# NCERT Solutions for Class 6 Maths Chapter 3

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.1, Exercise 3.2, Exercise 3.3, Exercise 3.4, Exercise 3.5, Exercise 3.6 and Exercise 3.7 in English and Hindi Medium for session 2020-2021.

Solutions of Prashnavali 3.1, Prashnavali 3.2, Prashnavali 3.3, Prashnavali 3.4, Prashnavali 3.5, Prashnavali 3.6 and Prashnavali 3.7 in Hindi Medium in PDF file format as well as study online without downloading. Video Format Solution with explanation of 6 Maths Chapter 3 Playing With Numbers is also available in English Medium free to download. Download Class 6 Maths App for Offline use.

## NCERT Solutions for Class 6 Maths Chapter 3

 Class: 6 Maths (English and Hindi Medium) Chapter 3: Playing With Numbers

NCERT Solutions for Class 6 Maths Chapter 3 in Hindi Medium and English Medium are given below, free to use. All the solutions are updated for new academic session 2020-2021 based on latest NCERT Books 2020-21.

• ### Class 6 Maths Chapter 3 Solutions in Hindi Medium

#### Class 6 Maths Chapter 3 Exercise 3.1 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.1 Explanation
Class 6 Maths Chapter 3 Exercise 3.1 Solutions

#### Class 6 Maths Chapter 3 Exercise 3.2 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.2 Explanation
Class 6 Maths Chapter 3 Exercise 3.2 Solutions

#### Class 6 Maths Chapter 3 Exercise 3.3 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.3 Explanation
Class 6 Maths Chapter 3 Exercise 3.3 Solutions

#### Class 6 Maths Chapter 3 Exercise 3.4 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.4 Explanation
Class 6 Maths Chapter 3 Exercise 3.4 Solutions

#### Class 6 Maths Chapter 3 Exercise 3.5 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.5 Explanation
Class 6 Maths Chapter 3 Exercise 3.5 Solutions

#### Class 6 Maths Chapter 3 Exercise 3.6 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.6 Explanation
Class 6 Maths Chapter 3 Exercise 3.6 Solutions

#### Class 6 Maths Chapter 3 Exercise 3.7 Solution and Explanation

Class 6 Maths Chapter 3 Exercise 3.7 Explanation
Class 6 Maths Chapter 3 Exercise 3.7 Solutions
##### About 6 Maths Chapter 3 Solutions

In 6 Maths Chapter 3 Playing With Numbers, we will study about the factors and multiples of numbers. A factor of a number is an exact divisor of that number. The following the some facts about a factor:
1 is a factor of every number.
Every number is a factor of itself.
Every factor of a number is an exact divisor of that number.
Every factor is less than or equal to the given number.
Number of factors of a given number are finite.
Following facts are related to a multiple of a number:
Every multiple of a number is greater than or equal to that number.
The number of multiples of a given number is infinite.
Every number is a multiple of itself.
Perfect Number: A number for which sum of all its factors is equal to twice the number is called a perfect number. The numbers 6 are perfect numbers as 6 = 1 + 2 + 3 and 28 = 1 + 2 + 4 + 7 + 14.

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### Important Questions on 6 Maths Chapter 3

Write all the factors of 24.
24
= 1 x 24
= 2 x 12
= 3 x 8
= 4 x 6
= 6 x 4

Factors of 24 = 1, 2, 3, 4, 6, 12, 24

निम्न के सार्व गुणनखंड ज्ञात कीजिए: 20 और 28
20 के गुणनखंड = 1, 2, 4, 5, 10, 20
28 के गुणनखंड = 1, 2, 4, 7, 14, 28

सार्व गुणनखंड = 1, 2, 4

Write first five multiples of 5.
5 x 1 = 5,
5 x 2 = 10,
5 x 3 = 15,
5 x 4 = 20,
5 x 5 = 25

First five multiples of 5 are 5, 10, 15, 20, 25.

Find all the multiples of 9 up to 100.
Multiples of 9 up to 100 are:

9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

बताइए कि किन्हीं दो संख्याओं का योग सम होता है या विषम होता है, यदि वे दोनों विषम संख्याएँ हों?
किन्हीं दो विषम संख्याओं का योग सम होता है।

उदाहरण:
1 + 3 = 4,
3 + 5 = 8

संख्या 13 और 31 अभाज्य संख्याएँ हैं। इन दोनों संख्याओं में दो अंक 1 और 3 हैं। 100 तक की संख्याओं में ऐसे अन्य सभी युग्म ज्ञात कीजिए।
17 और 71;
37 और 73;
79 और 97
Using divisibility test, determine whether 5445 is divisible by 11?
5445
Sum of the digits at odd places = 4 + 5 = 9
Sum of the digits at even places = 4 + 5 = 9
Difference of both sums = 9 – 9 = 0

Since the difference is 0, therefore, the number is divisible by 11.

Which factors are not included in the prime factorization of a composite number?
1 is the factor which is not included in the prime factorization of a composite number.
20 से छोटी सभी अभाज्य और भाज्य संख्याएँ अलग-अलग लिखिए।
अभाज्य संख्याएँ:
2, 3, 5, 7, 11, 13, 17, 19

भाज्य संख्याएँ:
4, 6, 8, 9, 10, 12, 14, 15, 16, 18

1 और 10 के बीच में सबसे बड़ी अभाज्य संख्या लिखिए?
1 और 10 के बीच में सबसे बड़ी अभाज्य संख्या ‘7’ है।
Write the greatest 4-digit number and express it in terms of its prime factors.
The greatest 4-digit number = 9999

The prime factors of 9999 are 3 × 3 × 11 × 101.

Write the smallest 5-digit number and express it in terms of its prime factors.
The smallest five digit number is 10000.
The prime factors of 10000 are 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5.
The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.

Example:
(i) 2 x 3 x 4 = 24
(ii) 4 x 5 x 6 = 120

अभाज्य संख्याओं के ऐसे तीन युग्म लिखिए जिनका अंतर 2 हो।
3 और 5;
5 और 7;
11 और 13
100 से छोटी सात क्रमागत भाज्य संख्याएँ लिखिए जिनके बीच में कोई अभाज्य संख्या नहीं हो।
100 से छोटी सात क्रमागत भाज्य संख्याएँ: 90, 91, 92, 93, 94, 95, 96
20 से छोटी अभाज्य संख्याओं के ऐसे पाँच युग्म लिखिए जिनका योग 5 से विभाज्य हो।
2 + 3 = 5;
7 + 13 = 20;
3 + 17 = 20;
2 + 13 = 15;
5 + 5 = 10
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 exact number of times.
For finding maximum weight, we have to find H.C.F. of 75 and 69.

Factors of 75 = 3 x 5 x 5
Factors of 69 = 3 x 69
H.C.F. = 3

Therefore the required weight is 3 kg.

The sum of two consecutive odd numbers is always divisible by 4. Verify this statement with the help of some examples.
3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.
I am the smallest number, having four different prime factors. Can you find me?
The smallest four prime numbers are 2, 3, 5 and 7.
Hence, the required number is 2 x 3 x 5 x 7 = 210
The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
The measurement of longest tape = H.C.F. of 825 cm, 675 cm and 450 cm.
Factors of 825 = 3 x 5 x 5 x 11
Factors of 675 = 3 x 5 x 5 x 3 x 3
Factors of 450 = 2 x 3 x 3 x 5 x 5
H.C.F. = 3 x 5 x 5 = 75 cm

Therefore, the longest tape is 75 cm.                                     