# NCERT Solutions for Class 6 Mathematics Chapter 3

The NCERT solutions for Class 6 Mathematics Chapter 3 ‘Playing with Numbers’ are helpful for test preparation. Mathematics develops logical reasoning, abstract thinking, and imagination among students.The subject matter experts have prepared these NCERT solutions to help students get a better understanding of the topic. Multiples, divisors, factors,  how to recognize factors and multiples and HCF and LCM, are covered in this chapter.

## NCERT Solutions for Class 6 Mathematics Chapter 3 - Playing with Numbers

When students enter Class 6, they are introduced to the NCERT solutions for Class 6 Mathematics Chapter 3 syllabus, which is quite diverse and covers a wide range of topics. Playing with Numbers, the third chapter of the Mathematics syllabus focuses on multiples and divisors. Students are introduced to subjects such as common factors and multiples, divisibility laws, best common factors, lowest common factors, and so on. Students will be able to understand prime and composite numbers and their differences, provided they have a thorough understanding of the chapter.

### NCERT Solutions for Class 6 Mathematics Chapter 3 - Playing with Numbers

The content of the NCERT Solutions for Class 6 Mathematics Chapter 3 - Playing with Numbers is made engaging and appealing to make learning enjoyable with examples, graphics to make it colourful and interesting.

NCERT Class 6 Mathematics Chapter 3

Students should study for their examinations with the help of NCERT solutions for Class 6 Mathematics Chapter 3. Various solved examples are provided to assist students in understanding how to calculate these sums. The solutions are available on the official website of Extramarks. The formulas studied and implemented in this chapter are primary equations that will help  to  answer  difficult equations later on.

NCERT Class 6 Mathematics Chapter 3 Exercises

NCERT solutions for Mathematics Class 6 from Extramarks can be accessed easily. The NCERT Class 6 Mathematics Chapter 3 in text exercises as well as exercises at the end of each unit in NCERT textbooks are designed to assist students in answering difficult  problems.

### Benefits of NCERT Class 6 Mathematics Chapter 3

Here are some of the advantages of using Extramarks NCERT solutions:

• The solutions have been written in the examination format to aid students in achieving high scores.
• All of the questions and answers are written in simple language and very well explained with examples.
• Every problem has a crisp and to-the-point response which students can easily comprehend.
• All the solutions help students to get the hang of basic Mathematics such as prime numbers, prime factors to solve HCF and LCM problems which form the base in higher classes.

Students can now strive for excellence in their exams with the help of Extramarks Class 6 Mathematics Chapter 3 solutions.

3.1 – Introduction

The objective is to give students a basic understanding of division and components. The rest of the study is centred  around this section.

3.2 – Factors and Multiples

The first set of tasks in NCERT answers for Class 6 Mathematics Chapter 3 – Playing with Numbers encourages students to recognize multiples and factors. According to the rule, every integer is also a multiple of itself.

3.3 – Prime and Composite Numbers

This portion of the textbook determines which numbers are prime and composite. The segment also explains why 2 is the smallest prime number and why one cannot be a composite and a prime number.

3.4 – Tests for Divisibility of Numbers

This section of Chapter 3 teaches students which integers are divisible by 10, 5, 2, 3, 6, 4, 8, 9, and 11. Find out how this divisibility test can help you.

3.5 – Common Factors and Common Multiples

Mastering the subject of numbers necessitates a thorough understanding of frequent multiples and factors. This exercise  shows how some numbers are multiples and factors of others.

3.6 – Some More Divisibility Rules

To obtain a thorough understanding of factors and multiples, the learner will need to know some divisibility rules. For an in-depth knowledge of the topic, consult the NCERT solutions Chapter 3.

3.7 – Prime Factorization

The prime factors are discussed in this section  of NCERT Class 6 Mathematics Chapter 3 and how they are multiplied to form the original number. In this segment,  students will learn   how to factorise prime  numbers.

3.8 – Highest Common Factor

Students must understand the basic concept about HCF or the highest common factor. It’s important from the exam point of view.

3.9 – Lowest Common Multiple

LCM (Lowest Common Multiple) is usually included in the tests. Once students grasp the equations and how they are framed, the LCM  calculation method is relatively simple.

3.10 – Problems of HCF and LCM

The learner must tackle both theoretical and practical problems in this phase. Students are given some examples to practice what they have learned  at the end of the chapter.

Q.1 Write all the factors of the following numbers:
(a) 24 (b) 15 (c) 21
(d) 27 (e) 12 (f) 20
(g) 18 (h) 23 (i) 36

Ans.

(a) Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
(b) Factors of 15 = 1, 3, 5, 15
(c) Factors of 21 = 1, 3, 7, 21
(d) Factors of 27 = 1, 3, 9, 27
(e) Factors of 12 = 1, 2, 3, 4, 6, 12
(f) Factors of 20 = 1, 2, 4, 5, 10, 20
(g) Factors of 18 = 1, 2, 3, 6, 9, 18
(h) Factors of 23 = 1, 23
(i) Factors of 36 = 1, 2, 3, 4, 6, 12, 18, 36

Q.2 Write first five multiples of:

(a) 5 (b) 8 (c) 9

Ans.

(a) First five multiples of 5 = 5, 10, 15, 20, 25
(b) First five multiples of 8 = 8, 16, 24, 32, 40
(c) First five multiples of 9 = 9, 18, 27, 36, 45

Q.3 Match the items in column 1 with the items in column 2.

 Column 1 Column 2 (i) 35 (a) Multiple of 8 (ii) 15 (b) Multiple of 7 (iii) 16 (c) Multiple of 70 (iv) 20 (d) Factor of 30 (v) 25 (e) Factor of 50 (f) Factor of 20

Ans.

(i) 35 (b) Multiple of 7

(ii) 15 (d) Factor of 30

(iii) 16 (a) Multiple of 8

(iv) 20 (f) Factor of 20

(v) 25 (e) Factor of 50

Q.4 Find all the multiples of 9 upto 100.

Ans.

Sol: Multiples of 9 upto 100 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

Q.5 What is the sum of any two
(a) Odd numbers? (b) Even numbers?

Ans.

(a) Sum of any two odd numbers is even number.
(b) Sum of any two even numbers is even number.

Q.6 State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.

Ans.

(a) The sum of three odd numbers is even. (F)
(b) The sum of two odd numbers and one even number is even. (T)
(c) The product of three odd numbers is odd. (T)
(d) If an even number is divided by 2, the quotient is always odd. (F)
(e) All prime numbers are odd. (F)
(f) Prime numbers do not have any factors. (F)
(g) Sum of two prime numbers is always even. (F)
(h) 2 is the only even prime number. (T)
(i) All even numbers are composite numbers. (F)
(j) The product of two even numbers is always even. (T)

Q.7 The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.

Ans. The pairs of prime numbers like 13 and 31 are: 17 and 71, 37 and 73, 79 and 97.

Q.8 Write down separately the prime and composite numbers less than 20.

Ans. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.

Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.

Q.9 What is the greatest prime number between 1and 10?

Ans. The greatest prime number between 1 and 10 is 7.

Q.10 Express the following as the sum of two odd primes. (a) 44 (b) 36 (c) 24 (d) 18

Ans. Expression of the sum of two prime numbers
(a) 44 = 3 + 41
(b) 36 = 5 + 31
(c) 24 = 5 + 19
(d) 18 = 5 + 13

Q.11 Give three pairs of prime numbers whose difference is 2.

Ans. Three pairs of prime numbers whose difference is 2 i.e., co-prime numbers are: 5 and 7, 17 and 19, 41 and 43.

Q.12 Which of the following numbers are prime?

(a) 23 (b) 51 (c) 37 (d) 26

Ans.

(a) 23 = 1 × 23, is a prime number.
(b) 51 = 1 × 3 × 17, is not a prime number.
(c) 37 = 1 × 37, is a prime number.
(d) 26 = 1 × 2 × 13, is not a prime number.

Q.13 Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

Ans. 7 consecutive composite numbers less than 100 according to given condition are: 90, 91, 92, 93, 94, 95, 96.

Q.14 Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61

Ans. Given numbers are expressed as the sum of three odd primes are given below:
(a) 21 = 3 + 5 + 13
(b) 31 = 7 + 11 + 13
(c) 53 = 13 + 17 + 23
(d) 61 = 7 + 23 + 31

Q.15 Write five pairs of prime numbers less than 20 whose sum is divisible by 5.

Ans. Pair of prime numbers less than 20 and divisible by 5 are: 2 and 3, 3 and 7, 7 and 13, 3 and 17, 13 and 17.

Q.16 Fill in the blanks:
(a) A number which has only two factors is called a ______.
(b) A number which has more than two factors is called a ______.
(c) 1 is neither ______ nor ______.
(d) The smallest prime number is ______.
(e) The smallest composite number is _____.
(f) The smallest even number is ______.

Ans.

(a) prime number.

(b) composite number.

(c) prime number nor composite number.

(d) 2.

(e) 4.

(f) 2.

Q.17 Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10 ; by 11 (say, yes or no):

 Numbers Divisible by 2 3 4 5 6 8 9 10 11 128 990 1586 275 6686 639210 429714 2856 3060 406839

Ans.

 Numbers Divisible by 2 3 4 5 6 8 9 10 11 128 Yes No Yes No No Yes No No No 990 Yes Yes No Yes Yes No Yes Yes Yes 1586 Yes No No No No No No No No 275 No No No Yes No No No No Yes 6686 Yes No No No No No No No No 639210 Yes Yes No Yes Yes No No Yes Yes 429714 Yes Yes No No Yes No Yes No No 2856 Yes Yes Yes No Yes Yes No Yes No 3060 Yes Yes Yes Yes Yes No Yes Yes No 406839 No Yes No No No No No No No

Q.18 Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:
(a) 572 (b) 726352 (c) 5500 (d) 6000
(e) 12159 (f) 14560 (g) 21084 (h) 31795072
(i) 1700 (j) 2150

Ans.

(a) 572
Since, last two digits i.e., 72 is divisible by 4.
So, 572 is divisible by 4.
572 is not divisible by 8.

(b) 726352
Since, last two digits i.e., 52 is divisible by 4.
So, 726352 is divisible by 4.
Last three digits 352 is divisible by 8, so given number is divisible by 8.

(c) 5500
Since, last two digits i.e., 00 is divisible by 4.
So, 5500 is divisible by 4.
Last three digits 500 is not divisible by 8, so given number is not divisible by 8.

(d) 6000
Since, last two digits i.e., 00 is divisible by 4.
So, 6000 is divisible by 4.
Last three digits 000 is divisible by 8, so given number is divisible by 8.

(e) 12159
Since, last two digits i.e., 59 is not divisible by 4.
So, 12159 are not divisible by 4.
Last three digits 159 are not divisible by 8. So 12159 is not divisible by 8.

(f) 14560
Since, last two digits i.e., 60 are divisible by 4.
So, 14560 is divisible by 4.
Last three digits 560 are divisible by 8. So 14560 is divisible by 8.

(g) 21084
Since, last two digits i.e., 84 are divisible by 4.
So, 21084 is divisible by 4.
Last three digits 084 are not divisible by 8. So 21084 is not divisible by 8.

(h) 31795072
Since, last two digits i.e., 72 are divisible by 4.
So, 31795072 is divisible by 4.
Last three digits 072 are divisible by 8. So 31795072 is divisible by 8.

(i) 1700
Since, last two digits i.e., 00 are divisible by 4.
So, 1700 is divisible by 4.
Last three digits 700 are not divisible by 8. So 1700 is not divisible by 8.

(j) 2150
Since, last two digits i.e., 50 are not divisible by 4. So, 2150 is not divisible by 4.
Last three digits 150 are not divisible by 8. So 2150 is not divisible by 8.

Q.19 Using divisibility tests, determine which of following numbers are divisible by 6:
(a) 297144 (b) 1258 (c) 4335 (d) 61233
(e) 901352 (f) 438750 (g) 1790184 (h) 12583
(i) 639210 (j) 17852

Ans.

(a) 297144
Since, the last digit is 4 which is divisible by 2.
So, 297144 is divisible by 2.
2 + 9 + 7 + 1 + 4 + 4 = 27 is divisible by 3.
So, 297144 is divisible by 3.
Since, 297144 is divisible by 2 and 3.
So, 297144 is divisible by 6.

(b) 1258
Since, the last digit is 58 which is divisible by 2.
So, 1258 is divisible by 2.
1 + 2 + 5 + 8 = 16 is not divisible by 3.
So, 1258 is not divisible by 3.
Since, 1258 is divisible by 2 but not by 3.
So,1258 is not divisible by 6.

(c) 4335
Since, the last digit is 35 which are not divisible by 2.
So, 4335 is not divisible by 2.
4 + 3 + 3 + 5 = 15 is divisible by 3.
So, 1258 is divisible by 3.
Since, 1258 is divisible by 3 but not by 2.
So, 1258 is not divisible by 6.

(d) 61233
Since, the last digit is 33 which are not divisible by 2.
So, 61233 is not divisible by 2.
6 + 1 + 2 + 3 + 3 = 15 is divisible by 3.
So, 61233 is divisible by 3.
Since, 61233 is divisible by 3 but not by 2.
So, 61233 is not divisible by 6.

(e) 901352
Since, the last digit is 52 which are divisible by 2.
So, 901352 is divisible by 2.
9 + 0 + 1 + 3 + 5 + 2 = 20 is not divisible by 3.
So, 901352 is not divisible by 3.
Since, 901352 is divisible by 2 but not by 3. So, 901352 is not divisible by 6.

(f) 438750
Since, the last digit is 50 which are divisible by 2.
So, 438750 is divisible by 2.
4 + 3 + 8 + 7 + 5 + 0 = 27 is divisible by 3.
So, 438750 is divisible by 3.
Since, 438750 is divisible by 2 and by 3. So, 901352 is divisible by 6.

(g) 1790184
Since, the last digit is 84 which are divisible by 2.
So, 1790184 is divisible by 2.
1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 is divisible by 3.
So, 1790184 is divisible by 3.
Since, 1790184 is divisible by 2 and by 3. So, 1790184 is divisible by 6.

(h) 12583
Since, the last digit is 83 which are not divisible by 2.
So, 12583 is not divisible by 2.
1 + 2 + 5 + 8 + 3 = 19 is not divisible by 3.
So, 12583 is divisible by 3.
Since, 12583 is divisible by 2 and by 3. So,12583 is divisible by 6.

Q.20 Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445 (b) 10824
(c) 7138965 (d) 70169308
(e) 10000001 (f) 901153

Ans.

 Number Sum of the digits ( at odd places ) from the right Sum of the digits ( at even places ) from the right Difference ( a ) 5445 5 + 4 = 9 4 + 5 = 9 9 − 9 = 0 ( b ) 1 0 824 4 + 8 + 1 = 13 2 + 0 = 2 13 − 2 = 11 ( c ) 7138965 5 + 9 + 3 + 7 = 24 6 + 8 + 1 = 15 24 − 15 = 9 ( d ) 70169308 8 + 3 + 6 + 0 = 17 0 + 9 + 1 + 7 = 17 17 − 17 = 0 ( e ) 10000001 1 + 0 + 0 +0 = 1 0 + 0 + 0 + 1 = 1 1 − 1 = 0 ( f ) 901153 3 + 1 + 0 = 4 5 + 1 + 9 = 15 15 − 4 = 11

If the difference is divisible by 11, the number is divisible by 11 . Here, 7138965 is not divisible by 11 .
Numbers (a) 5445 (b) 10824 (d) 70169308 (e) 10000001 and (f) 901153 are divisible by 11 .

Q.21 Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :

1. __ 6724
2. 4765 __ 2

Ans.

1.
The sum of given digits = 6 + 7 + 2 + 4 = 19
On adding 2 in 19, we get 21 which is divisible by 3. Therefore, the number formed by using the smallest digit is 26724.
On adding 8 in 19, we get 27 which is divisible by 3. Therefore, the number formed by using the greatest digit is 86724.

2.
The sum of given digits = 4 + 7 + 6 + 5 + 2 = 24
On adding 0 in 24, we get 24 which is divisible by 3. Therefore, the number formed by using the smallest digit is 476502.
On adding 9 in 24, we get 33 which is divisible by 3. Therefore, the number formed by using the greatest digit is 476592.

Q.22 Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 :

(a) 92 __ 389

(b) 8 __ 9484

Ans.

(a) 92 __ 389
The sum of digits at odd places from right
= 9 + 3 + 2
= 14
The sum of remaining digits from right
= 8 + 9
= 17
The difference = 17 – 14 = 3
The difference will be divisible by 11, if 8 is added in 3. Therefore, the missing digit is 8.

(b) 8 __ 9484
The sum of digits at even places from right
= 8 + 9 + 8
= 25
The sum of remaining digits at odd positions from right = 4 + 4
= 8
The difference = 8 – 25
= – 17
The difference will be divisible by 11, if 6 is added in – 17. Therefore, the missing digit is 6.
Therefore, the number is 869484.

Q.23 Find the common factors of:
(a) 20 and 28 (b) 15 and 25
(c) 35 and 50 (d) 56 and 120

Ans.

(a) Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 28 = 1, 2, 4, 7, 14, 28
Common factors of 20 and 28 = 1, 2, 4

(b) Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors of 15 and 25 = 1, 5

(c) Factors of 35 = 1, 5, 7, 35
Factors of 50 = 1, 5, 10, 25, 50
Common factors of 35 and 50 = 1, 5

(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 12, 24, 40, 60, 120
Common factors of 35 and 50 = 1, 2, 4, 8

Q.24 Find the common factors of: (a) 4, 8 and 12 (b) 5, 15 and 25

Ans.

(a) Factors of 4 = 1, 2, 4
Factors of 8 = 1, 2, 4, 8
Factors of 12 = 1, 2, 4, 6, 12
Common factors of 4, 8 and 12 = 1, 2, 4

(b)Factors of 5 = 1, 5
Factors of 15 = 1, 3, 5, 15
Factors of 25 = 1, 5, 25
Common factors of 5, 15 and 25 = 1, 5

Q.25 Find first three common multiples of:
(a) 6 and 8
(b) 12 and 18

Ans.

(a) Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, …
First three common multiples of 6 and 8 = 24, 48, 72

(b) Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, …
Multiples of 18 = 36, 54, 72, 90, 108, …
First three common multiples of 6 and 8 = 36, 72, 108

Q.26 Write all the numbers less than 100 which are common multiples of 3 and 4.

Ans.

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99
Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96
Common multiples of 3 and 4 = 12, 24, 36, 48, 60, 72, 84, 96.

Q.27 Which of the following numbers are co-prime?
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16

Ans.

(a) Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 35 = 1, 5, 7, 35
Common factor = 1
So, 18 and 35 are co-prime numbers.

(b) Factors of 15 = 1, 3, 5, 15
Factors of 37 = 1, 37
Common factor = 1
So, 15 and 37 are co-prime numbers.

(c) Factors of 30 = 1, 3, 5, 6, 15, 30
Factors of 415 = 1, 5, 83, 415
Common factor = 1, 5
So, 30 and 415 are not co-prime numbers.

(d) Factors of 17 = 1, 17
Factors of 68 = 1, 2, 4, 17, 34, 68
Common factor = 1
So, 17 and 68 are co-prime numbers.

(e) Factors of 216 = 1, 2, 3, 4, 6, 8, 36, 72, 108, 216
Factors of 215 = 1, 5, 43, 215
Common factor = 1
So, 216 and 215 are co-prime numbers.

(f) Factors of 81 = 1, 3, 9, 27, 81
Factors of 16 = 1, 2, 4, 8, 16
Common factor = 1
So, 81 and 16 are co-prime numbers.

Q.28 A number is divisible by both 5 and 12. By which other number will that number be always divisible?

Ans.

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, …
Multiples of 12 = 12, 24, 36, 48, 60, 72, …
Here, 60 is divisible by both 5 and 12.
60 is always divisible by 1, 2, 3, 4, 6, 10, 15, 30 too.

Q.29 A number is divisible by 12. By what other numbers will that number be divisible?

Ans. A number which is divisible by 12, can be divided by 1, 2, 3, 4 and 6.

Q.30 Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

Ans.

(a) False,
(b) True,
(c) False,
(d) True,
(e) False,
(f) False,
(g) True,
(h) True,
(i) False.

Q.31 Here are two different factor trees for 60. Write the missing numbers.
(a)

(b)

Ans.

Factor trees with missing numbers are given below:

(a)

(b)

Q.32 Which factors are not included in the prime factorisation of a composite number?

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

Q.33 Write the greatest 4-digit number and express it in terms of its prime factors.

Ans.

$\begin{array}{l}\text{The greatest 4}-\text{digit number}=\text{9999}\\ \text{}\text{}\text{}\text{}3\overline{)9999}\\ \text{}\text{}\text{}\text{}3\overline{)3333}\\ \text{}\text{}\text{}\text{}11\overline{)1111}\\ \text{}\text{}\text{}\text{}\text{}\text{}101\overline{)101}\\ \text{\hspace{0.17em}}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\overline{)1}\\ \text{Prime factor of 9999}=\text{3}×\text{3}×\text{11}×\text{1}0\text{1}\end{array}$

Q.34 Write the smallest 5-digit number and express it in the form of its prime factors.

Ans.

$\begin{array}{l}\text{The smallest 5}-\text{digit number}=\text{1}0000\\ \text{}\text{}\text{}\text{}2\overline{)10000}\\ \text{}\text{}\text{}\text{}2\overline{)5000}\\ \text{}\text{}\text{}\text{}2\overline{)2500}\\ \text{}\text{}\text{}\text{}2\overline{)1250}\\ \text{}\text{}\text{}\text{}5\overline{)625}\\ \text{}\text{}\text{}\text{}5\overline{)125}\\ \text{}\text{}\text{}\text{}5\overline{)25}\\ \text{}\text{}\text{}\text{}5\overline{)5}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\overline{)1}\\ \text{Prime factors of 1}0000\text{}=\text{2}×\text{2}×\text{2}×\text{2}×\text{5}×\text{5}×\text{5}×\text{5}\end{array}$

Q.35 Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.

Ans.

$\begin{array}{l}7\overline{)1729}\\ 13\overline{)247}\\ \text{\hspace{0.17em}}19\overline{)19}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\overline{)1}\\ \text{Prime factors of 1729}=7×13×19\\ \mathrm{The}\text{difference between two consecutive prime factors is 6.}\end{array}$

Q.36 The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Ans. Let us take three consecutive numbers as 2, 3, 4.

Product of three numbers = 2 × 3 × 4

= 24, which is divisible by 6.

Let us take another three numbers as 11, 12, 13.

Product of these three numbers = 11 × 12 × 13

= 1716, which is divisible by 6

In this way, we see that product of three consecutive numbers is always divisible by 6.

Q.37 The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Ans. Let us take two consecutive odd numbers as 3 and 5. Then, their sum = 3 + 5
= 8, which is divisible by 4.

Let us take another set of two consecutive odd numbers as 17 and 19.Then,

Sum of numbers = 17 + 19

= 36, which is divisible by 4.

Thus, the sum of two consecutive odd numbers is divisible by 4.

Q.38 In which of the following expressions, prime factorisation has been done?
(a) 24 = 2 × 3 × 4 (b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7 (d) 54 = 2 × 3 × 9

Ans. In option (b) and (c), prime factorisation has been done because 2, 5 and 7 are prime numbers. While 4 and 9 are not prime numbers which are given in option (a) and (d).

Q.39 Determine if 25110 is divisible by 45.

Ans. A number will be divisible by 45 if it is divisible by 5 and 9.
Given number 25110 is divisible by 5 because one’s digit is 0.
Now, 2 + 5 + 1 + 1 + 0 = 9 is divisible by 9. So, 25110 is also divisible by 9.
Therefore, we can say that 25110 is divisible by 5 and 9 together, so it is divisible by 45.

Q.40 18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.

Ans. A number which is divisible by 4 and 6 is not divisible by 24. There is a example to prove it. 12 is divisible by 4 and 6 but it is not divisible by 24.

Q.41 I am the smallest number, having four different prime factors. Can you find me?

Ans. Four smallest prime numbers are 2, 3, 5 and 7.

So, the smallest number = 2×3×5×7
= 210

Q.42 Find the HCF of the following numbers:
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75

Ans.

$\left(\mathrm{a}\right)\text{}18,\text{}48\text{ }\begin{array}{cc}2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\text{ }\begin{array}{cc}2& 48\\ 2& 24\\ 2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}$

$\begin{array}{l}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}18=2×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}48=2×2×2×2×3\\ \text{\hspace{0.17em}}\mathrm{HCF}=2×3=6\end{array}$ $\begin{array}{l}\left(\mathrm{b}\right)\text{}30,\text{}42\text{}\\ \text{}\begin{array}{cc}2& 30\\ 3& 15\\ 5& 5\\ & 1\end{array}\text{ }\begin{array}{cc}2& 42\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}30=2×3×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}42=2×3×7\\ \text{\hspace{0.17em}}\mathrm{HCF}=2×3\\ \text{\hspace{0.17em}\hspace{0.17em}}=6\\ \left(\mathrm{c}\right)\text{}18,\text{}60\text{}\\ \text{}\begin{array}{cc}2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\text{ }\begin{array}{cc}2& 60\\ 2& 30\\ 3& 15\\ 5& 5\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}18=2×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}60=2×2×3×5\\ \text{\hspace{0.17em}}\mathrm{HCF}=2×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ \left(\mathrm{d}\right)\text{}27,\text{}63\end{array}$ $\begin{array}{l}\text{}\begin{array}{cc}3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\text{ }\begin{array}{cc}3& 63\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}27=3×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}63=3×3×7\\ \text{\hspace{0.17em}}\mathrm{HCF}=3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9\\ \left(\mathrm{e}\right)\text{}36,\text{}84\\ \text{}\begin{array}{cc}2& 36\\ 2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\text{ }\begin{array}{cc}2& 84\\ 2& 42\\ 3& 21\\ 7& 7\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}36=2×2×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}84=2×2×3×7\\ \text{\hspace{0.17em}}\mathrm{HCF}=2×2×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=12\\ \left(\mathrm{f}\right)\text{}34,\text{}102\\ \text{}\begin{array}{cc}2& 34\\ 17& 17\\ & 1\end{array}\text{ }\begin{array}{cc}2& 102\\ 3& 51\\ 17& 17\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}34=3×17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}102=2×3×17\\ \text{\hspace{0.17em}}\mathrm{HCF}=3×17\text{\hspace{0.17em}}=34\end{array}$ $\begin{array}{l}\left(\mathrm{g}\right)\text{}70,\text{}105,\text{}175\\ \text{}\begin{array}{cc}2& 70\\ 5& 35\\ 7& 7\\ & 1\end{array}\text{ }\begin{array}{cc}3& 105\\ 5& 35\\ 7& 7\\ & 1\end{array}\text{ }\begin{array}{cc}5& 175\\ 5& 35\\ 7& 7\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}70=2×5×7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}105=3×5×7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}175=5×5×7\\ \text{\hspace{0.17em}}\mathrm{HCF}=5×7\text{\hspace{0.17em}}=35\\ \left(\mathrm{h}\right)\text{}91,\text{}112,\text{}49\\ \text{}\begin{array}{cc}7& 91\\ 13& 13\\ & 1\end{array}\text{ }\begin{array}{cc}2& 112\\ 2& 56\\ 2& 28\\ 2& 14\\ 7& 7\\ & 1\end{array}\text{ }\begin{array}{cc}7& 49\\ 7& 7\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}91=7×13\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}112=2×2×2×2×7\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}49=7×7\\ \text{\hspace{0.17em}}\mathrm{HCF}=7\\ \left(\mathrm{i}\right)\text{}18,\text{}54,\text{}81\\ \text{}\begin{array}{cc}2& 18\\ 3& 9\\ 3& 3\\ & 1\end{array}\text{ }\begin{array}{cc}2& 54\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\text{ \hspace{0.17em}}\begin{array}{cc}3& 81\\ 3& 27\\ 3& 9\\ 3& 3\\ & 1\end{array}\end{array}$ $\begin{array}{l}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}18=2×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}54=2×3×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}81=3×3×3×3\\ \text{\hspace{0.17em}}\mathrm{HCF}=3×3=9\\ \left(\mathrm{j}\right)\text{}12,\text{}45,\text{}75\\ \text{}\begin{array}{cc}2& 12\\ 2& 6\\ 3& 3\\ & 1\end{array}\text{ }\begin{array}{cc}3& 45\\ 3& 15\\ 5& 5\\ & 1\end{array}\text{ \hspace{0.17em}}\begin{array}{cc}3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12=2×3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}45=3×3×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}75=3×5×5\\ \text{\hspace{0.17em}}\mathrm{HCF}=3\end{array}$

Q.43 What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?

Ans. (a) HCF of two consecutive numbers is 1.
(b) HCF of two consecutive even numbers is 2.
(c) HCF of two consecutive odd numbers is 1.

Q.44 HCF of co-prime numbers 4 and 15 was found as follows by factorisation:
4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Ans. No, it is not correct. HCF of 4 and 15 is not 0. HCF of 4 and 15 is 1.

Q.45 Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Ans.

$\begin{array}{l}\mathrm{Weights}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{two}\text{\hspace{0.17em}}\mathrm{bags}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{fertilizers}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{kg}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}69\text{\hspace{0.17em}}\mathrm{kg}.\\ \mathrm{Then},\\ \begin{array}{cc}3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\begin{array}{cc}3& 69\\ 23& 23\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}75\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}5\text{\hspace{0.17em}}×\text{\hspace{0.17em}}5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}69\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}23\\ \mathrm{HCF}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}69\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\\ \mathrm{Thus},\text{\hspace{0.17em}}\mathrm{maximum}\text{\hspace{0.17em}}\mathrm{value}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{weight}\text{\hspace{0.17em}}\mathrm{which}\text{\hspace{0.17em}}\mathrm{can}\text{\hspace{0.17em}}\mathrm{measure}\text{\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{kg}\\ \mathrm{and}\text{\hspace{0.17em}}69\text{\hspace{0.17em}}\mathrm{kg}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\mathrm{kg}\end{array}$

Q.46 Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Ans.

$\begin{array}{l}\mathrm{Weights}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{two}\text{\hspace{0.17em}}\mathrm{bags}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{fertilizers}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{kg}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}69\text{\hspace{0.17em}}\mathrm{kg}.\\ \mathrm{Then},\\ \begin{array}{cc}3& 75\\ 5& 25\\ 5& 5\\ & 5\end{array}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\begin{array}{cc}3& 69\\ 23& 23\\ & 1\end{array}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}75\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}5\text{\hspace{0.17em}}×\text{\hspace{0.17em}}5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}69\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}23\\ \mathrm{HCF}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}69\text{\hspace{0.17em}}=\text{\hspace{0.17em}}3\\ \mathrm{Thus},\text{\hspace{0.17em}}\mathrm{maximum}\text{\hspace{0.17em}}\mathrm{value}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{weight}\text{\hspace{0.17em}}\mathrm{which}\text{\hspace{0.17em}}\mathrm{can}\text{\hspace{0.17em}}\mathrm{measure}\text{\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{kg}\\ \mathrm{and}\text{\hspace{0.17em}}69\text{\hspace{0.17em}}\mathrm{kg}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\mathrm{kg}\end{array}$

Q.47 Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Ans.

$\begin{array}{l}\mathrm{Weights}\text{of two bags of fertilizers is 75 kg and 69 kg.}\\ \text{Then,}\\ \begin{array}{cc}3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\text{ }\begin{array}{cc}3& 69\\ 23& 23\\ & 1\end{array}\\ \therefore 75=3×5×5\\ 69=3×23\\ \mathrm{HCF}\text{of 75 and 69}=3\\ \mathrm{Thus},\text{maximum value of weight which can measure 75 kg}\\ \text{and 69 kg}\mathrm{is}\text{3 kg.}\end{array}$

Q.48 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.

$\begin{array}{l}\mathrm{The}\text{step measures of first boy}=63\text{cm}\\ \mathrm{The}\text{step measures of second boy}=70\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{The}\text{step measures of third boy}=77\text{cm}\end{array}$ $\begin{array}{l}\begin{array}{cc}2& 63,70,77\\ 3& 63,35,77\\ 3& 21,35,77\\ 5& 7,35,77\\ 7& 7,7,77\\ 11& 1,1,11\\ & 1,1,1\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{LCM}\text{of 63, 70 and 77}=2×3×3×5×7×11\\ =6930\\ \mathrm{Thus},\text{the three boys will cover 6930 cm distance in}\\ \text{complete steps.}\end{array}$

Q.49 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.

Ans.

$\begin{array}{l}\mathrm{The}\text{length of a room}=825\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{The}\text{breadth of a room}=675\text{\hspace{0.17em}}\mathrm{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{The}\text{height of a room}=450\text{\hspace{0.17em}}\mathrm{cm}\\ \begin{array}{cc}3& 825\\ 5& 275\\ 5& 55\\ 11& 11\\ & 1\end{array}\text{ }\begin{array}{cc}3& 675\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\text{ }\begin{array}{cc}2& 450\\ 3& 225\\ 3& 75\\ 5& 25\\ 5& 5\\ & 1\end{array}\\ \therefore 825=3×5×5×11\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}675=3×3×3×5×5\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}450=2×3×3×5×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{HCF}=3×5×5=75\\ \mathrm{Therefore},\text{the length of the longest tape which can}\\ \text{measure the}\mathrm{}\text{three dimensions of the room exactly is 75 cm.}\end{array}$

Q.50 Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Ans.

$\begin{array}{l}\mathrm{Given}\text{numbers are 6, 8 and 12.}\\ \begin{array}{cc}2& 6,8,12\\ 2& 3,4,6\\ 2& 3,2,3\\ 3& 3,1,3\\ & 1,1,1\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{LCM}=2×2×2×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24\\ \mathrm{Multiples}\text{of 24}=24,48,72,96,120,...\\ \mathrm{Therefore},\text{the smallest 3-digits number which is}\\ \text{divisible by 6,8 and 12 is 120.}\end{array}$

Q.51 Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Ans.

$\begin{array}{l}\mathrm{Given}\text{numbers are 8, 10 and 12.}\\ \begin{array}{cc}2& 8,\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}12\\ 2& 4,\text{\hspace{0.17em}}5,6\\ 2& 2,5,3\\ 3& 1,5,3\\ 5& 1,5,1\\ & 1,1,1\end{array}\end{array}$ $\begin{array}{l}\mathrm{LCM}=2×2×2×3×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=120\\ \mathrm{The}\text{greatest 3-digit number}=\text{999}\\ 8\\ 120\overline{)999}\\ \underset{¯}{960}\\ 39\\ \mathrm{The}\text{greatest 3-digit number divisible by 120}\\ =999-39\\ =960\\ \mathrm{Therefore},\text{the greatest 3-digit number which is}\\ \text{divisible by 6,8 and 12 is 960.}\end{array}$

Q.52 The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Ans.

$\begin{array}{l}\text{The traffic lights at first road crossings change after time}\\ \text{\hspace{0.17em}}=\text{48 seconds}\\ \text{The traffic lights at second road crossings change after time}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{72 seconds}\\ \text{The traffic lights at third road crossings change after time}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{108 seconds}\end{array}$ $\begin{array}{l}\begin{array}{cc}\text{2}& \text{48,\hspace{0.17em}72,\hspace{0.17em}108}\\ \text{2}& \text{24,\hspace{0.17em}36,54}\\ \text{2}& \text{12,18,27}\\ \text{2}& \text{6,9,27}\\ \text{3}& \text{3,9,27}\\ \text{3}& \text{1,3,9}\\ \text{3}& \text{1,1,3}\\ & \text{1,1,1}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}LCM of 48, 72 and 108=\hspace{0.17em}2×2×2×2×3×3×3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=432}\\ \text{All the three lights will change after 432 seconds.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}432 seconds=7\hspace{0.17em}\hspace{0.17em}min\hspace{0.17em}\hspace{0.17em}12\hspace{0.17em}seconds}\\ \text{If three lights changes at 7 a.m., then}\end{array}$

$\begin{array}{l}\text{ }Hrs\text{ }Min\text{ }\mathrm{Sec}\\ \text{ } \text{ }7\text{ }00\text{ }00\\ \underset{¯}{+\text{ }0\text{ }07\text{ }12}\\ \text{ } \text{ }7\text{ }07\text{ }12\end{array}$

$\begin{array}{l}\text{They will change simultaneously again}\\ \text{at time = 7:07:12\hspace{0.17em}hrs.}\end{array}$

Q.53 Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Ans.

$\begin{array}{l}\mathrm{Diesel}\text{in first tank}=403\text{litres}\\ \mathrm{Diesel}\text{in second tank}=434\text{litres}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Diesel}\text{in third tank}=465\text{litres}\end{array}$ $\begin{array}{l}\begin{array}{cc}13& 403\\ 31& 31\\ & 1\end{array}\text{ }\begin{array}{cc}2& \text{\hspace{0.17em}}434\\ 7& \text{\hspace{0.17em}}217\\ 31& 31\\ & 1\end{array}\text{ }\begin{array}{cc}3& \text{\hspace{0.17em}}465\\ 5& \text{\hspace{0.17em}}155\\ 31& 31\\ & 1\end{array}\\ \therefore 403=13×31\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}434=2×7×31\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}465=3×5×31\\ \mathrm{H}.\mathrm{C}.\mathrm{F}.\text{of 48, 72 and 108}=31\\ \mathrm{Thus},\text{the maximum capacity of a container that can}\\ \text{measure the diesel of the three containers}=31\text{\hspace{0.17em}}\mathrm{litres}\end{array}$

Q.54 Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Ans.

$\begin{array}{l}\mathrm{Given}\text{divisors are 6,15 and 18.}\\ \text{Remainder in each case is 5.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\begin{array}{cc}2& 6,15,18\\ 3& 3,15,9\\ 3& 1,5,3\\ 5& 1,5,1\\ & 1,1,1\end{array}\\ \text{\hspace{0.17em}}\mathrm{LCM}\text{​ of 6, 15 and 18}=2×3×3×5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\\ \mathrm{The}\text{required number}=90+5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=95\end{array}$ $\begin{array}{l}\text{Therefore, when 95 is divided by 6, 15 and 18 leave}\\ \text{remainder 5 in each case.}\end{array}$

Q.55 Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Ans.

$\begin{array}{l}\mathrm{Given}\text{numbers are 18, 24 and 32.}\\ \begin{array}{cc}2& 18,\text{\hspace{0.17em}}24,\text{\hspace{0.17em}}32\\ 2& 9,\text{\hspace{0.17em}}12,16\\ 2& 9,6,8\\ 2& 9,3,4\\ 2& 9,3,2\\ 3& 9,3,1\\ 3& 3,1,1\\ & 1,1,1\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{LCM}\text{of 18, 24 and 32}=2×2×2×2×2×3×3\\ =288\\ \mathrm{The}\text{smallest number of 4-digits}\\ =\text{1000}\end{array}$ $\begin{array}{l}3\\ 288\overline{)1000}\\ \underset{¯}{\text{\hspace{0.17em}\hspace{0.17em}}864}\\ 136\end{array}$ $\begin{array}{l}\mathrm{The}\text{smallest 4-digits number divisible by 288}\\ \text{ }=1000+\left(288-136\right)\\ \text{ }=1000+152\\ \text{ }=1152\end{array}$ $\begin{array}{l}\mathrm{Therefore},\text{the smallest 4-digits number which is}\\ \text{divisible by 18,24 and 32 is 1152.}\end{array}$

Q.56 Find the LCM of the following numbers:
(a) 9 and 4 (b) 12 and 5
(c) 6 and 5 (d) 15 and 4
Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{ Prime factors of 9 = 3×3}\\ \text{ Prime factors of 4 = 2×2}\\ \text{ LCM of 9 and 4 = 2×2×3×3}\\ \text{ = 36}\\ \left(\text{b}\right)\text{ Prime factors of 12 = 2×2×3}\\ \text{ Prime factors of 5 = 1×5}\\ \text{ LCM of 12 and 5 = 2×2×3×5}\\ \text{ = 60}\\ \left(\text{c}\right)\text{ Prime factors of 6 = 2×3}\\ \text{ Prime factors of 5 =1×5}\\ \text{ LCM of 6 and 5 = 2×3×5}\\ \text{ = 30}\\ \left(\text{d}\right)\text{ Prime factors of 15 = 3×5}\\ \text{ Prime factors of 4 = 2×2\hspace{0.17em}}\\ \text{ LCM of 15 and 5 = 2×2×3×5}\\ \text{ = 60}\end{array}$

Yes, LCM is the product of two numbers in each case.

Q.57 Find the LCM of the following numbers in which one number is the factor of the other.
(a) 5, 20 (b) 6, 18 (c) 12, 48 (d) 9, 45 What do you observe in the results obtained?

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{ Prime factors of 5 = 1×5}\\ \text{Prime factors of 20 = 2×2×5}\\ \text{LCM of 5 and 20 = 2×2×5}\\ \text{ = 20}\end{array}$

$\begin{array}{l}\left(\text{b}\right)\text{ Prime factors of 6 = 2×3}\\ \text{Prime factors of 18 = 2×3×3}\\ \text{LCM of 6 and 18 = 2×3×3}\\ \text{ = 18}\end{array}$

$\begin{array}{l}\left(\text{c}\right)\text{ Prime factors of 12 = 2×2×3}\\ \text{Prime factors of 48 = 2×2×2×2×3}\\ \text{LCM of 12 and 48 = 2×2×2×2×3}\\ \text{ = 48}\end{array}$

$\begin{array}{l}\left(\text{d}\right)\text{ Prime factors of 9 = 3×3}\\ \text{Prime factors of 45 = 3×3×5}\\ \text{LCM of 9 and 45 = 3×3×5}\\ \text{ = 45}\end{array}$

We observe that LCM is equal to greater number in given numbers in each case.

1. What is covered in NCERT Mathematics Chapter 3 for Class 6?

The importance of comprehending factors and multiples of numbers is covered in the NCERT answer for Class 6 Mathematics Chapter 3. The chapter delves into several topics such as: Understanding the laws of divisibility, prime numbers, composite numbers, prime factors, LCM, HCF, common factors, and common multiples and so on.

This chapter’s unique examples will help the student to learn the concepts and principles of Mathematics. Understanding how each number is a factor and multiple and how prime factors multiply to generate the original number is also required. The main focus of this chapter is on HCF and LCM, which are frequently tested in the  exams.

2. Why should you use the NCERT Chapter 3 Mathematics Class 6 solutions?

Students can consult the NCERT Chapter 3 solutions for Class 6 Mathematics on Extramarks official website. The reason is that these solutions contain enough exercises and examples  that aid students’ comprehension. The examples have been arranged from simple to advanced level so that the student has no difficulty solving  them. Class 6 CBSE is a chapter designed to assist students in  comprehending and developing their fundamental knowledge of numbers. The solution will assist students in resolving any doubts or problems they may have regarding  numbers.

3. What's the difference between multiples and factors?

A factor is one or more numbers that can divide a given number without leaving a remainder. For instance, 6 x 5 = 30. The elements of 30, in this case, are 6 and 5. A multiple is a number that may be divided by another number several times without leaving a residue. For instance, 3 x 5 = 15. 15 is a multiple of 3 and 5 in this case.

4. Why is Chapter 3 of Class 6 Mathematics so vital for further education?

Class 6 Mathematics Chapter 3 ‘Playing With Numbers’ is vital for higher classes since it contains many topics or concepts useful in subsequent courses. It will reinforce your roots in Mathematics and make it easier for you to apply these concepts at higher levels if you understand topics like prime and composite numbers, factors and multiples, divisibility tests, HCF, and LCM. Subject experts recommend the Extramarks Mobile app for better understanding.

5. How can I improve the fundamental concepts of Chapter 3?

Basic ideas are crucial for students to understand, and they should not be overlooked at any cost. There are many important topics in NCERT Class 6 Mathematics Chapter 3 solutions, and students should pay good attention to them. These core ideas will help you in exams and they will also help you in the next class. As a result, one should concentrate on mastering these concepts and practising them regularly. You may obtain extra study material for the same at Extramarks.