NCERT Solutions and Notes for Class 6 Maths Chapter 3: Playing With Numbers – Download Free PDF

In this chapter, students will understand the concept of numbers in different mathematical operations. The students will get to know about the factors and multiples, prime and composite numbers, divisibility of numbers, prime factorization, and more in an interactive manner. Read through for CBSE Class 6 NCERT Maths Chapter 3 Playing With Numbers Notes, Exercise, and Important Questions.

CBSE Class 6 NCERT Maths Chapter 3 Playing With Numbers Notes Free PDF Download

Explore all the Chapters of Class 6 Mathematics:-

Chapter 1	Chapter 2	Chapter 3	Chapter 4
Chapter 5	Chapter 6	Chapter 7	Chapter 8
Chapter 9	Chapter 10	Chapter 11	Chapter 12

CBSE Class 6 NCERT Maths Chapter 3 Playing With Numbers Notes – Free PDF Download

Below we have given topic-wise notes for the CBSE Class 6 NCERT Maths Chapter 3 Playing With Numbers. We have also provided a downloadable free PDF at the end of these notes so you can download and take a printout to study later when you need quick revision before going to the exam hall. 

Topic 1: Factors and Multiples

• Factor: A factor is an exact divisor of a number. A factor is either lesser than or equal to the given number. 

For eg, 4 × 1 = 4, 2 × 2 = 4

Here, 1, 2 and 4 are the exact divisors of 4. 

• Multiple: A number is a multiple of each of its factors. In the above case, 4 is a multiple of 1, 2 and 4. 

Key Takeaways: 

1. A number has finite factors only. 
2. 1 is a factor of every number. 
3. Every number is a factor of itself too (as given above, 1 × 4 = 4; 1 and 4 are factors of 4).
4. Every multiple of a number is either greater than or equal to the given number. 

For eg, let’s find the multiples of 7

7 × 1 = 7

7 × 2 = 14 

7 × 3 = 21 

7 × 4 = 28; and so on.

Here, 7, 14, 21, and 28 are the multiples of 7 and all of these numbers are either equal or greater than 7. 

5. Multiples of a number are infinite. 
6. Every number is a multiple of itself.

For eg, 7 × 1 = 7; so, as we can see, 7 is a factor as well as the multiple of itself. 

Topic 2: Prime and Composite Numbers

• Prime Numbers: The numbers that have only two factors, i.e. 1 and themselves. 

For eg, 2, 3, 5, 7, 11,…..

The above numbers can only be divided either by themselves or by 1. Hence, they have only two factors. These are called prime numbers.   

• Composite Numbers: Unlike the prime numbers, the composite numbers have more than two factors. 

For eg, 1 × 4 = 4, 2 × 2 = 4; so, 4 has three factors, i.e. 1, 2, and 4. 

So, 4 is a composite number. 

 Exception: 1 is neither a prime number nor a composite number. 

Also Read: Different Branches of Mathematics

Topic 3: Even and Odd Numbers

• Even Numbers:  A number that is divisible by 2 is called an even number. Or, we can say – a number that has 0, 2, 4, 6, and 8 at the one’s place (or we can say a number ending in 0, 2, 4, 6, and 8) is an even number.

For eg, 2560, 6158, 3336, 4952

We see that the above numbers end in 0, 8, 6, and 2. This means that these numbers are divisible by 2. So, the given numbers are even numbers.

• Odd Numbers: All the numbers that are not divisible by 2 are called odd numbers. Or we can say the numbers that do not end with 0, 2, 4, 6 or 8 are called the odd numbers. 

For eg, 3531, 4573, 6697, 1599

The above numbers are not divisible by 2. So, the above numbers are the odd numbers. 

Key Takeaways:

1. 2 is the smallest even prime number. 
2. 3 is the smallest odd prime number.  
3. All the prime numbers other than 2 are odd numbers. 

Topic 4: Tests for Divisibility of Numbers

Let us understand some rules by which we will understand which number can be easily divisible by another given number. To check for the divisibility of a number, let us understand certain rules.

• Divisibility by 10: A number that ends in 0 or we can say – a number that has 0 at its one’s place is divisible by 10. 

For eg, 100, 250, 350, 460, 1000 – all these numbers end with 0 and so, they are divisible by 10. 

• Divisibility by 5: A number that ends with either 0 or 5 is divisible by 5. 

For eg, 555, 4535, 1000, 10, 250 – all these numbers end with 0 and 5 and so, they are divisible by 5. 

• Divisibility by 2: As we have learned above, any number that ends with 0, 2, 4, 6, or 8 is divisible by 2. 

For eg, 222, 2458, 3628, and 2528 – these all numbers are divisible by 2. 

• Divisibility by 3: The rule of divisibility of numbers by 3 is a bit different than the rest. In this, we check if the sum of the digits of a given number is a multiple of 3 then, the number is divisible by 3. 

For eg, 333, 54, 21.

Now, let us add the digits in the above numbers:   

333: 3 + 3 + 3 = 9; 9 is a multiple of 3. Hence, 333 is divisible by 3.

54: 5 + 4 = 9; 9 is a multiple of 3. Hence, 54 is divisible by 3.

21: 2 + 1 = 3; 3 is a multiple of 3. Hence, 21 is divisible by 3.

• Divisibility by 6: A number that is divisible by both 2 and 3, will be divisible by 6 as well. 

For eg, 6, 12, 27

6: we see that 6 is divisible by both 2 and 3. So, 6 is divisible by 6.

12: 1 + 2 = 3, the number 12 ends with 2 and the sum of its digits is 3. So the number is divisible by both 2 and 3. Hence, 12 is divisible by 6.

27: 2 + 7 = 9, we see that the number 27 is divisible by 3 but not by 2. Hence, 27 is not divisible by 6.

• Divisibility by 4: The divisibility of 1 and 2-digit numbers by 4 must be checked by the actual division. There is no trick to determine whether a single or double-digit number is divisible by 4 other than dividing the number by 4 itself. 

However, the divisibility of numbers with 3 or more than 3 digits can be checked by observing the last two digits in the number. If the number formed by the last two digits of a given number is divisible by 4, then, the whole number will be divisible by 4.

For eg, 212, 336, 528

In 212, the last 2 digits make the number: 12; as 12 is divisible by 4. We assume that the whole number 212 is divisible by 4.

Similarly, in 336 and 528, the numbers formed by the last two digits are 36 and 28. We know that both 36 and 28 are divisible by 4. Hence, the numbers 336 and 528 are also divisible by 4. 

• Divisibility by 8: Just like in the case of divisibility by 4, the divisibility of 1, 2 and 3-digit numbers by 8 must be checked by the actual division. 

However, 4 or more digit numbers must be checked by their last 3 digits, i.e. the digits at their ones, tens and thousands places. 

If the number formed by the last 3 digits of a 4 or more digits number is divisible by 8, then, the whole number will be divisible by 8. 

For eg, 1000, 2104, 1416

The numbers formed by the last 3 digits of each of the given numbers are 000, 104 and 416. Each of these numbers is divisible by 8 (checked by actual division). Hence the numbers 1000, 2104 and 1416 are divisible by 8. 

• Divisibility by 9: If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.

For eg, 36, 45, 54

The sum of all the above numbers is 9, we know that 9 is divisible by 9. So, all the above numbers are divisible by 9.

• Divisibility by 11: To check the divisibility of a number by 11, the rule is, to find the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number. If the difference is either 0 or divisible by 11, then the number is divisible by 11.

For eg, in 5081, the digits at the odd places from the right are 1 and 0 and the digits at the even place from the right are 8 and 5. 

Hence, their difference: (5 + 8) – (0 + 1) = 13 – 1 = 12

We know that 12 is not divisible by 11. So, the number 5081 is not divisible by 11.

Also Read: Maths Formulas for Class 10

Topic 5: Common Factors and Common Multiples 

• Common Factors: When two different numbers have more than 1 number as their common factors.

For eg, 8 and 10

Factors of 8 ⇒ 1 × 8, 2 × 4. So, 1, 2, 4 and 8 are the factors of 8.

Factors of 10 ⇒ 1 × 10, 2 × 5. So, 1, 2, 5 and 10 are the factors of 10. 

So, here, 1 and 2 are the common factors of 8 and 10. 

• Co-prime Numbers: Two numbers having only 1 as a common factor are called co-prime numbers.

For eg, 4 and 15.

Factors of 4: 1 × 4, 2 × 4

Factors of 15: 1 × 15, 3 × 5

So, the numbers 4 and 15 have only 1 as a common factor among them. So the numbers 4 and 15 are the co-prime numbers. 

• Common Multiples: When two numbers have more than 1 multiples in common. 

For eg, the multiples of 2 are 2, 4, 6, 8, 10, 12,……

The multiples of 4 are 4, 8, 12, 16,…

 We can see that the numbers 2 and 4 have 4, 8, 12 and so on as their common multiples.  

Key Takeaways:

1. If a number is divisible by another number then it is divisible by each of the factors of that number.
2. If a number is divisible by two co-prime numbers then it is divisible by their product also.
3. If two given numbers are divisible by a number, then their sum is also divisible by that number.
4. If two given numbers are divisible by a number, then their difference is also divisible by that number.

Topic 6: Prime Factorisation

• Prime Factors: When a number is expressed as a product of its factors we say that the number has been factorized. Thus, when we write 24 = 3×8, we say that 24 has been factorized. This is only one of the factorizations of 24. The others are:

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

In all the above factorisations of 24, we ultimately arrive at only one factorisation 2 × 2 × 2 × 3. In this factorization, the only factors 2 and 3 are prime numbers. Such a factorization of a number is called a prime factorization.

In the above case, the prime factorization of 24 is 2 × 2 × 2 × 3. i.e. the only prime

factorisation of 24.

Also Read: Maths Books for Competitive Exams

Topic 7: Highest Common Factor

• Highest Common Factors (HCF): The Highest Common Factor (HCF) of two or more given numbers is the highest (or greatest) of their common factors. It is also known as the Greatest Common Divisor (GCD).

For eg, 20, 28, 36

The HCF of 20, 28 and 36 can also be found by prime factorization of these

numbers as follows:

Thus, 20 = 2, 2, 5

28 = 2, 2, 7

36 = 2, 2, 3, 3

The common factor of 20, 28 and 36 is 2 (occurring twice). Thus, the HCF of 20, 28 and 36 is 2 × 2 = 4.

Topic 8: Lowest Common Multiple

• Lowest Common Multiple (LCM): The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples.

For eg, 12 and 18

As we calculated the HCF of two numbers separately by finding their prime factors, we will find the prime factors of the given numbers together in one table to find out their Lowest Common Multiple. 

Hence, the lowest common multiple of 12 and 18 comes out to be ⇒ 2 × 2 × 3 × 3 = 36

CBSE Class 6 NCERT Maths Chapter 3 Playing With Numbers Notes Free PDF Download

Explore all the Chapters of Class 6 Mathematics:-

Chapter 1	Chapter 2	Chapter 3	Chapter 4
Chapter 5	Chapter 6	Chapter 7	Chapter 8
Chapter 9	Chapter 10	Chapter 11	Chapter 12

Important Questions in NCERT Class 6 Maths Chapter 3: Free PDF Download

Below we have provided some important exercise questions and their solutions from Chapter 3 of CBSE NCERT Maths for Class 6. 

Exercise 3.1

Q 1. Write all the factors of the following numbers:

1. 24
2. 15
3. 21

Ans. The factors of the given numbers are:

1. 24 = 1 × 24, 2 × 12, 3 × 8, 4 × 6; so all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. 
2. 15 = 1 × 15, 3 × 5; so all the factors of 15 are 1, 3, 5 and 15. 
3. 21 = 1 × 21, 3 × 7; so all the factors of 21 are 1, 3, 7 and 21. 

Q 2. Write the first five multiples of 5. 

Ans. The first 5 multiples of 5 are 5, 10, 15, 20, and 25.  

Q 3. Find all the multiples of 9 up to 100.

Ans. The multiples of 9 up to 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, and 99.

Exercise 3.2

Q 1. What is the sum of any two 

1. Odd numbers? 
2. Even numbers?

Ans. Let us solve the given question by taking 2-2 examples of each.

1. Let us first take two same odd numbers, say 1. 

The sum, 1 + 1 = 2

Now, let’s take two different numbers, 5 and 7.

The sum, 5 + 7 = 12

So, we see that the sum of odd numbers results in an even number. 

2. Let us first take two same odd numbers, say 2. 

The sum, 2 + 2 = 4

Now, let’s take two different numbers, 4 and 6.

The sum, 4 + 6 = 10

So, we see that the sum of two even numbers is also an even number. 

Q 2. State whether the following statements are True or False:

1. The sum of three odd numbers is even.
2. The sum of two odd numbers and one even number is even.
3. The product of three odd numbers is odd.
4. All prime numbers are odd.
5. Prime numbers do not have any factors.
6. The sum of two prime numbers is always even.
7. All even numbers are composite numbers.

Ans. Let us see each statement below.

1. Let us add three randomly picked odd numbers: 3, 5 and 7.

Now, 3 + 5 + 7 = 15

Let’s take another example, 7, 9, 15

So, 7 + 9 + 15 = 31

Hence, we see that the sum of 3 odd numbers results in an odd number. So, the given statement is False.

2. We know that the sum of two odd numbers gives an even number. Now, when we add an even number with another even number, the result is again an even number. So, the given statement is True.
3. Let us solve this by taking an example: 3, 3, 5

The product of three numbers is ⇒ 3 × 3 × 5 = 45

Let’s take another example: 3, 5, 7

The product of three numbers is ⇒ 3 × 5 × 7 = 105

So, the given statement is True. 

4. False, 2 is an even prime number.
5. False, all numbers have at least two factors, i.e. 1 and the number itself.
6. False, the sum of 2 with any other prime number comes out to be an odd number.
7. False, 2 is an even number but has only two factors. So, it is not a composite number.

Q 3. Express each of the following numbers as the sum of three odd primes:

1. 21 
2. 31 
3. 53 

Ans. The 3 odd prime numbers of each are given below:

1. 3 + 5 + 13 = 21
2. 3 + 5 + 23 = 31
3. 13 + 17 + 23 = 53  

Q 4. 12. Fill in the blanks : 

1. A number which has only two factors is called a ______. 
2. A number which has more than two factors is called a ______. 
3. 1 is neither ______ nor ______. 
4. The smallest prime number is ______. 
5. The smallest composite number is _____. 
6. The smallest even number is ______.

Ans. The answers are given below.

1. Prime Number
2. Composite Number
3. Prime Number, Composite Number
4. 2
5. 4
6. 2

Also Read: Word Problems in Arithmetic Operation

Exercise 3.3

Q 1. Using divisibility tests, determine which of the following numbers are divisible by

4; by 8:

1. 572 
2. 726352 

Ans. For divisibility by 4, we check whether the last 2 digits of the given number are divisible by 4. Similarly, for the divisibility of 4-digit numbers by 8, we check if the last 3 digits are divisible by 8. 

1. 572: for divisibility by 4, we check if 72 is divisible by 4. Since 72 is divisible by 4, 572 is also divisible by 4.

However, to check the divisibility by 8, we check by direct division. So, 572 is not divisible by 8.

2. 726352: for divisibility by 4, we check if 52 is divisible by 4. Since 52 is divisible by 4, so, 726352 is divisible by 4.

Now, for divisibility by 8, we check if 352 is divisible by 8. Since 352 is divisible by 8. So, 726352 is also divisible by 8.

Q 2. Using divisibility tests, determine which of the following numbers are divisible by 11:

1. 7138965 
2. 70169308 
3. 10000001

Ans. The solutions for each of the given numbers are:

1. The digits at the odd places from the right are 5, 9, 3, and 7 and the digits at the even place from the right are 6, 8, and 1. 

Hence, their difference: (7 + 3 + 9 + 5) – (1 + 8 + 6) = 24 – 15 = 9

We know that 9 is not divisible by 11. So, the number 7138965 is not divisible by 11.

2. The digits at the odd places from the right are 8, 3, 6, and 0 and the digits at the even place from the right are 0, 9, 1, and 7. 

Hence, their difference: (7 + 1 + 9 + 0) – (3 + 8 + 6 + 0) = 17 – 17 = 0

We know that 0 is divisible by 11. So, the number 70169308 is divisible by 11.

3. The sum of digits at odd places = 1

The sum of digits at even places = 1

Difference = 1 – 1 = 0

We know that 0 is divisible by 11. So, the number 10000001 is divisible by 11.

Exercise 3.4

Q 1. Find the common factors of :

1. 20 and 28 
2. 15 and 25

Ans. The common factors are

1. The factors of 20 are 1, 2, 4, 5, 10 and 20.

And the factors of 28 are 1, 2, 4, 7, 14, and 28. 

So, the common factors are 1, 2, and 4.

2. The factors of 15 are 1, 3, 5 and 15.

And the factors of 28 are 1, 5, and 25. 

So, the common factors are 1 and 5.

Q 2.  Find the first three common multiples of 6 and 8.

Ans. The multiples of 6 are 6, 12, 18, 24, 30,… and that of 8 are 8, 16, 24, 32,…

So, the first three common multiples of 6 and 8 are 24, 48, and 72.

Q 3. A number is divisible by 12. By what other numbers will that number be divisible?

Ans. As per the rules of divisibility, if a number is divisible by another number then it is divisible by each of the factors of that number. So, if a number is divisible by 12, it will also be divisible by each of its factors. So, the number will also be divisible by 1, 2, 3, 4, and 6.

Q 4. A number is divisible by both 5 and 12. By which other number will that number be

always divisible?

Ans. We know that if a number is divisible by two co-prime numbers then it is divisible by their product also.

Now, 5 and 12 are two co-prime numbers. So, a number is divisible by both 5 and 12 will also be divisible by 60, and the factors of 60, i.e. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60. 

Also Read: Maths Tricks to Ace GMAT Quant Section

Exercise 3.5

Q 1. Which of the following statements is true?

1. If a number is divisible by 3, it must be divisible by 9.
2. If a number is divisible by 9, it must be divisible by 3.
3. If a number is divisible by 9 and 10, then it must be divisible by 90.
4. If two numbers are co-primes, at least one of them must be prime.
5. If a number exactly divides two numbers separately, it must exactly divide their sum.
6. If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

Ans. The correct answers for the given statements are given below.

1. False, 6 is divisible by 3 but not by 9. 
2. True, as 9 = 3 × 3. Hence, if a number is divisible by 9, it will also be divisible by 3.
3. True, (read key takeaways properties in the common factors topic)  
4. False, for eg, both 15 and 32 are co-prime and composite numbers.
5. True, for eg, 2 divides 4 and 8, but it also divides their sum, i.e. 12 (4 + 8 = 12).
6. False, since 2 divides 12 but it does not divide 7 and 5.

Q 2. Which factors are not included in the prime factorization of a composite number?

Ans. The number itself and 1 are not included in the prime factorization of a composite number.

Q 3. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Ans. Let us check this by the following examples.

1. 2 × 3 × 4 = 24 which is divisible by 6.
2. 5 × 6 × 7 = 210 which is divisible by 6.

Hence, the given statement is verified. 

Exercise 3.6

Q 1.  Find the HCF of 18 and 48.

Ans. Let us find the prime factors of each of the given numbers:

Hence, the common prime factors among 18 and 48 are 2 and 3. So, their HCF is 2 × 3 = 6.

Q 2. What is the HCF of two consecutive numbers?

Ans. The HCF of two consecutive numbers is 1. For eg, the HCF of 2 and 3 is 1.  

Exercise 3.7

Q 1. Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer an exact number of times.

Ans. To calculate the maximum weight that can measure the weights of fertilisers the exact number of times = HCF of the two given weights

Hence, HCF of 75 and 69:

The highest common factor between the two numbers is 3. So, the maximum value of weight which can measure the weight of the fertilizer an exact number of times is 3 kg.

Q 2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm, and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Ans. The steps of the three boys measure 63, 70, and 77 cm respectively. So, to calculate the minimum distance that each boy should cover so that they all cover the distance in complete steps, we must calculate the LCM.

The LCM comes out to be ⇒ 2 × 3 × 3 × 5 × 7 × 11 = 6930

Hence, each boy should cover 6930 cm so that they complete the complete distance in complete steps. 

Q 3. The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm respectively. Find the longest tape that can measure the three dimensions of the room exactly.

Ans. To find the longest tape that can measure the length of three dimensions of the room exactly can be calculated by finding the HCF. 

So, the HCF comes out to be = 3 × 5 × 5 = 75 cm.

Therefore, the longest tape that can measure the three dimensions of the room should be 75 cm. 

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