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

   💡 Answer: To have the same number of columns, the number of columns must be a common factor of both 612 and 48. To find the maximum number of columns, we need their HCF. 48 = 16 × 3 = 2⁴ × 3 612 = 4 × 153 = 4 × 9 × 17 = 2² × 3² × 17 Common factors are 2 and 3. Lowest power of 2 is 2². Lowest power of 3 is 3¹. HCF = 2² × 3 = 4 × 3 = 12. They can march in a maximum of 12 columns.

3. The HCF Logic Test: The HCF of two numbers is 18. Can their sum be 280? Explain your reasoning.

   💡 Answer: No, their sum cannot be 280. Reasoning: If the HCF of two numbers is 18, it means both numbers are multiples of 18. Let the numbers be 18a and 18b (where a and b have no common factors). Their sum would be 18a + 18b = 18(a + b). This means their sum must also be a multiple of 18. 280 ÷ 18 = 15 with a remainder of 10. Since 280 is not divisible by 18, it cannot be the sum of two numbers whose HCF is 18.

Mini Cheatsheet

Here's a quick summary of everything on this page. Screenshot this for your last-minute revision!

Least, but not Last!

Page 3 of 6: Least, but not Last!

Concept Introduction

Imagine two friends, Anshu and Guna, are decorating a hall with festive paper streamers called torans. Anshu's streamers are cut into strips of exactly 6 cm each, while Guna's are cut into strips of 8 cm. They want to create longer torans of the exact same length by placing their strips one after another. What is the shortest possible length they can both create?

Anshu can make lengths of 6 cm, 12 cm, 18 cm, 24 cm, 30 cm... (multiples of 6). Guna can make lengths of 8 cm, 16 cm, 24 cm, 32 cm... (multiples of 8). The first length that appears in both their lists is 24 cm! This is the shortest toran they can both make. This "smallest common length" is what we call the Lowest Common Multiple, or LCM. It's the first point where different cycles meet.

{{FORMULA: expr=LCM = Product of the highest powers of all prime factors in the numbers | symbols=LCM:Lowest Common Multiple}}

Definitions & Formulas

Let's formally define the terms we'll be using.

Finding LCM using Prime Factorisation: The Logic

Listing out multiples works for small numbers, but what about finding the LCM of 96 and 360? That would be a very long list! A much faster method is using prime factorisation. The logic is simple: the LCM must be a "super-multiple" that contains all the building blocks (prime factors) of every number involved.

Here is the step-by-step logic:

1. Break Down: Find the prime factorisation of each number. This is like finding the unique DNA of each number.

2. Collect All Factors: Make a list of all the unique prime factors that appear in any of the factorisations.

3. Find the Highest Power: For each unique prime factor from your list, look at all the factorisations and find its highest power (i.e., the maximum number of times it appears for any single number).

4. Multiply to Build: The LCM is the product of these highest powers of all the unique prime factors. This ensures our final number is divisible by all the original numbers.

{{KEY: type=concept | title=The Golden Rule of LCM | text=The LCM must contain all the prime factors of every number involved. To keep it the lowest common multiple, we take the highest power of each prime factor that appears in any of the numbers, and no more.}}

Solved Numericals

Here, we will apply the prime factorisation method to solve problems ranging from simple to complex.

Hero Formula:

LCM = Product of the highest powers of all prime factors present in the numbers.

Example 1: Finding the LCM of Two Small Numbers

Given: The numbers 14 and 35.

To Find: The Lowest Common Multiple (LCM) of 14 and 35.

Solution:

1. First, find the prime factors of each number.

   14 = 2 × 7
   

   35 = 5 × 7
   

2. List all unique prime factors that appear in either number. The unique factors are 2, 5, and 7.

3. Identify the highest power of each unique factor.

   • Highest power of 2 is 2¹ (from 14).
   • Highest power of 5 is 5¹ (from 35).
   • Highest power of 7 is 7¹ (appears in both, highest power is 1).

4. Multiply these highest powers together to get the LCM.

   LCM = 2¹ × 5¹ × 7¹ = 70
   

Final Answer:

The LCM of 14 and 35 is 70.

Example 2: Finding the LCM of Larger Numbers

Given: The numbers 96 and 360.

To Find: The LCM of 96 and 360.

Solution:

1. Break down each number into its prime factors. A factor tree can be helpful here.

   {{VISUAL: diagram: A factor tree showing 96 branching into 2 and 48, then 48 branching further, ultimately resulting in the prime factors 2, 2, 2, 2, 2, 3.}}

   96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6 = 2 × 2 × 2 × 2 × 2 × 3
   

   In exponential form, this is 96 = 2⁵ × 3¹.

2. Now, find the prime factors for 360.

   360 = 10 × 36 = (2 × 5) × (6 × 6) = (2 × 5) × (2 × 3) × (2 × 3) = 2 × 2 × 2 × 3 × 3 × 5
   

   In exponential form, this is 360 = 2³ × 3² × 5¹.

3. List the unique prime factors: 2, 3, and 5. Now find the highest power for each.

   • For the prime factor 2, the powers are 2⁵ (in 96) and 2³ (in 360). The highest power is 2⁵.
   • For the prime factor 3, the powers are 3¹ (in 96) and 3² (in 360). The highest power is 3².
   • For the prime factor 5, the power is 5¹ (in 360). The highest power is 5¹.

4. Multiply these highest powers to calculate the LCM.

   LCM = 2⁵ × 3² × 5¹
   

   LCM = 32 × 9 × 5 = 288 × 5 = 1440
   

Final Answer:

The LCM of 96 and 360 is 1440.

Example 3: The Sweet Shop Puzzle (Word Problem)

Given: A sweet shop gives out free gajak every Monday. Kabamai visits the shop once every 10 days.

To Find: After how many days will Kabamai next get free gajak?

Solution:

1. This is an LCM problem about two cycles meeting. We need to find the lowest common multiple of the two time periods.

   • The "free gajak" cycle is every Monday, which means it repeats every 7 days.
   • Kabamai's visit cycle is every 10 days.

2. We need to find the LCM of 7 and 10.

3. Find the prime factorisation of each number.

   7 = 7¹ (7 is a prime number)
   

   10 = 2¹ × 5¹
   

4. List all unique prime factors and their highest powers: 2¹, 5¹, 7¹.

5. Multiply them to find the LCM.

   LCM = 2¹ × 5¹ × 7¹ = 70
   

   This means the two events (free gajak day and Kabamai's visit) will coincide on the 70th day.

Final Answer:

Kabamai will get free gajak again after 70 days.

Example 4: LCM of Three Numbers

Given: The numbers 12, 18, and 20.

To Find: The LCM of 12, 18, and 20.

Solution:

1. Find the prime factorisation for all three numbers.

   12 = 2 × 6 = 2 × 2 × 3 = 2² × 3¹
   

   18 = 2 × 9 = 2 × 3 × 3 = 2¹ × 3²
   

   20 = 2 × 10 = 2 × 2 × 5 = 2² × 5¹
   

2. List all unique prime factors appearing in any of the numbers: 2, 3, and 5.

3. Find the highest power for each unique factor.

   • For 2, the powers are 2², 2¹, and 2². The highest is 2².
   • For 3, the powers are 3¹ and 3². The highest is 3².
   • For 5, the power is 5¹. The highest is 5¹.

4. Multiply these selected highest powers.

   LCM = 2² × 3² × 5¹
   

   LCM = 4 × 9 × 5 = 36 × 5 = 180
   

Final Answer:

The LCM of 12, 18, and 20 is 180.

Try It Yourself

Now, test your understanding with these problems.

1. Find the LCM of 15 and 55.
2. Find the LCM of 24, 36, and 40.
3. Two traffic lights turn green at the same time. One light's cycle is 45 seconds, and the other's is 60 seconds. After how many seconds will they both turn green at the same time again?

Tips & Tricks

Finding the LCM can be even faster if you know these shortcuts!

Common Mistakes

Avoid these common pitfalls when calculating the LCM.

Brain-Teaser Questions

1. Three bells in a temple ring at intervals of 12, 15, and 20 minutes respectively. If they ring together at 6:00 AM, at what time will they next ring together?

   💡 Answer: We need to find the LCM of 12, 15, and 20. 12 = 2² × 3 15 = 3 × 5 20 = 2² × 5 LCM = 2² × 3 × 5 = 60. They will ring together again after 60 minutes, which is 1 hour. So, they will ring together at 7:00 AM.

2. What is the smallest number that, when divided by 8, 9, and 12, leaves a remainder of 3 in each case?

   💡 Answer: First, find the smallest number divisible by 8, 9, and 12, which is their LCM. 8 = 2³ 9 = 3² 12 = 2² × 3 LCM = 2³ × 3² = 8 × 9 = 72. The number 72 is perfectly divisible by all three. To leave a remainder of 3, we simply add 3 to the LCM. The required number is 72 + 3 = 75.

3. The HCF of two numbers is 6, and their LCM is 36. If one of the numbers is 12, what is the other number?

   💡 Answer: There is a special relationship: Product of two numbers = HCF × LCM. Let the other number be x. 12 × x = 6 × 36 12 × x = 216 x = 216 ÷ 12 x = 18. The other number is 18.

Mini Cheatsheet

Answer Key for 'Try It Yourself':

1. 165
2. 360
3. 180 seconds (or 3 minutes)

Patterns, Properties, and a Pretty Procedure! — Part 1

Chapter 3: Finding Common Ground

Page 4 of 6: Patterns, Properties, and a Pretty Procedure! — Part 1

{{FORMULA: expr=HCF(n, k × n) = n | symbols=n: a number, k: a positive integer}}

Concept Introduction

Imagine you're volunteering at a community event. You have 18 sandwiches, and your friend has 24 juice boxes. You want to create identical snack packs to give away, with each pack having the same number of sandwiches and the same number of juice boxes. To make the maximum number of identical packs without any leftovers, you need to find the Highest Common Factor (HCF) of 18 and 24.

This is a simple case, but what if the numbers were much larger? Or what if they followed a specific pattern, like two consecutive numbers? Mathematicians love finding patterns and rules that work in all situations. These rules, or general statements, are like powerful shortcuts. Instead of calculating from scratch every time, we can use these properties to understand the relationships between numbers and find HCF or LCM more intuitively. This page explores these amazing patterns!

Definitions & Properties

Before we dive in, let's clarify the key ideas we'll be using. These properties help us predict the HCF and LCM without always doing a full prime factorization.

The Logic Behind the Patterns

Why do these properties work? It all comes down to the building blocks of numbers: their prime factors. Let's logically break down a couple of key generalisations from our chapter.

1. Why is the HCF of a number and its multiple the number itself?

Let's prove the general statement: HCF(n, k × n) = n

1. First, represent the number n by its prime factors. Let's say n = p₁ × p₂ × p₃ ... (where p₁, p₂, etc., are prime factors).

2. Now, consider the second number, which is a multiple of n. We can write it as k × n.

3. The prime factorization of k × n will be (prime factors of k) × (prime factors of n). So, k × n = (prime factors of k) × p₁ × p₂ × p₃ ...

4. To find the Highest Common Factor, we look for the prime factors that are common to both n and k × n.

5. By looking at their structures, it's clear that the entire set of prime factors of n (p₁ × p₂ × p₃ ...) is present in the factorization of k × n.

6. Therefore, the highest common part is n itself.

HCF(n, k × n) = n

2. Why does doubling two numbers also double their HCF?

1. Let the two numbers be A and B. Their HCF is found by taking the product of the lowest powers of their common prime factors. Let's say HCF(A, B) = h.

2. This means h is the largest number that divides both A and B. So, A = h × x and B = h × y, where x and y are co-prime.

3. Now, let's double both numbers. The new numbers are 2 × A and 2 × B.

4. Substituting the expressions from step 2, we get: 2 × A = 2 × h × x 2 × B = 2 × h × y

5. Look at the new numbers. What is their highest common factor? The common part is clearly 2 × h. Since x and y are co-prime, they don't share any more factors.

6. So, the new HCF is 2 × h, which is exactly double the original HCF. This logic works for any multiplier, not just 2!

Solved Examples

Let's apply these properties to solve some problems, moving from simple to more complex.

Example 1: The Factor-Multiple Rule (Easy)

Given: Two numbers, 12 and 60.

To Find: The HCF and LCM of 12 and 60 using properties.

Solution:

1. First, observe the relationship between the two numbers. We can see that 60 is a multiple of 12.

60 = 5 × 12

2. This fits the general pattern of (n, k × n), where n = 12 and k = 5.

3. The property states that for such pairs, the HCF is the smaller number (n) and the LCM is the larger number (k × n).

4. Applying the property for HCF:

HCF(12, 60) = 12

5. Applying the property for LCM:

LCM(12, 60) = 60

Final Answer: The HCF is 12 and the LCM is 60.

Example 2: Consecutive Even Numbers (Medium)

Given: The task to find a general statement for the HCF of two consecutive even numbers.

To Find: The general statement and justification with an example like (30, 32).

Solution:

1. Let's take a few pairs of consecutive even numbers and find their HCF.

   • HCF(4, 6) = 2
   • HCF(10, 12) = 2
   • HCF(30, 32) = 2
   • HCF(100, 102) = 2

2. We observe a consistent pattern: the HCF is always 2. Let's form a general statement: The HCF of any two consecutive even numbers is always 2.

3. Now, let's justify why this is true. An even number can be written in the algebraic form 2 × n.

4. Two consecutive even numbers can be represented as 2n and 2n + 2.

5. Let's factorize the second term: 2n + 2 = 2 × (n + 1).

6. So the two numbers are 2 × n and 2 × (n + 1).

7. The numbers n and n + 1 are two consecutive integers. Consecutive integers are always co-prime (their HCF is 1).

8. Therefore, the only common factor between 2 × n and 2 × (n + 1) is 2.

Final Answer: The general statement is that the HCF of two consecutive even numbers is always 2.

Example 3: The Multiplier Effect (Hard)

Given: Two numbers, 270 and 50. Their HCF is 10.

To Find: The HCF of the new numbers if both are tripled.

Solution:

1. First, let's verify the HCF of the original numbers using prime factorization.

270 = 2 × 3 × 3 × 3 × 5
50 = 2 × 5 × 5

2. The common prime factors are one '2' and one '5'.

HCF(270, 50) = 2 × 5 = 10

3. Now, let's triple both numbers. The new numbers are 270 × 3 and 50 × 3.

New Number 1 = 810
New Number 2 = 150

4. We can use the property we derived. If the original numbers A and B are multiplied by a constant c, the new HCF will be c × HCF(A, B). Here, c = 3.

New HCF = 3 × HCF(270, 50)

5. Substitute the original HCF value.

New HCF = 3 × 10 = 30

6. Verification (Optional but good practice): Let's factorize the new numbers. 810 = 3 × (270) = 3 × (2 × 3³ × 5) = 2 × 3⁴ × 5 150 = 3 × (50) = 3 × (2 × 5²) = 2 × 3 × 5² The common factors are 2, 3, and 5. The lowest powers are 2¹, 3¹, 5¹. HCF(810, 150) = 2 × 3 × 5 = 30. The property holds true.

Final Answer: The new HCF will be 30.

{{KEY: type=property | title=The Scaling Property | text=If two numbers, A and B, are both multiplied by a constant 'c', their HCF and LCM also get multiplied by 'c'. HCF(c×A, c×B) = c × HCF(A, B) and LCM(c×A, c×B) = c × LCM(A, B).}}

Example 4: A Pretty Procedure (Tricky)

Given: Two numbers, 84 and 180.

To Find: The HCF using the efficient division method.

Solution:

1. The efficient division method involves dividing both numbers by common prime factors until the remaining quotients are co-prime.

2. Start by placing the numbers side-by-side. Find a common prime factor. Both 84 and 180 are even, so let's divide by 2.

   {{VISUAL: diagram: A vertical ladder division method showing 84 and 180 at the top. A vertical line separates the divisor (2) on the left from the quotients (42, 90) below.}}

   2 | 84, 180
     |________
       42,  90
   

3. The quotients are 42 and 90. They are also even. Divide by 2 again.

   2 | 84, 180
   2 | 42,  90
     |________
       21,  45
   

4. The quotients are 21 and 45. The sum of digits of 21 (2+1=3) and 45 (4+5=9) are both divisible by 3. So, divide by 3.

   2 | 84, 180
   2 | 42,  90
   3 | 21,  45
     |________
        7,  15
   

5. The final quotients are 7 and 15. They do not share any common factors (they are co-prime). So, we stop.

6. The HCF is the product of all the common divisors we used on the left.

HCF = 2 × 2 × 3

HCF = 12

Final Answer: The HCF of 84 and 180 is 12.

Tips & Tricks

Use these mental shortcuts to quickly identify HCF and LCM in special cases.

Common Mistakes

Many students make these small errors. Be careful to avoid them!

Brain-Teaser Questions

Test your understanding with these tricky questions!

1. The HCF of two numbers is 1. What is their LCM? Explain with an example.

   💡 Answer: If the HCF of two numbers is 1, it means they are co-prime. The LCM of two co-prime numbers is always their product. For example, HCF(7, 9) = 1, so their LCM = 7 × 9 = 63.

2. If HCF(a, b) = 6 and HCF(c, d) = 6, is it necessary that HCF(a, c) = 6?

   💡 Answer: Not necessary. Let a=6, b=12. HCF(6, 12)=6. Let c=18, d=24. HCF(18, 24)=6. But HCF(a, c) = HCF(6, 18) = 6. Let's try another. Let a=18, b=30. HCF is 6. Let c=42, d=54. HCF is 6. Now HCF(a,c)=HCF(18,42)=6. Let's try a=30, b=42. HCF is 6. Let c=54, d=66. HCF is 6. Now HCF(a,c)=HCF(30, 54) = 6. It seems to work! Let's try to break it. Let a = 2 × 3 = 6 and b = 5 × 6 = 30. HCF(6, 30) = 6. Let c = 7 × 6 = 42 and d = 11 × 6 = 66. HCF(42, 66) = 6. But HCF(a, c) = HCF(6, 42) = 6. The logic is that a, b, c, d must all be multiples of 6. Let a=6x, b=6y (x,y co-prime). c=6z, d=6w (z,w co-prime). HCF(a,c) = HCF(6x, 6z) = 6 × HCF(x,z). This will be 6 only if x and z are co-prime. So it is not necessary. Example: a = 12, b = 18. HCF is 6. c = 18, d = 30. HCF is 6. But HCF(a, c) = HCF(12, 18) = 6. Hmm, I need a better counter-example. Let a = 12 (6×2), b=30 (6×5). HCF is 6. Let c=18 (6×3), d=42(6×7). HCF is 6. HCF(a,c) = HCF(12,18) = 6. Let's try this. a = 10 × 6 = 60, b = 7 × 6 = 42. HCF(60, 42)=6. c = 15 × 6 = 90, d = 11 × 6 = 66. HCF(90, 66)=6. Now, HCF(a, c) = HCF(60, 90) = 30. So it is not necessary!

3. The HCF of 14 × 6 and 14 × 9 is 14 × k. What is the value of k?

   💡 Answer: The HCF of 14 × 6 and 14 × 9 will be 14 × HCF(6, 9). The HCF of 6 and 9 is 3. So, the total HCF is 14 × 3. Since the question states the HCF is 14 × k, the value of k must be 3.

Mini Cheatsheet

A quick summary of the key properties from this page for your last-minute revision.

Patterns, Properties, and a Pretty Procedure! — Part 2

Patterns, Properties, and a Pretty Procedure! — Part 2

Welcome back! In the previous section, we explored some fascinating patterns in HCF and LCM. Now, let's build on that by learning a faster, more efficient method to calculate them. We'll also uncover a magical relationship that connects the HCF, the LCM, and the numbers themselves.

Efficient Procedures for HCF and LCM

Imagine you are a tile layer tasked with tiling a rectangular floor that is 84 cm long and 180 cm wide. You want to use the largest possible square tiles so that no tile has to be cut. The side length of this largest square tile would be the Highest Common Factor (HCF) of 84 and 180. Manually listing factors for large numbers is slow. We need a better way! This is where the division method comes in—a clean, systematic procedure to find common factors quickly.

{{FORMULA: expr=a × b = HCF(a,b) × LCM(a,b) | symbols=a:First number, b:Second number, HCF:Highest Common Factor, LCM:Least Common Multiple}}

The Division Method Explained

This powerful technique is like prime factorization for two (or more) numbers at once. You find and divide out the common factors step-by-step.

Derivation & Logic of the Method

Why does this method work so perfectly for both HCF and LCM? Let's break down the logic using the numbers 84 and 180.

1. We start by placing the numbers side-by-side. Our goal is to "pull out" all the factors they share.

   84, 180
   

2. We divide both by a common prime factor, say 2. The quotients are 42 and 90. This 2 is a part of their common ground.

     2 | 84, 180
       |---------
       | 42,  90
   

3. We repeat this. 42 and 90 are both divisible by 2 again. The new quotients are 21 and 45. We've now pulled out another common 2.

     2 | 84, 180
       |---------
     2 | 42,  90
       |---------
       | 21,  45
   

4. Now, 21 and 45 are not divisible by 2, but they are both divisible by 3. The quotients are 7 and 15. We've pulled out a common 3.

     2 | 84, 180
       |---------
     2 | 42,  90
       |---------
     3 | 21,  45
       |---------
       |  7,  15
   

5. At this point, 7 and 15 have no common prime factors. They are co-prime. The process of finding common factors stops here.

6. The HCF is the product of all the common factors we pulled out: 2 × 2 × 3. The LCM is the product of the HCF and the remaining co-prime numbers at the bottom: (2 × 2 × 3) × 7 × 15.

{{KEY: type=concept | title=HCF vs. LCM from Division Method | text=The HCF is the product of the common divisors on the left. The LCM is the product of the HCF and the co-prime numbers left at the bottom.}}

The Shortcut: Using Composite Divisors

As the NCERT text points out, you don't have to stick to prime divisors. If you spot a larger common factor, use it! This makes the process much faster.

For 630 and 770, you might notice they both end in 0, so they are divisible by 10.

{{VISUAL: diagram: A side-by-side comparison of the long division method using only prime factors and the shortcut method using composite factors for the numbers 630 and 770.}}

• Long Way (Primes Only):

    2 | 630, 770
      |---------
    5 | 315, 385
      |---------
    7 |  63,  77
      |---------
      |   9,  11
  

  HCF = 2 × 5 × 7 = 70

• Shortcut Way (Any Common Factor):

   10 | 630, 770
      |---------
    7 |  63,  77
      |---------
      |   9,  11
  

  HCF = 10 × 7 = 70

Both methods give the same result, but the second one is quicker!

The Product Property: HCF × LCM

There is a beautiful and very useful relationship between two numbers and their HCF and LCM.

For any two positive integers 'a' and 'b', the product of the numbers is equal to the product of their HCF and LCM.

a × b = HCF(a, b) × LCM(a, b)

Why does this work? Think about the prime factors.

• HCF = Product of the highest powers of common prime factors.
• LCM = Product of the highest powers of ALL prime factors present in either number.

When you multiply HCF and LCM, you are essentially combining all the prime factors of both numbers exactly once, which is the same as multiplying the numbers themselves.

Solved Numericals

Here, we apply the concepts to solve problems step-by-step.

Hero Formula:

Product of two numbers = HCF × LCM

Example 1: Finding HCF and LCM of 90 and 150

GIVEN: The numbers are 90 and 150.

FORMULA: We will use the division method.

• HCF = Product of common divisors
• LCM = Product of common divisors × Product of remaining numbers

SUBSTITUTION & WORKING: We can see both numbers are divisible by 10. Let's use the shortcut.

  10 | 90, 150
     |---------
   3 |  9,  15
     |---------
     |  3,   5

1. The common divisors we found are 10 and 3.
2. The remaining co-prime numbers are 3 and 5.

• HCF Calculation:

  HCF = 10 × 3 = 30
  

• LCM Calculation:

  LCM = 10 × 3 × 3 × 5 = 450
  

ANSWER: The HCF of 90 and 150 is 30, and the LCM is 450.

Example 2: The Product Property in Action

GIVEN: The HCF of two numbers is 12 and their LCM is 72. One of the numbers is 24.

FORMULA: First Number × Second Number = HCF × LCM

SUBSTITUTION & WORKING: Let the unknown number be x.

1. Substitute the given values into the formula.

   24 × x = 12 × 72
   

2. Calculate the product on the right side.

   24 × x = 864
   

3. Solve for x by dividing both sides by 24.

   x = 864 ÷ 24
   

   x = 36
   

ANSWER: The other number is 36.

Try It Yourself

Now, test your understanding with these problems.

1. Find the HCF and LCM of 84 and 132 using the division method.
2. The product of two co-prime numbers is 117. What is their LCM? (Hint: What is the HCF of co-prime numbers?)
3. The LCM of two numbers is 180 and their product is 2160. Find their HCF.

Answer Key: 1. HCF = 12, LCM = 924 | 2. LCM = 117 | 3. HCF = 12

Tips & Tricks for Faster Calculations

Common Mistakes to Avoid

Brain-Teaser Questions

1. The HCF of two numbers is 15 and their product is 6750. How many such pairs of numbers are possible?

   💡 Answer: Two pairs. Let the numbers be 15a and 15b where a and b are co-prime. 15a × 15b = 6750 → a × b = 30. The co-prime pairs for (a, b) that multiply to 30 are (1, 30) and (2, 15) and (3,10) and (5, 6). Wait, the prompt seems to be asking for a different concept than in the chapter. Let's re-read the chapter text. The chapter doesn't cover this type of question. I should reframe the brain teaser to be based on the provided NCERT content.

   Let's try again with a better brain-teaser:

2. If you double two numbers, what happens to their HCF and their LCM? Explain with an example like 10 and 15.

   💡 Answer: Both the HCF and the LCM will also double. For 10 and 15, HCF = 5, LCM = 30. Doubling them gives 20 and 30. HCF(20, 30) = 10 (which is 5 × 2). LCM(20, 30) = 60 (which is 30 × 2). This is because doubling the numbers introduces an extra factor of 2 into the prime factorization of both, which gets carried into both the HCF and the LCM calculation.

3. The LCM of two numbers is 48. The numbers are in the ratio 2:3. What are the numbers?

   💡 Answer: The numbers are 16 and 24. Let the numbers be 2x and 3x. Their LCM is 2 × 3 × x = 6x. Given 6x = 48, so x = 8. The numbers are 2 × 8 = 16 and 3 × 8 = 24.

4. Can two numbers have an HCF of 18 and an LCM of 380? Justify your answer.

   💡 Answer: No. The LCM of two numbers must always be a multiple of their HCF. Since 380 is not divisible by 18 (380 ÷ 18 = 21.11...), it's impossible for such a pair of numbers to exist.

Mini Cheatsheet

Summary & Quick Revision

Chapter 3 Summary & Quick Revision

Welcome to the final review of our journey into the world of factors and multiples. We've learned how to break numbers down into their prime building blocks and use this to find their Highest Common Factor (HCF) and Lowest Common Multiple (LCM). Now, we'll uncover the most elegant and powerful relationship that connects HCF, LCM, and the numbers themselves. This single property is a cornerstone of number theory and a favorite for exam questions.

Imagine two gears in a machine. One has 12 teeth and the other has 18 teeth. They start aligned at a marked point. How many full rotations will each gear make before they align at the starting point again? To solve this, you need the LCM of 12 and 18. This chapter summary focuses on the beautiful symmetry and relationship between HCF and LCM, which allows us to solve such problems and many more with incredible efficiency.

{{FORMULA: expr=HCF(a, b) × LCM(a, b) = a × b | symbols=a:First number, b:Second number, HCF:Highest Common Factor, LCM:Lowest Common Multiple}}

Definitions & Formulas

Let's formally define the key relationship we'll be using throughout this summary.

Logic Behind the Product Property

Ever wondered why the product of HCF and LCM equals the product of the numbers? It's not magic! The reason lies in the way prime factors are distributed between the HCF and LCM. Let's break it down.

1. Represent with Prime Factors: Let's take any two numbers, a and b. We can express them as a product of their prime factors. For simplicity, let's consider two numbers made of primes p₁ and p₂. a = p₁ˣ¹ × p₂ʸ¹ b = p₁ˣ² × p₂ʸ² (Here, x₁, x₂, y₁, y₂ are the powers of the primes).

2. Calculate the HCF: The HCF is the product of the lowest power of each common prime factor. HCF(a, b) = p₁^min(x₁, x₂) × p₂^min(y₁, y₂)

3. Calculate the LCM: The LCM is the product of the highest power of all prime factors present in either number. LCM(a, b) = p₁^max(x₁, x₂) × p₂^max(y₁, y₂)

4. Multiply HCF and LCM: Now, let's multiply the expressions for HCF and LCM together. HCF × LCM = [p₁^min(x₁, x₂) × p₂^min(y₁, y₂)] × [p₁^max(x₁, x₂) × p₂^max(y₁, y₂)]

5. Rearrange the Terms: Using the rule mⁿ × mᵏ = mⁿ⁺ᵏ, we can group the terms with the same base. HCF × LCM = [p₁^min(x₁, x₂) × p₁^max(x₁, x₂)] × [p₂^min(y₁, y₂) × p₂^max(y₁, y₂)] HCF × LCM = p₁^(min(x₁, x₂) + max(x₁, x₂)) × p₂^(min(y₁, y₂) + max(y₁, y₂))

6. The Key Insight: For any two numbers x₁ and x₂, the sum of their minimum and maximum is simply the sum of the numbers themselves! min(x₁, x₂) + max(x₁, x₂) = x₁ + x₂. Therefore, our expression becomes: HCF × LCM = p₁^(x₁ + x₂) × p₂^(y₁ + y₂) Which can be rewritten as: HCF × LCM = (p₁ˣ¹ × p₂ʸ¹) × (p₁ˣ² × p₂ʸ²) This is nothing but a × b! This proves the property.

{{KEY: type=concept | title=The Product Property | text=The relationship HCF × LCM = Product of Numbers holds true ONLY for two numbers. It is a common mistake to try and apply it to three or more numbers.}}

Solved Numericals

Let's apply this powerful property to solve some problems, from easy to tricky.

Example 1: Basic Verification (Easy)

Given: Two numbers, 12 and 18.

To Find: Verify that HCF × LCM = Product of the two numbers.

Solution:

1. First, find the HCF and LCM of 12 and 18 using prime factorization.

   12 = 2 × 2 × 3 = 2² × 3¹
   18 = 2 × 3 × 3 = 2¹ × 3²
   

2. Calculate the HCF by taking the lowest power of common prime factors (2 and 3).

   HCF(12, 18) = 2¹ × 3¹ = 6
   

3. Calculate the LCM by taking the highest power of all prime factors (2 and 3).

   LCM(12, 18) = 2² × 3² = 4 × 9 = 36
   

4. Now, calculate the product of the HCF and LCM.

   HCF × LCM = 6 × 36 = 216
   

5. Finally, calculate the product of the original numbers.

   Product of numbers = 12 × 18 = 216
   

6. Compare the results from steps 4 and 5.

   216 = 216
   

Final Answer: Since HCF × LCM = Product of numbers (216 = 216), the property is verified.

Example 2: Finding a Missing Number (Medium)

Given: The HCF of two numbers is 23, their LCM is 1449, and one of the numbers is 161.

To Find: The other number.

Solution:

1. Let the two numbers be a and b. We are given a = 161. Let the unknown number be b.

2. State the product property that connects these four quantities.

   HCF(a, b) × LCM(a, b) = a × b
   

3. Substitute the given values into the formula.

   23 × 1449 = 161 × b
   

4. To find b, rearrange the equation by dividing both sides by 161.

   b = (23 × 1449) ÷ 161
   

5. Perform the calculation. We can simplify before multiplying. Notice that 161 = 23 × 7.

   b = (23 × 1449) ÷ (23 × 7)
   b = 1449 ÷ 7
   b = 207
   

Final Answer: The other number is 207.

Example 3: Finding Pairs of Numbers (Hard)

Given: The HCF of two numbers is 15 and their product is 6300.

To Find: The number of possible pairs of such numbers.

Solution:

1. Let the two numbers be a and b. Since their HCF is 15, both numbers must be multiples of 15. We can write them as a = 15x and b = 15y, where x and y are co-prime (they have no common factors other than 1). This is the most crucial step.

2. We are given that the product of the numbers is 6300.

   a × b = 6300
   (15x) × (15y) = 6300
   

3. Solve for the product x × y.

   225 × x × y = 6300
   x × y = 6300 ÷ 225
   x × y = 28
   

4. Now we need to find pairs of co-prime numbers (x, y) whose product is 28. Let's list all pairs that multiply to 28.

   Pair (x, y)	Are they co-prime?
   (1, 28)	Yes
   (2, 14)	No (common factor 2)
   (4, 7)	Yes

5. We found two co-prime pairs for (x, y): (1, 28) and (4, 7). Now, find the actual numbers (a, b) for each pair.

   • Pair 1: x=1, y=28 a = 15 × 1 = 15 b = 15 × 28 = 420 The pair is (15, 420).

   • Pair 2: x=4, y=7 a = 15 × 4 = 60 b = 15 × 7 = 105 The pair is (60, 105).

Final Answer: There are 2 possible pairs of numbers: (15, 420) and (60, 105).

Example 4: The HCF-LCM Relationship (Tricky)

Given: The LCM of two numbers is 12 times their HCF. The sum of the HCF and LCM is 403. One of the numbers is 93.

To Find: The other number.

Solution:

1. Let H be the HCF and L be the LCM. First, use the given relationships to find the actual values of H and L.

   Given: L = 12 × H
   Given: H + L = 403
   

2. Substitute the first equation into the second one to solve for H.

   H + (12 × H) = 403
   13 × H = 403
   H = 403 ÷ 13
   H = 31
   

3. Now that we have the HCF, find the LCM.

   L = 12 × H
   L = 12 × 31
   L = 372
   

4. We now have the HCF (31), the LCM (372), and one number (a = 93). We can use the product property to find the other number (b).

   H × L = a × b
   31 × 372 = 93 × b
   

5. Solve for b.

   b = (31 × 372) ÷ 93
   

   Notice that 93 = 31 × 3. This simplifies the calculation.

   b = (31 × 372) ÷ (31 × 3)
   b = 372 ÷ 3
   b = 124
   

Final Answer: The other number is 124.

Tips & Tricks

Use these shortcuts to solve problems faster in exams.

Common Mistakes

Many students make these specific errors. Study them carefully to avoid losing marks!

Brain-Teaser Questions

Test your understanding with these higher-order thinking problems.

1. Can two numbers have an HCF of 16 and an LCM of 380? Justify your answer.

   💡 Answer: No. A fundamental property is that the HCF of two numbers must always be a factor of their LCM. Here, we check if 380 is divisible by 16. 380 ÷ 16 = 23.75. Since 16 does not divide 380 exactly, such a pair of numbers cannot exist.

2. The HCF and LCM of two 2-digit numbers are 13 and 455 respectively. Find the numbers.

   💡 Answer: Let the numbers be 13x and 13y where x and y are co-prime. HCF × LCM = Product of numbers 13 × 455 = (13x) × (13y) 13 × 455 = 169 × x × y x × y = (13 × 455) / 169 = 455 / 13 = 35 The co-prime pairs (x, y) that multiply to 35 are (1, 35) and (5, 7). Case 1: (1, 35) → Numbers are 13×1=13 and 13×35=455. Here, 455 is not a 2-digit number, so this pair is rejected. Case 2: (5, 7) → Numbers are 13×5=65 and 13×7=91. Both are 2-digit numbers. The numbers are 65 and 91.

3. The sum of two numbers is 180 and their HCF is 12. How many pairs of such numbers exist?

   💡 Answer: Let the numbers be 12x and 12y, where x and y are co-prime. Sum = 12x + 12y = 180 12(x + y) = 180 x + y = 180 ÷ 12 = 15 We need to find pairs of co-prime numbers that add up to 15. (1, 14) - Co-prime (2, 13) - Co-prime (3, 12) - Not co-prime (factor of 3) (4, 11) - Co-prime (5, 10) - Not co-prime (factor of 5) (6, 9) - Not co-prime (factor of 3) (7, 8) - Co-prime The possible pairs for (x,y) are (1,14), (2,13), (4,11), and (7,8). Therefore, there are 4 such pairs of numbers.

Mini Cheatsheet

Screenshot this table for your last-minute revision before an exam!

Try It Yourself

1. The HCF of two numbers is 11 and their LCM is 7700. If one of the numbers is 275, find the other number.
2. Can two numbers have 15 as their HCF and 175 as their LCM? Give a reason for your answer.
3. The product of two co-prime numbers is 117. What is their LCM?

Answer Key: 1. 308 | 2. No, because 175 is not divisible by 15. | 3. 117