Playing with Numbers

Playing with Numbers: CBSE Class 6 Maths Notes - Complete Guide 2024-25

Introduction to Playing with Numbers

The Playing with Numbers chapter is one of the most fundamental and engaging topics in CBSE Class 6 Mathematics. This chapter builds the foundation for understanding how numbers work, their relationships, and their properties. From finding factors and multiples to determining the HCF and LCM of numbers, this chapter equips students with essential mathematical skills that are used throughout their academic journey.

Numbers can be expressed as products of different numbers in various ways.

For example:

• 6 = 1 × 6 = 2 × 3
• 12 = 1 × 12 = 2 × 6 = 3 × 4

Understanding these relationships helps us explore fascinating number properties and solve real-world problems efficiently.

Important Concepts Covered in This Chapter

1. Factors

A factor of a number is an exact divisor of that number. When a number divides another number completely without leaving any remainder, it is called a factor.

Example: Finding factors of 24

• 24 = 1 × 24
• 24 = 2 × 12
• 24 = 3 × 8
• 24 = 4 × 6

Therefore, the factors of 24 are: 1, 2, 3, 4, 6, 8, 12, and 24

Important Properties of Factors:

• 1 is a factor of every number
• Every number is a factor of itself
• The number of factors of any number is finite
• The smallest factor of any number is 1
• The largest factor of any number is the number itself

2. Multiples

A multiple of a number is obtained by multiplying that number by any whole number. Each number is a multiple of its factors.

Example: Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21...

Important Properties of Multiples:

• Every number is a multiple of itself
• Every multiple of a number is greater than or equal to the number
• The number of multiples of any number is infinite
• The smallest multiple of a number is the number itself

3. Perfect Numbers

A perfect number is a number whose sum of all factors (excluding the number itself) equals the number.

Example: 6 is a perfect number

• Factors of 6: 1, 2, 3, 6
• Sum of factors (excluding 6): 1 + 2 + 3 = 6 ✓

Example: 28 is a perfect number

• Factors of 28: 1, 2, 4, 7, 14, 28
• Sum of factors (excluding 28): 1 + 2 + 4 + 7 + 14 = 28 ✓

4. Prime Numbers

Prime numbers are numbers that have exactly two factors: 1 and the number itself.

Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47...

Facts About Prime Numbers:

• 2 is the smallest prime number
• 2 is the only even prime number
• 1 is neither prime nor composite

5. Composite Numbers

Composite numbers are numbers that have more than two factors. All whole numbers greater than 1 that are not prime are composite.

Examples: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20...

Illustration: Separating prime and composite numbers from 2, 5, 6, 7, 17, 19, 20, 23, 33, 35, 39, 41

• Prime: 2, 5, 7, 17, 19, 23, 41
• Composite: 6, 20, 33, 35, 39

6. Even and Odd Numbers

Type	Definition	Examples
Even Numbers	Multiples of 2	2, 4, 6, 8, 10, 12...
Odd Numbers	Numbers not divisible by 2	1, 3, 5, 7, 9, 11...

7. Prime Factorization

Prime factorization is the process of expressing a number as a product of its prime factors.

Example: Prime factorization of 80

• 80 = 2 × 40
• 80 = 2 × 2 × 20
• 80 = 2 × 2 × 2 × 10
• 80 = 2 × 2 × 2 × 2 × 5
• 80 = 2⁴ × 5

More Examples:

• 36 = 2 × 2 × 3 × 3 = 2² × 3²
• 52 = 2 × 2 × 13 = 2² × 13
• 99 = 3 × 3 × 11 = 3² × 11
• 100 = 2 × 2 × 5 × 5 = 2² × 5²

8. Co-Prime Numbers

Two numbers are co-prime (or coprime) if they have only 1 as their common factor.

Example: Are 21 and 25 co-prime?

• Factors of 21: 1, 3, 7, 21
• Factors of 25: 1, 5, 25
• Common factor: 1 only
• Yes, 21 and 25 are co-prime.

Note: Co-prime numbers need not be prime numbers themselves.

Divisibility Rules Explained

Understanding divisibility rules helps quickly determine if a number is divisible by another without performing actual division.

Complete Divisibility Tests Table

Divisor	Rule	Examples
2	The ones digit is 0, 2, 4, 6, or 8	10, 12, 100, 248
3	Sum of digits is divisible by 3	39 (3+9=12), 63, 135
4	Last two digits form a number divisible by 4	96, 108, 224
5	Ones digit is 0 or 5	105, 255, 350, 675
6	Divisible by both 2 and 3	24, 48, 96, 108
8	Last three digits form a number divisible by 8	512, 9216, 1024
9	Sum of digits is divisible by 9	81, 927, 1089
10	Ones digit is 0	100, 210, 250
11	Difference of sum of digits at odd and even places is 0 or multiple of 11	2431, 4263556

Divisibility by 11 – Detailed Example

Is 4263556 divisible by 11?

Step 1: Identify digits at odd places (from right): 6, 5, 6, 4 → Sum = 21
Step 2: Identify digits at even places (from right): 5, 3, 2 → Sum = 10
Step 3: Find difference: 21 - 10 = 11
Step 4: Since 11 is divisible by 11, 4263556 is divisible by 11

Additional Divisibility Rules

1. If a number is divisible by another number, it is divisible by each factor of that number.
2. If a number is divisible by two co-prime numbers, it is divisible by their product.
3. If two numbers are divisible by a number, their sum and difference are also divisible by that number.

Highest Common Factor (HCF)

The Highest Common Factor (HCF), also known as Greatest Common Divisor (GCD), is the largest number that divides two or more numbers exactly.

Finding HCF by Prime Factorization

Example 1: Find HCF of 14, 20, and 50

• 14 = 2 × 7
• 20 = 2 × 2 × 5
• 50 = 2 × 5 × 5
• HCF = 2 (common factor with lowest power)

Example 2: Find HCF of 18, 24, and 36

• 18 = 2 × 3 × 3
• 24 = 2 × 2 × 2 × 3
• 36 = 2 × 2 × 3 × 3
• HCF = 2 × 3 = 6

Lowest Common Multiple (LCM)

The Lowest Common Multiple (LCM) is the smallest number that is a multiple of two or more given numbers.

Finding LCM by Prime Factorization

Example 1: Find LCM of 12 and 24

• 12 = 2 × 2 × 3
• 24 = 2 × 2 × 2 × 3
• LCM = 2 × 2 × 2 × 3 = 24 (take highest power of each prime)

Example 2: Find LCM of 24, 30, and 36

• 24 = 2³ × 3
• 30 = 2 × 3 × 5
• 36 = 2² × 3²
• LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Division Method for LCM

2 | 24, 30, 36
2 | 12, 15, 18
2 | 6, 15, 9
3 | 3, 15, 9
3 | 1, 5, 3
5 | 1, 5, 1
 1, 1, 1
LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360

Important Formula: Relationship Between HCF and LCM

For any two numbers a and b:

HCF(a, b) × LCM(a, b) = a × b

This formula is extremely useful for solving problems where either HCF or LCM is unknown.

Formulas

Formula/Concept	Mathematical Representation	Explanation
HCF-LCM Relationship	HCF(a,b) × LCM(a,b) = a × b	Product of HCF and LCM equals product of numbers
Number of Factors	If n = p^a × q^b × r^c, then factors = (a+1)(b+1)(c+1)	Formula to find total number of factors
Sum of Factors	Formula using prime factorization	Used for perfect number verification
LCM of Co-primes	LCM(a,b) = a × b (if HCF = 1)	For co-prime numbers, LCM is their product
Divisibility by 3	Sum of digits ÷ 3	Check if digit sum is multiple of 3
Divisibility by 9	Sum of digits ÷ 9	Check if digit sum is multiple of 9
Divisibility by 11	(Sum of odd place digits) - (Sum of even place digits) = 0 or 11k	Alternating sum rule

How to Solve Letter Substitution Puzzles Step by Step

Letter substitution puzzles (cryptarithmetic puzzles) are fascinating problems where letters represent digits. Here's a systematic approach:

Step-by-Step Method

Step 1: Identify Constraints

• Each letter represents a unique digit (0-9)
• The leading digit of any number cannot be 0
• Standard arithmetic rules apply

Step 2: Analyze Carry-Overs

• Look for columns that might produce carries
• Consider the maximum and minimum possible values

Step 3: Use Logical Deduction

• Start with columns that have the fewest possibilities
• Use divisibility rules where applicable
• Test systematic possibilities

Step 4: Verify Your Solution

• Substitute all values back into the puzzle
• Check that the arithmetic is correct
• Ensure no letter has the same digit as another

Sample Puzzle Solved

Puzzle: Find A and B if AB + BA = 121

Solution:

• AB represents a 2-digit number = 10A + B
• BA represents a 2-digit number = 10B + A
• AB + BA = (10A + B) + (10B + A) = 11A + 11B = 11(A + B)
• Given: 11(A + B) = 121
• Therefore: A + B = 11

Possible combinations where A + B = 11:

• A = 2, B = 9 → 29 + 92 = 121 ✓
• A = 3, B = 8 → 38 + 83 = 121 ✓
• A = 4, B = 7 → 47 + 74 = 121 ✓
• A = 5, B = 6 → 56 + 65 = 121 ✓

All these combinations are valid solutions!

Solved Problems from Playing with Numbers

Problem 1: Real-Life Application - Milk Measurement

Question: A milkman has two cans of 40 litres and 60 litres full of milk. Find the maximum capacity of the pot that can measure milk of both cans exactly.

Solution: Since we need the maximum capacity, we find the HCF of 40 and 60.

• 40 = 2 × 2 × 2 × 5
• 60 = 2 × 2 × 3 × 5
• HCF = 2 × 2 × 5 = 20 litres

Answer: A pot of 20 litres capacity is required.

Problem 2: Finding Numbers with Specific Remainders

Question: Find the least number which when divided by 14, 16, 18, and 24 leaves a remainder of 9 in each case.

Solution: Step 1: Find LCM of 14, 16, 18, and 24

2 | 14, 16, 18, 24
2 | 7, 8, 9, 12
2 | 7, 4, 9, 6
2 | 7, 2, 9, 3
3 | 7, 1, 9, 3
3 | 7, 1, 3, 1
7 | 7, 1, 1, 1
 1, 1, 1, 1

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 1008

Step 2: Add the required remainder Required number = LCM + Remainder = 1008 + 9 = 1017

Problem 3: Finding the Greatest Divisor

Question: Find the greatest number that divides 615 and 963, leaving a remainder of 6 in each case.

Solution: Step 1: Subtract the remainder from each number

• 615 - 6 = 609
• 963 - 6 = 957

Step 2: Find HCF of 609 and 957

• 609 = 3 × 7 × 29
• 957 = 3 × 11 × 29
• HCF = 3 × 29 = 87

Answer: The greatest number is 87.

Download NCERT Solutions for Playing with Numbers Exercise 16.1

For comprehensive practice and detailed solutions to all NCERT textbook exercises, students should refer to:

• Exercise 16.1: Basic problems on factors and multiples
• Exercise 16.2: Prime and composite numbers, prime factorization
• Exercise 16.3: Divisibility rules and applications
• Exercise 16.4: HCF and LCM problems
• Exercise 16.5: Word problems and real-life applications

Note: The NCERT textbook for Class 6 Mathematics includes this chapter with various exercises designed to reinforce the concepts progressively.

Watch Video Lessons for Better Understanding

Visual learning significantly enhances comprehension of mathematical concepts. For the Playing with Numbers chapter, video lessons typically cover:

1. Introduction to Factors and Multiples – Understanding the basic relationship
2. Prime Factorization Methods – Factor tree and division methods explained visually
3. Divisibility Rules – Quick tricks and memory techniques
4. HCF and LCM Calculation – Step-by-step demonstration
5. Problem-Solving Techniques – Application-based questions

Practice Problems

Subjective Questions

1. Write all prime factors of 104, 115, and 225.
2. Find the HCF of 396 and 1080 by prime factorization method.
3. The HCF of two numbers is 16 and their product is 3072. Find their LCM.
4. Find the greatest number which divides 615 and 963, leaving remainder 6 in each case.
5. Find the least number divisible by each of 15, 20, 24, 32, and 36.
6. If a and b are coprimes, find their LCM.
7. The length, breadth, and height of a box are 60cm, 55cm, and 40cm. Find the longest tape that can measure the three dimensions exactly.
8. The HCF of two numbers is 2 and LCM is 84. If one number is 14, find the other number.
9. The circumferences of four wheels are 50cm, 60cm, 75cm, and 100cm. What is the least distance they should cover so each wheel makes complete revolutions?
10. Find the greatest number that divides 445, 572, and 599 leaving remainders 4, 5, and 6 respectively.

Objective Questions

1. Which of the following numbers does not have more than two factors?

(a) 91

(b) 95

(c) 113

(d) 105

Answer: (c) 113

2. Sum of an even and odd number is:

(a) always even

(b) always odd

(c) may be even or odd

(d) cannot say

Answer: (b) always odd

3. A number divisible by 63 is also:

(a) divisible by 3 and 4

(b) divisible by 3 and 5

(c) divisible by 3 and 7

(d) divisible by 9 and 12

Answer: (c) divisible by 3 and 7

4. The largest even prime number is:

(a) 2

(b) 1024

(c) 2 × 10²⁴

(d) Cannot be determined

Answer: (a) 2

5. A number divisible by 11, 13, and 17 is also divisible by:

(a) 2413

(b) 2143

(c) 2431

(d) 2341

Answer: (c) 2431 (since 11 × 13 × 17 = 2431)

Conclusion

The Playing with Numbers chapter forms the cornerstone of number theory in CBSE Class 6 Mathematics. Mastering factors, multiples, prime numbers, divisibility rules, HCF, and LCM provides students with powerful tools for solving complex mathematical problems. These concepts not only appear in higher classes but also have practical applications in everyday life, from scheduling to resource allocation.

Regular practice of the exercises provided, along with understanding the underlying concepts, will help students excel in this chapter and build a strong mathematical foundation for future learning.