NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers  FREE PDF Download
NCERT Class 6 Maths Chapter 3 Solutions  Playing with Numbers. This chapter is all about exploring the world of numbers in a fun and engaging way. You’ll learn how to find factors and multiples, understand prime and composite numbers, and play with different number patterns. In Class 6 Maths NCERT Solutions, we break down complex concepts into simple steps, making it easy for you to understand and apply them in your maths problems. Vedantu’s NCERT Class 6 Maths Chapter 3 PDF Solutions provides detailed explanations and stepbystep solutions to all the important questions, helping students understand the CBSE Class 6 Maths Syllabus effectively.
 5.1Exercise 3.1
 5.2Exercise 3.2
 5.3Exercise 3.3
 5.4Exercise 3.5
 5.5Exercise 3.6
 5.6Exercise 3.7
 6.1Class 6 Maths Chapter 3: Exercises Breakdown
Glance on Maths Chapter 3 Class 6  Playing with Numbers
To determine the factors of a number. Factors are numbers that can be multiplied to get the original number.
Finding multiples of a number. Multiples are numbers that are products of the original number and any integer.
Prime Factorization, which means breaking down a composite number into its prime factors using factor trees or repeated division.
Methods to find Highest Common Factor(H.C.F) and Least Common Multiple(L.C.M).
This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 3  Playing with Numbers, which you can download as PDFs.
There are seven exercises (53 fully solved questions) in class 6th Maths chapter 3 Playing with Numbers.
Access Exercise Wise NCERT Solutions for Chapter 3 Maths Class 6
Exercises Under NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers
Exercise 3.1 – Introduction
This exercise 4 contains questions and solutions. Purpose of this exercise is to provide students with the basic knowledge of what can be classified as a divisor and what is a factor. This part is what the entire chapter is mostly based on.
Exercise 3.2 – Factors and Multiples
This exercise 12 contains questions and solutions, this contains set of questions that asks students to spot multiples and factors. It also states that every number is a multiple and factor of itself.
Exercise 3.3 – Prime and Composite Numbers
This exercise 6 contains questions and solutions. This section of the chapter elaborates on which numbers can be defined as prime numbers and which as composite numbers. Segment also specifies how 2 is the smallest of all the prime numbers, and 1 can neither be a composite nor a prime number.
Exercise 3.4 – Tests for Divisibility of Numbers
This exercise 7 contains questions and solutions. Main focus of this part of Chapter 3 in NCERT Maths book Class 6 is to make students understand which numbers are divisible by 10, 5, 2, 3 6, 4, 8 9, and 11. Learn more about how this divisibility test is helpful.
Exercise 3.5 – Common Factors and Common Multiples
This exercise 10 contains questions and solutions. Understanding common multiples and common factors are mandatory for gaining a grasp over the subjectmatter of numbers. Focus of this exercise is to teach how certain numbers are the multiple and factor of more than one number.
Exercise 3.6 – Some More Divisibility Rules
This exercise 3 contains questions and solutions. There are certain divisibility rules that the student will need to know to gain a proper grasp over the subject of factors and multiples. Refer to Chapter 3 solutions from NCERT to gain more detailed knowledge on the topic.
Exercise 3.7 – Prime Factorization
This exercise 11 contains questions and solutions. This section of the NCERT Class 6 Maths Chapter 3 talks about prime factors and how which number when multiplied forms the original number. Learn about prime factor number play in this segment.
Access NCERT Solutions for Class 6 Maths Chapter 3 – Playing with Numbers
Exercise 3.1
1. Write all the factors of the following numbers.
$\mathbf{24}$
Ans: Given: number $24$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $24$ as,
$ 1 \times 24 $
$ 2 \times 12 $
$ 3 \times 8 $
$ 4 \times 6 $
Therefore, the factors of $24$ will be $1,2,3,4,6,8,12,24$.
$\mathbf{15}$
Ans:
Given: number $15$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $15$ as,
$ 1 \times 15 \\$
$ 3 \times 5 \\$
Therefore, the factors of $15$ will be $1,3,5,15$.
$\mathbf{21}$
Ans:
Given: number $21$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $21$ as,
$1 \times 21 $
$ 3 \times 7 $
Therefore, the factors of $21$ will be $1,3,7,21$.
$\mathbf{27}$
Ans:
Given: number $27$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $27$ as,
$1 \times 27\\$
$3 \times 9\\ $
Therefore, the factors of $27$ will be $1,3,9,27$.
$\mathbf{12}$
Ans:
Given: number $12$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $12$ as,
$1 \times 12\\$
$2 \times 6\\$
$3 \times 4\\ $
Therefore, the factors of $12$ will be $1,2,3,4,6,12$.
$\mathbf{20}$
Ans:
Given: number $20$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $20$ as,
$ 1 \times 20\\$
$2 \times 10\\$
$4 \times 5\\ $
Therefore, the factors of $20$ will be $1,2,4,5,10,20$.
$\mathbf{18}$
Ans:
Given: number $18$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $18$ as,
$ 1 \times 18\\$
$ 2 \times 9\\$
$ 3 \times 6\\ $
Therefore, the factors of $18$ will be $1,2,3,6,9,18$.
$23$
Ans:
Given: number $23$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $23$ as,
$1 \times 23$
Therefore, the factors of $23$ will be $1,23$.
$\mathbf{36}$
Ans:
Given: number $36$
We need to write all the factors of the given number.
A factor is a number that divides a given integer completely without leaving any reminder.
We can write $36$ as,
$ 1 \times 36\\$
$ 2 \times 18\\$
$3 \times 12\\$
$4 \times 9\\$
$6 \times 6\\ $
Therefore, the factors of $36$ will be $1,2,3,4,6,9,12,18,36$.
2. Write first five multiples of:
$\mathbf{5}$
Ans:
Given: number $5$
We need to write the first five multiples of the given number.
The product of a number and a counting number is a multiple of that number.
Thus,
$ 1 \times 5 = 5$
$ 2 \times 5 = 10$
$3 \times 5 = 15$
$4 \times 5 = 20$
$5 \times 5 = 25$
Therefore, the first five multiples of $5$ will be $5,10,15,20,25$.
$\mathbf{8}$
Ans:
Given: number $8$
We need to write the first five multiples of the given number.
The product of a number and a counting number is a multiple of that number.
Thus,
$1 \times 8 = 8$
$ 2 \times 8 = 16$
$3 \times 8 = 24$
$4 \times 8 = 32$
$5 \times 8 = 40$
Therefore, the first five multiples of $8$ will be $8,16,24,32,40$.
$\mathbf{9}$
Ans:
Given: number $9$
We need to write the first five multiples of the given number.
The product of a number and a counting number is a multiple of that number.
Thus,
$ 1 \times 9 = 9$
$ 2 \times 9 = 18$
$3 \times 9 = 27$
$ 4 \times 9 = 36$
$5 \times 9 = 45$
Therefore, the first five multiples of $9$ will be $9,18,27,36,45$.
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:
Given: two columns having different options
We need to match the correct items in column $1$ with the items in column $2$.
(i) $\mathbf{35}$
We can write $35$ as,
$7 \times 5$
So, $35$ is a multiple of $7$(b).
(ii) $\mathbf{15}$
We know that a factor of $30$ is $15$.
So, the correct option is (d).
(iii) $\mathbf{16}$
We can write $16$ as,
$8 \times 2$
So, $16$ is a multiple of $8$(a).
(iv) $\mathbf{20}$
We can write $20$ as,
$20 \times 1$
So, $20$ is a factor of \[20\].
So, the correct option is (f).
(v) $\mathbf{25}$
We know $25 \times 2 = 50$
So, $25$ is a factor of $50$(e).
Therefore,
Column $1$  Column $2$

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 $50$ 
4. Find all the multiples of $9$ up to $100$.
Ans:
Given: number $9$
We need to write all the multiples of $9$ up to $100$.
We know that the product of a number and a counting number is a multiple of that number.
Therefore,
$9 \times 1 = 9$
$9 \times 2 = 18$
$9 \times 3 = 27$
$9 \times 4 = 36$
$9 \times 5 = 45$
$9 \times 6 = 54$
$9 \times 7 = 63$
$9 \times 8 = 72$
$9 \times 9 = 81$
$9 \times 10 = 90$
$9 \times 11 = 99$
Therefore, all the multiples of $9$ up to $100$ are $9,18,27,36,45,54,63,72,81,90,99$.
Exercise 3.2
1. What is the sum of any two:
Odd numbers
Ans:
Given: two odd numbers
We need to find the sum of two odd numbers.
We know that odd numbers are the numbers which cannot be divided by $2$ completely.
Consider,
$ 1,3\\$
$ \Rightarrow 1 + 3\\$
$ = 4\\ $
Now consider,
$ 13,45\\$
$ \Rightarrow 13 + 45\\$
$ 58\\ $
Therefore, we can say that the sum of two odd numbers is always an even number.
Even Numbers
Ans:
Given: two even numbers
We need to find the sum of two even numbers.
We know that even numbers are the numbers which can be divided by $2$ completely.
Consider, two even numbers
$ 2,4\\$
$ \Rightarrow 2 + 4\\$
$ = 6\\ $
Now, consider two more even numbers
$ 98,40\\$
$ \Rightarrow 98 + 40\\$
$ = 138\\ $
Therefore, we can say that the sum of two even numbers is always an even number.
2. State whether the following statements are true or false:
The sum of three odd numbers is even.
Ans:
Given: the sum of three odd numbers is even
We need to state whether the statement is true or false.
Consider,
$ 1,29,99\\$
$ \Rightarrow 1 + 29 + 99\\$
$ = 129\\ $
This is odd. Therefore, the given statement is False.
The sum of two odd numbers and one even number is even.
Ans:
Given: the sum of two odd numbers and one even number is even
We need to state whether the statement is true or false.
Consider,
$ 1,95,56\\$
$ \Rightarrow 1 + 95 + 56\\$
$ = 152\\ $
This is even. Therefore, the given statement is True.
The product of three odd numbers is odd.
Ans:
Given: the product of three odd numbers is odd
We need to state whether the statement is true or false.
Consider,
$ 1,99,105\\$
$\Rightarrow 1 \times 99 \times 105\\$
$ = 10395\\ $
This is odd. Therefore, the given statement is True.
If an even number is divided by $2$, the quotient is always odd.
Ans: Given: If an even number is divided by $2$, the quotient is always odd.
We need to state whether the statement is true or false.
Consider, number $44$
Divide by $2$, we will get
$44 \div 2$
$= 22$
Thus, the quotient is not odd. Therefore, the given statement is False.
All prime numbers are odd.
Ans:
Given: All prime numbers are odd
We need to state whether the statement is true or false.
We know that $2$ is a prime number which is not odd. Therefore, the given statement is False.
Prime numbers do not have any factors.
Ans:
Given: Prime numbers do not have any factors.
We need to state whether the statement is true or false.
We know that a factor is a number that divides a given integer completely without leaving any reminder.
Consider, $23$
The factors of $23$ are $1,23$. Therefore, all prime numbers have factors itself and $1.$ Therefore, the statement is False.
Sum of two prime numbers is always even.
Ans:
Given: Sum of two prime numbers is always even.
We need to state whether the statement is true or false.
We know that all prime numbers are odd except $2.$ So, if we add two add numbers then the sum of those numbers is even.
Consider, $2,5$
Adding the numbers, we get
$2 + 5\\$
$= 7\\ $
It is not even. Therefore, the given statement is False.
2 is the only even prime number.
Ans:
Given: 2 is the only even prime number.
We need to state whether the statement is true or false.
We know that $2$ is the only prime number. All other prime numbers are odd. Therefore, the given statement is True.
All even numbers are composite numbers.
Ans:
Given: All even numbers are composite numbers
We need to state whether the statement is true or false.
We know that the composite numbers are the numbers having factors other than $1$ and the number itself.
Consider, $2$
This is an even number. But it does not have any factors other than $1$ and itself.
Therefore, $2$ is not a composite number. Therefore, the given statement is False.
The product of two even numbers is always even.
Ans:
Given: The product of two even numbers is always even.
We need to state whether the statement is true or false.
Consider, two even numbers $2,4$
Product of these numbers will be
$ 2 \times 4 \\$
$ = 8 \\ $
It is even.
Now consider, $48,98$
Product of these numbers will be
$48 \times 98 \\$
$= 4704 \\ $
It is also even.
Therefore, the statement is True.
3. The numbers $13$ and $31$ are prime numbers. Both these numbers have the same digits $1$ and $3$. Find such pairs of prime numbers up to $100$.
Ans:
Given: The numbers $13$ and $31$ are prime numbers. Both these numbers have same digits $1$ and $3$
We need to find such pairs of prime numbers up to $100$.
The prime numbers up to $100$ are
$2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97$
Therefore, the pairs of prime number will be
$17\;{\text{and }}71.$
4. Write down separately the prime and composite numbers less than $20$.
Ans:
Given: $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20$
We need to write separately the prime and composite numbers less than $20$.
We know that the composite numbers are the numbers having factors other than $1$ and the number itself.
Prime numbers are the numbers having factors $1$ and itself.
Therefore, prime numbers will be
$2,3,5,7,11,13,17,19$
The composite numbers will be
$4,6,8,9,10,12,14,15,16,18$
5. What is the greatest prime number between $1$ and $10$?
Ans:
Given: prime numbers $2,3,5,7$
We need to find the greatest prime number between $1$ and $10$.
Therefore, we can see that the greatest prime number among the given numbers is $7.$
6. Express the following as the sum of two odd numbers:
$44$
Ans:
Given: $44$
We need to express the given numbers as the sum of two odd numbers.
We can write the given number $44$ as
$ 5 + 39 \\$
$ = 44 \\ $
$36$
Ans:
Given: $36$
We need to express the given numbers as the sum of two odd numbers.
We can write the given number $36$ as
$ 9 + 27 \\$
$= 36 \\ $
$24$
Ans:
Given: $24$
We need to express the given numbers as the sum of two odd numbers.
We can write the given number $24$ as
$3 + 21 \\$
$ = 24 \\ $
$18$
Ans:
Given: $18$
We need to express the given numbers as the sum of two odd numbers.
We can write the given number $18$ as
$5 + 13 \\$
$= 18 \\ $
7. Give three pairs of prime numbers whose difference is $2$. Remark: Two prime numbers whose difference is $2$ are called twin primes.
Ans:
Given: prime numbers
We need to find three pairs of prime numbers whose difference is $2$.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, the three pairs of prime numbers whose difference is $2$ will be
$5{\text{and }}7 $
$ 11{\text{ and }}13 $
$17{\text{and }}19 $
8. Which of the following numbers are prime?
$\mathbf{23}$
Ans:
Given: $23$
We need to find if the given number is prime.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, $23$ can be written as $1 \times 23$.
Thus, it is a prime number.
$\mathbf{51}$
Ans:
Given: $51$
We need to find if the given number is prime.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, $51$ can be written as $3 \times 17$.
Thus, $51$ is not a prime number.
$\mathbf{37}$
Ans:
Given: $37$
We need to find if the given number is prime.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, $37$ can be written as $1 \times 37$.
Thus, it is a prime number.
$\mathbf{26}$
Ans:
Given: $26$
We need to find if the given number is prime.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, $26$ can be written as $2 \times 13$.
Thus, $26$ is not a prime number.
9. Write seven consecutive composite numbers less than $100$ so that there is no prime number between them.
Ans:
Given: Numbers less than $100$
We need to write seven consecutive composite numbers less than $100$ so that there is no prime number between them.
We know that prime numbers less than $100$ are $2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97$.
Therefore, seven consecutive composite numbers less than $100$ so that there is no prime number between them will be
$90,91,92,93,94,95,96$.
10. Express each of the following numbers as the sum of three odd primes:
$\mathbf{21}$
Ans:
Given: $21$
We need to express the given number as the sum of three odd primes.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, we can write $21$ as
$21 = 5 + 7 + 11$
$\mathbf{31}$
Ans:
Given: $31$
We need to express the given number as the sum of three odd primes.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, we can write $31$ as
$31 = 5 + 7 + 19$
$\mathbf{53}$
Ans:
Given: $53$
We need to express the given number as the sum of three odd primes.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, we can write $53$ as
$53 = 11 + 13 + 29$
$\mathbf{61}$
Ans:
Given: $61$
We need to express the given number as the sum of three odd primes.
We know that prime numbers are the numbers having factors $1$ and itself.
Therefore, we can write $61$ as
$61 = 3 + 5 + 53$
11. Write five pairs of prime numbers less than $20$ whose sum is divisible by $5$.
Ans:
Given: $2,3,5,7,11,13,17,19$
We need to write five pairs of prime numbers less than $20$ whose sum is divisible by $5$.
We know that prime numbers are the numbers having factors $1$ and itself.
A number is divisible by $5$ if the unit place digit is either $0$ or $5.$
Therefore, the pairs will be
$3 + 7 = 10$
$2 + 3 = 5$
$13 + 2 = 15$
$13 + 7 = 20$
$3 + 17 = 20$
$5 + 5 = 10$
12. Fill in the blanks:
A number which has only two factors is called a __________.
Ans:
We need to fill in the blanks with appropriate numbers or words.
A number which has only two factors is called a prime number.
A number which has more than two factors is called a _________.
Ans:
We need to fill in the blanks with appropriate numbers or words.
A number which has more than two factors is called a composite number.
$\mathbf{1}$ neither ________ nor _________.
Ans:
We need to fill in the blanks with appropriate numbers or words.
$1$ neither prime nor composite number.
The smallest prime number is ________.
Ans:
We need to fill in the blanks with appropriate numbers or words.
The smallest prime number is $2.$
The smallest composite number is ________.
Ans:
We need to fill in the blanks with appropriate numbers or words.
The smallest composite number is $4$.
The smallest even number is __________.
Ans:
We need to fill in the blanks with appropriate numbers or words.
The smallest even number is $2$.
Exercise 3.3
1. Using divisibility test, determine which of the following numbers are divisible by $2;$by $3;$ by $3;$ by $5;$ by $6;$ by $8;$ by $9;$ by $10;$ by $11.$(say yes or no)
Number  Divisible by  
$2$  $3$  $4$  $5$  $6$  $8$  $9$  $10$  $11$  
$128$  
$990$  
$1586$  
$275$  
$6686$  
$639210$  
$429714$  
$2856$  
$3060$  
$406839$ 
Ans:
We need to determine that which of the given numbers are divisible by \[2;\] by $3;$ by $4;$ by $5;$ by $6;$ by $8;$ by $9;$ by $10;$ by $11.$
Number  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  No  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  Yes  No  No  No  No 
$639210$  Yes  No  No  Yes  Yes  No  No  Yes  Yes 
$429714$  Yes  Yes  No  No  Yes  No  Yes  No  No 
$2856$  Yes  Yes  No  No  Yes  Yes  No  No  No 
$3060$  Yes  Yes  Yes  Yes  Yes  No  Yes  Yes  No 
$406839$  No  Yes  No  No  No  No  No  No  No 
2. Using divisibility test, determine which of the following numbers are divisible by 4; by $8:$
$\mathbf{572}$
Ans: Given: $572$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are divisible by $4;$ hence the number is divisible by $4.$
The given number is not divisible by $8,$ as the last three digits are not divisible by
$8.$
726352
Ans: Given: \[726352{\text{ }}\]
We need to determine whether the given number is divisible by $4;$ by $8.$
As the last two digits of the given number are divisible by $4,$ hence the given number is divisible by $4.$
The given number is divisible by$8,$ as the last three digits are divisible by $8.$
5500
Ans: Given:$5500$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are divisible by $4;$ hence the number is divisible by $4.$
The given number is not divisible by $8,$ as the last three digits are not divisible by
$8.$
6000
Ans: Given: \[6000\]
We need to determine whether the given number is divisible by $4;$ by $8.$
As the last two digits of the given number are divisible by $4,$ hence the given number is divisible by $4.$
The given number is divisible by $8,$ as the last three digits are divisible by $8.$
12159
Ans: Given: $12159$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are not divisible by $4;$ hence the number is not divisible by $4.$
The given number is not divisible by $8,$ as the last three digits are not divisible by
$8.$
14560
Ans: Given: $12159$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are divisible by $4;$ hence the number is divisible by $4.$
The given number is divisible by $8,$ as the last three digits are divisible by
$8.$
21084
Ans:Given: $21084$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are divisible by $4;$ hence the number is divisible by $4.$
The given number is not divisible by $8,$ as the last three digits are not divisible by
$8.$
31795072
Ans: Given: $31795072$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are divisible by $4;$ hence the number is divisible by $4.$
The given number is divisible by $8,$ as the last three digits are divisible by
$8.$
1700
Ans: Given: $1700$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are divisible by $4;$ hence the number is divisible by $4.$
The given number is not divisible by $8,$ as the last three digits are not divisible by
$8.$
2150
Ans: Given: $2150$
We need to determine whether the given number is divisible by $4;$ by $8.$
Here the last two digits of the number are not divisible by $4;$ hence the number is not divisible by $4.$
The given number is not divisible by $8,$ as the last three digits are not divisible by
$8.$
3. Using divisibility test, determine which of the following numbers are divisible by $6:$
297144
Ans:
Given: $297144$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is divisible by $3$ as the sum of digits is divisible by $3.$
As the number is divisible by both $2$ and $3,$hence the given number is divisible by $6.$
1258
Ans: Given: $1258$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is not divisible by $3$ as the sum of digits $( = 16)$ is not divisible by $3.$
As the number is divisible by $2$ but not by $3,$ hence the given number is not divisible by $6.$
4335
Ans:
Given: $4335$
We need to find whether the given number is divisible by $6.$
The units place digit is an odd number; hence the given number is not divisible by $2.$
The given number is divisible by $3$ as the sum of digits $( = 15)$ is divisible by $3.$
As the number is not divisible by both $2$ and $3,$ hence the given number is not divisible by $6.$
61233
Ans: Given: $61233$
We need to find whether the given number is divisible by $6.$
The units place digit is an odd number; hence the given number is not divisible by $2.$
The given number is divisible by $3$ as the sum of digits$( = 15)$ is divisible by $3.$
As the number is not divisible by both $2$ and $3,$ hence the given number is not divisible by $6.$
901352
Ans: Given: $901352$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is not divisible by $3$ as the sum of digits $( = 20)$ is not divisible by $3.$
As the number is not divisible by both $2$ and $3,$ hence the given number is not divisible by $6.$
438750
Ans: Given: $438750$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is divisible by $3$ as the sum of digits $( = 27)$ is divisible by $3.$
As the number is divisible by both $2$ and $3,$ hence the given number is divisible by $6.$
1790184
Ans: Given: $1790184$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is divisible by $3$ as the sum of digits $( = 30)$ is divisible by $3.$
As the number is divisible by both $2$ and $3,$ hence the given number is divisible by $6.$
12583
Ans: Given: $12583$
We need to find whether the given number is divisible by $6.$
The units place digit is an odd number; hence the given number is not divisible by $2.$
The given number is not divisible by $3$ as the sum of digits $( = 19)$ is not divisible by $3.$
As the number is not divisible by both $2$ and $3,$ hence the given number is not divisible by $6.$
639210
Ans: Given: $438750$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is divisible by $3$ as the sum of digits $( = 21)$ is divisible by $3.$
As the number is divisible by both $2$ and $3,$ hence the given number is divisible by $6.$
17852
Ans: Given: $17852$
We need to find whether the given number is divisible by $6.$
The units place digit is an even number; hence the given number is divisible by $2.$
The given number is not divisible by $3$ as the sum of digits $( = 23)$ is not divisible by $3.$
As the number is divisible by $2$ but not divisible by $3,$ hence the given number is not divisible by $6.$
4. Using divisibility test, determine which of the following numbers are divisible by $11:$
5445
Ans:
Given: $5445$
We need to find whether the given number is divisible by $11$ or not.
The sum of digits at odd places is $4 + 5 = 9$ and sum of digits at even places is $4 + 5 = 9$
Difference of both sums is $9  9 = 0.$
As the difference is $0,$ therefore, the number is divisible by $11.$
10824
Ans: Given: $10824$
We need to find whether the given number is divisible by $11$ or not.
The sum of digits at odd places is $4 + 8 + 1 = 13$ and sum of digits at even places is $2 + 0 = 2$
Difference of both sums is $13  2 = 11$
As the difference is $11,$ therefore, the number is divisible by $11.$
7138965
Ans: Given: $7138965$
We need to find whether the given number is divisible by $11$ or not.
The sum of digits at odd places is $5 + 9 + 3 + 7 = 24$ and sum of digits at even places is $6 + 8 + 1 = 15$
Difference of both sums is $24  15 = 9$
As the difference is $9,$ therefore, the number is not divisible by $11.$
70169308
Ans: Given: $70169308$
We need to find whether the given number is divisible by $11$ or not.
The sum of digits at odd places is $8 + 3 + 6 + 0 = 17$ and sum of digits at even places is $0 + 9 + 1 + 7 = 17$
Difference of both sums is $17  17 = 0.$
As the difference is $0,$ therefore, the number is divisible by $11.$
10000001
Ans: Given: $10000001$
We need to find whether the given number is divisible by $11$ or not.
The sum of digits at odd places is $1 + 0 + 0 + 0 + = 1$ and sum of digits at even places is $0 + 0 + 0 + 1 = 1.$
Difference of both sums is $1  1 = 0.$
As the difference is $0,$ therefore, the number is divisible by $11.$
901153
Ans: Given: $5445$
We need to find whether the given number is divisible by $11$ or not.
The sum of digits at odd places is $3 + 1 + 0 = 4$ and sum of digits at even places is $5 + 1 + 9 = 15.$
Difference of both sums is $15  4 = 11.$
As the difference is $11,$ therefore, the number is divisible by $11.$
5. Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisible by \[3:\]
_6724
Ans: Given: $\_6724$
We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by $3.$
A number is divisible by $3$ if the sum of digits is divisible by $3.$
Hence, Smallest digit is \[2{\text{ }} \to {\text{ }}26724{\text{ }} = {\text{ }}2{\text{ }} + {\text{ }}6{\text{ }} + {\text{ }}7{\text{ }} + {\text{ }}2{\text{ }} + {\text{ }}4{\text{ }} = {\text{ }}21\]
And Largest digit is \[{\text{ }}8{\text{ }} \to {\text{ }}86724{\text{ }} = {\text{ }}8{\text{ }} + {\text{ }}6{\text{ }} + {\text{ }}7{\text{ }} + {\text{ }}2{\text{ }} + {\text{ }}4{\text{ }} = {\text{ }}27{\text{ }}\]
4765_2
Ans: Given: $4765\_2$
We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by $3.$
A number is divisible by $3$ if the sum of digits is divisible by $3.$
Hence, Smallest digit is \[{\text{ }}0{\text{ }} \to {\text{ }}476502{\text{ }} = {\text{ }}4{\text{ }} + {\text{ }}7{\text{ }} + {\text{ }}6{\text{ }} + {\text{ }}5{\text{ }} + {\text{ }}0{\text{ }} + {\text{ }}2{\text{ }} = {\text{ }}24\]
And Largest digit is \[9{\text{ }} \to {\text{ }}476592{\text{ }} = {\text{ }}4{\text{ }} + {\text{ }}7{\text{ }} + {\text{ }}6{\text{ }} + {\text{ }}5{\text{ }} + {\text{ }}0{\text{ }} + {\text{ }}2{\text{ }} = {\text{ }}33\]
6. Write the smallest digit and the largest digit in the blanks space of each of the following numbers so that the number formed is divisible by $11:$
$\mathbf{92\_\_\_\_389}$
Ans: Given: $92\_\_\_\_389$
We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by $11.$
Let this number is $x$
A number is divisible by $11$ if the difference of the sum of digits at odd and even places is either or $11.$
Hence, \[92x389\text{ }\!\!~\!\!\text{ }\to \]
Sum of digits at even places \[=\text{ }9\text{ }+\text{ }x\text{ }+\text{ }8\text{ }=\text{ }17\text{ }+\ x\text{ }\]
Sum of digits at odd places \[= {\text{ }}2{\text{ }} + {\text{ }}3{\text{ }} + {\text{ }}9{\text{ }} = {\text{ }}14\]
Number will be divisible by 11 if
\[\text{ }17\text{ }+\ x\text{ }14\ =\ 11\]
$ \Rightarrow 3+\ x\ =\ 11 \\ $
$ \Rightarrow \ x\ =\ 113\ =8 \\$
Also
$ \text{ }17\text{ }+\ x\text{ }14\ =\ 0 $
$ \Rightarrow \ 3+x\ =\ 0 $
$ \Rightarrow x\ =\ 03\ =\,3 $
Which is not possible
So the only digit that can be placed in this blank space is $8$
So number is \[\text{928389}\]
$\mathbf{8\_\_\_\_9484}$
Ans: Given: $8\_\_\_\_9484$
We need to find the smallest digit and the largest digit in the blank space so that the number formed is divisible by $11.$
Let this number is $x$
A number is divisible by $11$ if the difference of the sum of digits at odd and even places is either or $11.$
Hence, \[8x9484\text{ }\!\!~\!\!\text{ }\to \]
Sum of digits at even places \[=\text{ }8\text{ }+\text{ }9\text{ }+\text{ }8\text{ }= {\text{ }}25\]
Sum of digits at odd places \[= {\text{ }}x{\text{ }} + {\text{ }}4{\text{ }} + {\text{ }}4{\text{ }} = \text{ }8\text{ }+\ x\text{ }\]
Number will be divisible by $11$ if
$ 25\ (8+x)\ =\ 11 $
$ 258\ \ x\ =\ 11 $
$ \Rightarrow \ 17\ \ x\ =\ 11 $
$ \Rightarrow x\ =\ 1117 $
$ \Rightarrow x\ =\ 6 $
$ \Rightarrow x\ =\ 6 $
So the only digit that can be placed in this blank space is $6$
So number is \[\text{869484}\]
Exercise 3.5
1. Here are two different factors trees for \[60\]. Write the missing numbers.
(a)
Ans: Let’s first consider the number missing in left branch:
Prime factorisation of \[6\] is \[2 \cdot 3\]. \[2\] is given but \[3\]is missing , therefore number missing in left branch is \[3\] .
Now let’s consider the number missing in right branch:
Prime factorisation of \[10\] is \[2 \cdot 5\]. \[5\] is given but \[2\] is missing in the right branch, therefore number missing in right branch is \[2\].
(b)
Ans: Let consider the first branch:
Factor pair of \[60\] which includes \[30\] is \[60 = 2 \cdot 30\] . Because \[30\] is given but \[2\]is missing , therefore number missing in the first branch is \[2\].
Now let’s consider the second branch:
Factor pair of \[30\] which includes \[10\]is \[30 = 3 \cdot 10\] . Because \[10\]is given but \[3\] is missing , therefore number missing in the second branch is \[3\].
Now let’s consider the third branch:
Prime factorization of \[10\] is \[2 \cdot 5\] therefore \[2\]and \[5\]are the missing numbers in the third branch.
2. Which factors are not included in the prime factorization of a composite number?
Ans: Composite number is not a prime number because composite number is a number with more than two factors, i.e., \[1\]and the number itself. \[1\] is not a prime number. Because prime factorisation involves only prime factors, \[1\] and the composite number itself are not included in the prime factorisation of a composite number.
3. Write the greatest 4digit number and express it in terms of its prime factors.
Ans: Greatest 4 digit number is \[9999\].
We divide the number \[9999\]by \[2,3,5,7\] etc. in this order repeatedly so long as the quotient becomes \[1\] to get its prime factorization.
$ 3\left \!{\underline {\, {9999} \,}} \right. \\ 3\left \!{\underline {\, {3333} \,}} \right. \\ 11\left \!{\underline {\, {1111} \,}} \right. \\ 101\left \!{\underline {\, {101} \,}} \right. \\ \,\,\,\,\,\,\,\,\, {1} $
Therefore \[9999\] in terms of prime factors is \[{3^2} \cdot 11 \cdot 101\]
4. Write the smallest 5digit number and express it in the form of its prime factors.
Ans: Smallest 5 digit number is\[10000\].
We divide the number \[10000\]by \[2,3,5,7\] etc. in this order repeatedly so long as the quotient becomes \[1\] to get its prime factorization.
$ 2\left \!{\underline {\, {10000} \,}} \right. \\ 2\left \!{\underline {\, {5000} \,}} \right. \\ 2\left \!{\underline {\, {2500} \,}} \right. \\ 5\left \!{\underline {\, {1250} \,}} \right. \\ 5\left \!{\underline {\, {625} \,}} \right. \\ 5\left \!{\underline {\, {125} \,}} \right. \\ 5\left \!{\underline {\, {25} \,}} \right. \\ 5\left \!{\underline {\, {5} \,}} \right. \\ \,\,\,\,\,\,\, {1} $
Therefore, \[10000\] in terms of prime factors is \[{2^4} \cdot {5^4}\].
5. 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: We divide the number \[1729\]by \[2,3,5,7\] etc. in this order repeatedly so long as the quotient becomes \[1\] to get its prime factorization.
$ 7\left\!{\underline {\, {1729} \,}} \right. \\ 13\left\!{\underline {\, {247} \,}} \right. \\ 19\left \!{\underline{\, {19} \,}} \right.\\ \,\,\,\,\,\,\,\, {1} $
Therefore, \[1729\] in terms of prime factors is \[7 \cdot 13 \cdot 19\].
Prime factors of \[1729\]in ascending order are \[7,13,19\].
One can observe that \[13  7 = 19  13 = 6\],i.e., difference between two consecutive prime factors is \[6\]. Therefore, relation between two consecutive prime factors of \[1729\]is that the difference between them is a constant, \[6\].
6. The product of three consecutive numbers is always divisible by \[6\] . Verify this statement with the help of some examples.
Ans: Consider \[4,5\] and \[6\]. Product of\[4,5\] and \[6\] is \[120\].
\[\frac{{120}}{6} = 20\] ,i.e., \[120\]is divisible by 6.
Consider \[5,6\]and \[7\]. Product of\[5,6\]and \[7\]is \[210\].
\[\frac{{210}}{6} = 35\] , i.e., \[210\]is divisible by 6
7. The sum of two consecutive odd numbers is divisible by \[4\]. Verify this statement with the help of some examples.
Ans: Consider \[3\]and\[5\]. The sum of \[3\]and\[5\]is \[8\].
\[\frac{8}{4} = 2\],i.e., \[8\]is divisible by \[4\].
Consider \[7\]and \[9\] . The sum of \[7\]and \[9\] is \[16\].
\[\frac{{16}}{8} = 2\],i.e., \[16\]is divisible by \[4\].
8. In which of the following expressions, prime factorisation has been done?
(a) \[24 = 2 \cdot 3 \cdot 4\]
Ans: In the given expression\[4\]is not factorised further into its prime factors ,i.e,. Therefore, in the given expression, prime factorisation has not been done.
(b) \[56 = 7 \cdot 2 \cdot 2 \cdot 2\]
Ans: In the given expression, all factors are prime, therefore in the expression prime factorisation has been done.
(c) \[70 = 2 \cdot 5 \cdot 7\]
Ans: In the given expression, all factors are prime , therefore in the expression prime factorisation has been done.
(d) \[54 = 2 \cdot 3 \cdot 9\]
Ans: In the given expression \[9\]is not factorised further into its prime factors ,i.e,\[3 \cdot 3\]. Therefore in the given expression, prime factorisation has not been done.
9. \[18\] is divisible by both \[2\] and \[3\]. It is also divisible by \[2 \cdot 3 = 6\]. Similarly, a number is divisible by both \[4\]and\[6\]. Can we say that the number must also be divisible by \[4 \cdot 6 = 24\]? If not, give an example to justify your answer.
Ans: No, we can not say that if a number is divisible by both \[4\]and \[6\]must be divisible by \[4 \cdot 6 = 24\].
For example, consider \[12\] :
\[12\]is divisible by both \[4\]and \[6\]but is not divisible by \[24\].
10. I am the smallest number, having four different prime factors. Can you find me?
Ans: The smallest number with four different prime factors must be the product of the first four prime numbers ,i.e., \[2,3,5\]and \[7\].
Therefore smallest number with four different prime factors is :
\[2 \cdot 3 \cdot 5 \cdot 7 = 210\]
Exercise 3.6
1.Find the H.C.F. of the following numbers:
\[\mathbf{18,{\text{ }}48}\]
Ans:
Given: \[18,{\text{ }}48\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[18\] will be
\[{\text{ = 2 x 3 x 3}}\]
Factors of \[48\]will be
\[{\text{ = 2 x 2 x 2 x 2 x 3}}\]
Take common factors of \[18\]and \[48\], we get
H.C.F. of \[18,{\text{ }}48\]
$2 \times 3 = 6 \\ $
\[\mathbf{{\mathbf{30}},{\text{ }}{\mathbf{42}}}\]
Ans:
Given: \[30,42\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[30\] will be
\[{\text{ = 2 x 3 x 5}}\]
Factors of \[42\] will be
\[{\text{ = 2 x 3 x 7}}\]
Take common factors of \[30\] and \[42\], we get
H.C.F of \[30\] and \[42\]
$2 \times 3 = 6 \\ $
\[{\mathbf{18}},{\text{ }}{\mathbf{60}}\]
Ans:
Given: \[18,60\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[18\] will be
\[{\text{ = 2 x 3 x 3}}\,\]
Factors of \[60\] will be
\[{\text{ = 2 x 2 x 3 x 5}}\]
Take common factors of \[18\] and \[60\] , we get
H.C.F. of \[18\], \[60\]
$ {{\text{ = 2 x 3}}} \\ $
$ {{\text{ = 6}}} $
\[{\mathbf{27}},{\text{ }}{\mathbf{63}}\]
Ans:
Given: \[27,63\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[27\] will be
\[{\text{ = 3 x 3 x 3}}\]
Factors of \[63\] will be
\[{\text{ = 3 x 3 x 7}}\]
Take common factors of \[27\] and \[63\],we get
H.C.F. of \[27\], \[63\]
$ {{\text{ = 3 x 3}}} \\ $
$ {{\text{ = 9}}} $
\[{\mathbf{36}},{\mathbf{84}}\]
Ans:
Given: \[36,84\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[36\] will be
\[{\text{ = 2 x 2 x 3 x 3}}\]
Factors of \[84\] will be
\[{\text{ = 2 x 2 x 3 x 7}}\]
Take common factors of \[36\] and \[84\] , we get
H.C.F. of \[36\],\[84\]
$ {{\text{ = 2 x 2 x 3}}} \\$
${{\text{ = 12}}} $
\[{\mathbf{34}},{\mathbf{102}}\]
Ans:
Given: \[34,102\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[\;34\] will be
\[{\text{ = 2 x 17}}\]
Factors of \[102\] will be
\[{\text{ = 2 x 3 x 17}}\]
Take common factors of \[\;34\] and \[102\] ,we get
H.C.F. of \[\;34,102\]
${{\text{ = 2 x 17}}} \\ $
$ {{\text{ = 34}}} $
\[{\mathbf{70}},{\text{ }}{\mathbf{105}},{\text{ }}{\mathbf{175}}\]
Ans:
Given: \[70,{\text{ }}105,{\text{ }}175\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[70\]will be
= 2 x 5 x 7
Factors of \[105\] will be
= 3 x 5 x 7
Factors of \[175\]will be
= 5 x 5 x 7
Take common factors of \[70\],\[{\text{105}}\] and \[175\], we get
H.C.F. of \[70,{\text{ 105}},{\text{ }}175\]
${{\text{ = 7 x 5}}} \\$
$ {{\text{ = 35}}} $
\[{\mathbf{91}},{\text{ }}{\mathbf{112}},{\text{ }}{\mathbf{49}}\]
Ans:
Given:\[91,{\text{ }}112,{\text{ }}49\]
We need to find H.C.F of the given numbers .
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[91\] will be
\[{\text{ = 7 x 13}}\]
Factors of \[112\] will be
\[{\text{ = 2 x 2 x 2 x 2 x 7}}\]
Factors of \[{\text{ }}49\] will be
\[{\text{ = 7 x 7}}\]
Take common factors of \[91,{\text{ }}112\] and \[{\text{ }}49\], we get
H.C.F. of \[91,{\text{ }}112,{\text{ }}49\]
$ {\text{ = 1}} \times {\text{7}}\\$
$ {\text{ = 7}}\\ $
\[{\mathbf{18}},{\text{ }}{\mathbf{54}},{\text{ }}{\mathbf{81}}\]
Ans:
Given: \[18,{\text{ }}54,{\text{ }}81\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[18\] will be
\[{\text{ = 2 x 3 x 3}}\]
Factors of \[54\] will be
\[ = 2 \times 3 \times 3 \times 3\]
Factors of \[{\text{ }}81\] will be
\[{\text{ = 3 x 3 x 3 x 3}}\]
Take common factors of \[18,{\text{ }}54\] and \[{\text{ }}81\], we get
H.C.F. of \[18,{\text{ }}54,{\text{ }}81\]
$ {{\text{ = 3 x 3}}} \\ $
$ {{\text{ = 9}}} $
\[{\mathbf{12}},{\text{ }}{\mathbf{45}},{\text{ }}{\mathbf{75}}\]
Ans:
Given: \[12,{\text{ }}45,{\text{ }}75\]
We need to find H.C.F of the given numbers.
We know that the bigger factor that splits two or more numbers is the highest common factor.
Thus,
Factors of \[12\] will be
\[{\text{ = 2 x 2 x 3}}\]
Factors of \[45\] will be
\[{\text{ = 3 x 3 x 5}}\]
Factors of \[{\text{ }}75\] will be
\[{\text{ = 3 x 5 x 5}}\]
Take common factors of \[12,{\text{ }}45\]and \[75\],we get
H.C.F. of \[12,{\text{ }}45,{\text{ }}75\]
${{\text{ = 1 x 3}}} \\ $
$ {{\text{ = 3}}} \\ $
2. What is the H.C.F. of two consecutive:
Numbers?
Ans:
We have to find H.C.F. of two consecutive numbers:
We know that the bigger factor that splits two or more numbers is the highest common factor.
Let us take consecutive numbers \[2\] and \[3\],
Thus,
Factors of \[\;2\]
\[{\text{ = 2 x 1}}\]
Factors of \[3\]
\[{\text{ = 3 x 1}}\]
Hence, the H.C.F. of two consecutive numbers is \[1\].
Even Numbers?
Ans: We have to find H.C.F. of two even consecutive numbers:
We know that the bigger factor that splits two or more numbers is the highest common factor.
Let us take even consecutive numbers \[2\] and \[4\].
Factors of \[2\] will be
\[ = 1 \times 2\]
Factors of \[4\] will be
\[ = 2 \times 2\]
Take common factors of \[2\]and \[4\], we get
H.C.F. of \[2\] and \[4\]
\[ = 2\]
Hence, H.C.F. of two even consecutive numbers is \[2\].
Odd Numbers?
Ans: We have to find H.C.F. of two odd consecutive numbers:
We know that the bigger factor that splits two or more numbers is the highest common factor.
Let us take odd consecutive numbers\[3\]and \[5\],
Factors of \[3\] will be
\[ = 1 \times 3\]
Factors of \[5\]will be
\[ = 1 \times 5\]
Take common factors of \[3\] and \[5\], we get
H.C.F. of \[3\]and \[5\]
\[ = 1\]
Hence, H.C.F. of two consecutive odd numbers is \[1\].
3. H.C.F. of coprime numbers \[{\mathbf{4}}\] and \[{\mathbf{15}}\] was found as follows by factorization:
\[{\mathbf{4}} = {\mathbf{2}}{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{2}}\] and \[{\mathbf{15}} = {\mathbf{3}}{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{5}}\] since there is no common prime factor, so H.C.F. of \[4\] and \[15\] is \[0\]. Is the answer correct? If not, what is the correct H.C.F.?
Ans: Given:
Factors of \[4\]
\[ = 2 \times 2\]
Factors of \[15\]
\[ = 3 \times 5\]
H.C.F. of coprime numbers \[4\]and \[15\] is \[0\].
But this statement is wrong.
\[\because \]\[1\] is the common factor of each and every number.
Factors of \[4\]will be
\[ = 1 \times 2 \times 2\]
Factors of \[15\]will be
\[ = 1 \times 3 \times 5\]
Hence, H.C.F of coprime numbers \[4\] and \[15\] is \[1\]
Exercise 3.7
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 exact number of times.
Ans: It is given renu purchases two bags of fertilizer of weights 75kg and 69kg.
To find the maximum weight which can measure the weight of the fertilizer an exact number of times, we need to find the Highest Common Factor of 75 and 69.
Now list the factors of 75 and 69.
Factors of \[75 = 5 \times 3 \times 5\]
Factors of \[69 = 23 \times 3\]
We can see that the Highest Common Factor of 75 and 69 is 3.
Therefore the Maximum required weight is 3kg.
2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the maximum distance each should cover so that all can cover the distance in complete steps?
Ans: The step measure of the three boys is given as 63cm, 70cm, and 77cm.
To find the maximum distance each should cover so that all the three boys can cover the distance in complete steps, we need to find the Least Common Multiple of 63, 70 and 77.
Now let us find the LCM(63,70,77).
$ 2\left \!{\underline {\, {63,70,77} \,}} \right. \\ 3\left \!{\underline {\, {63,35,77} \,}} \right. \\ 3\left \!{\underline {\, {21,35,77} \,}} \right. \\ 5\left \!{\underline {\, {7,35,77} \,}} \right. \\ 7\left \!{\underline {\, {7,7,77} \,}} \right. \\ 11\left \!{\underline {\, {1,1,11} \,}} \right. \\ 1\left \!{\underline {\, {1,1,1} \,}} \right. $
LCM (63,70, 77) = \[7 \times 9 \times 10 \times 11 = 6930cm\]
Therefore, the maximum distance each of the boys should cover is 6930cm.
3. 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: It is given that,
Length of the room = 825cm.
Breadth of the room = 675cm.
Height of the room = 450cm.
To find the measurement of the tape we need to find the Highest Common Factor of 825, 675, 450.
Factors of \[825 = 5 \times 5 \times 3 \times 11\]
Factors of \[675 = 5 \times 5 \times 3 \times 3 \times 3\]
Factors of \[450 = 2 \times 3 \times 3 \times 5 \times 5\]
Therefore the HCF (825, 675, 450) =\[3 \times 5 \times 5 = 75\]
So, the length of the tape that can measure the three dimensions of the room exactly is 75cm.
4. Determine the smallest 3digit number which is exactly divisible by 6, 8 and12.
Ans: To find the smallest 3digit number that is exactly divisible by 6, 8 and 12 , first we need to find the Least Common Multiple of 6, 8 and 12.
$ 2\left \!{\underline {\, {6,8,12} \,}} \right. \\ 2\left \!{\underline {\, {3,4,6} \,}} \right. \\ 2\left \!{\underline {\, {3,2,3} \,}} \right. \\ 3\left \!{\underline {\, {3,1,3} \,}} \right. \\ 1\left \!{\underline {\, {1,1,1} \,}} \right. $
Therefore the LCM (6,8,12) = \[2 \times 2 \times 2 \times 3 = 24\]
The smallest three digit number divisible by 24 is the number that is exactly divisible by 6, 8 and 12.
120 is the smallest 3digit number divisible by 24.
Hence, the smallest 3digit number that is exactly divisible by 6,8,12 is 120.
5. Determine the largest 3digit number which is exactly divisible by 8, 10 and12.
Ans: To find the largest 3digit number that is exactly divisible by 8, 10 and 12, we need to find the Least Common Multiple of 8, 10 and 12.
$ 2\left \!{\underline {\, {8,10,12} \,}} \right. \\ 2\left \!{\underline {\, {4,5,6} \,}} \right. \\ 2\left \!{\underline {\, {2,5,3} \,}} \right. \\ 3\left \!{\underline {\, {1,5,3} \,}} \right. \\ 5\left \!{\underline {\, {1,5,1} \,}} \right. \\ 1\left \!{\underline {\, {1,1,1} \,}} \right. $
Therefore the LCM (8,10,12) = \[2 \times 2 \times 2 \times 3 \times 5 = 120\]
The largest 3digit number divisible by 120 is the number that is exactly divisible by 8, 10 and 12.
960 is the largest 3digit number divisible by 120
So, the largest 3digit number that is exactly divisible by 8, 10 and 12 is 960.
6. 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: It is given that 3 different traffic lights change after every 48 seconds, 72 seconds and 108 seconds.
In order to find after how many seconds the traffic light changes simultaneously we need to find the Least Common Multiple of 48, 72 and 108.
$ 2\left \!{\underline {\, {48,72,108} \,}} \right. \\ 2\left \!{\underline {\, {24,36,54} \,}} \right. \\ 2\left \!{\underline {\, {12,18,27} \,}} \right. \\ 3\left \!{\underline {\, {6,9,27} \,}} \right. \\ 3\left \!{\underline {\, {2,3,9} \,}} \right. \\ 2\left \!{\underline {\, {2,1,3} \,}} \right. \\ 3\left \!{\underline {\, {1,1,3} \,}} \right. \\ 1\left \!{\underline {\, {1,1,1} \,}} \right. $
Therefore the LCM (48,72,108) =\[2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 3 = 432\]sec.
After 432sec the 3 different traffic signals will change simultaneously.
i.e. if the change simultaneously at 7.am then,
432 sec = 7 minutes 12 seconds.
Therefore the time they will change simultaneously again = 7.am+7minutes 12 seconds
= 7:07:12am
The time the 3 different traffic signal changes simultaneously is 7:07:12am
7. 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 three containers the exact number of times.
Ans: It is given that 3 tankers can contain a diesel of 403 litres, 434 litres and 465 litres. In order to find the maximum capacity of the container we need to find the Highest Common Factor of 403, 434 and 465 litres.
Factors of \[403 = 31 \times 13\]
Factors of \[434 = 7 \times 2 \times 31\]
Factors of \[465 = 31 \times 3 \times 5\]
Therefore the HCF (403, 434, 465) = 31.
Thus 31 litres is the maximum capacity of a container that can measure the diesel of three containers an exact number of times.
8. Find the least number which when divided by 6, 15 and 18, leave the remainder 5 in each case.
Ans: To find the least number divided by 6, 15 and 18, we need to find the Least Common Multiple of 6, 15 and 18.
$ 2\left \!{\underline {\, {6,15,18} \,}} \right. \\ 3\left \!{\underline {\, {3,15,9} \,}} \right. \\ 3\left \!{\underline {\, {1,5,3} \,}} \right. \\ 5\left \!{\underline {\, {1,5,1} \,}} \right. \\ 1\left \!{\underline {\, {1,1,1} \,}} \right. $
The LCM (6, 15, 18) = \[2 \times 3 \times 3 \times 5 = 90\]
90 is the least number that is divided by 6, 15 and 18.
Therefore, the least number which when divided by 6, 15, and 18 leaves remainder 5 is 90+5 = 95.
9. Find the smallest 4digit number which is divisible by 18, 24 and 32.
Ans: To find the smallest 4digit number divisible by 18, 24 and 32 we need to find the Least Common Multiple of 18, 24 and 32.
$ 2\left \!{\underline {\, {18,24,32} \,}} \right. \\ 2\left \!{\underline {\, {9,12,16} \,}} \right. \\ 2\left \!{\underline {\, {9,6,8} \,}} \right. \\ 2\left \!{\underline {\, {9,3,4} \,}} \right. \\ 2\left \!{\underline {\, {9,3,2} \,}} \right. \\ 3\left \!{\underline {\, {9,3,1} \,}} \right. \\ 3\left \!{\underline {\, {3,1,1} \,}} \right. \\ 1\left \!{\underline {\, {1,1,1} \,}} \right. $
The LCM (18, 24, 32)=\[2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 288\]
The smallest 4digit number that is divisible by 288 is the number that is exactly divisible by 18, 24, and 32.
1152 is the smallest 4digit number divisible by 288.
So the smallest 4digit number that is divisible by 18, 24 and 32 is 1152.
10. Find the L.C.M. of the following numbers:
Observe a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?
9 and 4
Ans: LCM of 9 and 4:
$ 2\left \!{\underline {\, {9,4} \,}} \right. \\ 2\left \!{\underline {\, {9,2} \,}} \right. \\ 3\left \!{\underline {\, {9,1} \,}} \right. \\ 3\left \!{\underline {\, {3,1} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
LCM (9, 4) = \[2 \times 2 \times 3 \times 3 = 36\]
Therefore LCM (9, 4) = 36
12 and 5
Ans: LCM of 12 and 5:
$ 2\left \!{\underline {\, {12,5} \,}} \right. \\ 2\left \!{\underline {\, {6,5} \,}} \right. \\ 3\left \!{\underline {\, {3,5} \,}} \right. \\ 5\left \!{\underline {\, {1,5} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
LCM (12, 5) = \[2 \times 2 \times 3 \times 5 = 60\]
Therefore the LCM (12, 5) = 60
6 and 5
Ans: LCM of 6 and 5
$ 2\left \!{\underline {\, {6,5} \,}} \right. \\ 3\left \!{\underline {\, {3,5} \,}} \right. \\ 5\left \!{\underline {\, {1,5} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
The LCM (6, 5) = \[2 \times 3 \times 5 = 30\]
Therefore the LCM (6, 5) = 30
15 and 4
Ans: The LCM of 15 and 4:
$ 2\left \!{\underline {\, {15,4} \,}} \right. \\ 2\left \!{\underline {\, {15,2} \,}} \right. \\ 3\left \!{\underline {\, {15,1} \,}} \right. \\ 5\left \!{\underline {\, {5,1} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
LCM (15, 4) = \[2 \times 2 \times 3 \times 5 = 60\]
LCM(15, 4) = 60
The common property obtained in all the LCMs is that all the LCM is the multiple of 3.
Yes, the LCMs are the product of two numbers in each case.
11. Find the L.C.M of the following numbers in which one number is the factor of other:
What do you observe in the result obtained?
5, 20
Ans: LCM of 5 and 20:
$ 2\left \!{\underline {\, {5,20} \,}} \right. \\ 2\left \!{\underline {\, {5,10} \,}} \right. \\ 5\left \!{\underline {\, {5,5} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
The LCM (5, 20) =\[2 \times 2 \times 5 = 20\]
LCM (5, 20) = 20
6, 18
Ans: LCM of 6 and 18:
$ 2\left \!{\underline {\, {6,18} \,}} \right. \\ 3\left \!{\underline {\, {3,9} \,}} \right. \\ 3\left \!{\underline {\, {1,3} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
The LCM (6, 18) = \[2 \times 3 \times 3 = 18\]
Therefore LCM (6, 18) = 18
12, 48
Ans: LCM of 12 and 48:
$ 2\left \!{\underline {\, {12,48} \,}} \right. \\ 2\left \!{\underline {\, {6,24} \,}} \right. \\ 2\left \!{\underline {\, {3,12} \,}} \right. \\ 2\left \!{\underline {\, {3,6} \,}} \right. \\ 3\left \!{\underline {\, {3,3} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
The LCM (12, 48) = \[2 \times 2 \times 2 \times 2 \times 3 = 48\]
Therefore the LCM (12, 48) = 48
9, 45
Ans: The LCM of 9 and 25:
$ 3\left \!{\underline {\, {9,45} \,}} \right. \\ 3\left \!{\underline {\, {3,15} \,}} \right. \\ 5\left \!{\underline {\, {1,5} \,}} \right. \\ 1\left \!{\underline {\, {1,1} \,}} \right. $
The LCM (9, 45) = \[3 \times 3 \times 5 = 45\]
Therefore LCM (9, 45) = 45
From the answers of the LCM we can clearly observe that the LCM of the two given numbers is equal to that of the larger number given.
Overview of Deleted Syllabus for CBSE Class 6 Maths Playing with Numbers
Chapter  Dropped Topics 
Playing with Numbers  3.6 Some more Divisibility rules Page Number (5962) 
Class 6 Maths Chapter 3: Exercises Breakdown
Exercise  Number of Questions 
Exercise 3.1  4 Questions and Solutions 
Exercise 3.2  12 Questions and Solutions 
Exercise 3.3  6 Questions and Solutions 
Exercise 3.4  7 Questions and Solutions 
Exercise 3.5  10 Questions and Solutions 
Exercise 3.6  3 Questions and Solutions 
Exercise 3.7  11 Questions and Solutions 
Conclusion
NCERT Solutions for Class 6 Chapter 3 Playing with Numbers by Vedantu provides a thorough understanding of the key concepts, such as factors, multiples, prime and composite numbers, and divisibility rules. It emphasizes the importance of mastering these foundational topics to build a strong mathematical base. Key points to focus on include understanding prime factorization, finding the highest common factor (HCF), and the lowest common multiple (LCM). These concepts are crucial for solving various mathematical problems efficiently. Vedantu's solutions are studentfriendly, offering clear explanations and stepbystep solutions to ensure comprehension. Practising these problems regularly will help reinforce the concepts and improve problemsolving skills. In previous years' exams, around 4–6 questions from this chapter have been asked, focusing on prime factorization, HCF, and LCM. Consistent practice and attention to these areas will aid in performing well in exams.
Other Study Material for CBSE Class 6 Maths Chapter 3
S.No.  Important Links for Chapter 3 Playing with Numbers 
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ChapterSpecific NCERT Solutions for Class 6 Maths
Given below are the chapterwise NCERT Solutions for Class 6 Maths. Go through these chapterwise solutions to be thoroughly familiar with the concepts.
S.No.  NCERT Solutions Class 6 ChapterWise Maths PDF 
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FAQs on NCERT Solutions Class 6 Maths Chapter 3 Playing with Numbers
1. What is there in NCERT class 6 Maths Chapter 3?
The NCERT solution for Class 6 Maths Chapter 3 covers the importance of understanding the factors and multiples of numbers. The chapter elaborates different aspects of the subject matter. This includes various examples such as understanding the laws of divisibility, prime numbers, composite numbers, prime factors, LCM, HCF, common factors and common multiples.
The unique examples present in this chapter will allow the student to get a better grasp over numbers. Understanding how every number is a factor and multiple is also necessary as well as how prime factors when multiplied form the original number. The main highlight of this chapter is HCF and LCM, questions on which mostly appear in examinations.
2. Why avail the Chapter 3 Maths Class 6 NCERT Solutions?
The Class 6 Maths NCERT Chapter 3 solutions available on Vedantu should be referred by students. The reason for that is that these solutions consist of tothepoint answers that help students understand better. The examples have been framed from simple to advanced so that the student doesn’t face problems while understanding them.
Playing with NumbersClass 6 CBSE is a chapter that has been constructed to help students focus their minds on understanding and developing their basic knowledge of numbers. The solution will help the student to clear their doubt related to the subject of numbers. Understand your numbers without any problem with solutions framed by Vedantu that help you clear your examinations without any problems.
3. What is the difference between Factors and Multiples?
One or more numbers that can divide a certain number without leaving a remainder is called a factor. For example 6 x 5 = 30. Here, 6 and 5 are the factors of 30. A multiple is a number that may be divided by another number without leaving a remainder in a specified number of times. For example: 3 x 5= 15. Here, 15 is a multiple of 3 and 5.
4. Explain Prime and Composite numbers?
A prime number is a whole number that comprises only two divisors, which are 1 and itself. For example 2, 3, 5, 7, 11, 13, etc. (1 is not a prime number because it has only one divisor). A composite number is a whole number that comprises two or more divisors.
For example:
4 = 1,2 and 4 are the divisors.
10= 1, 2, 5, and 10 are the divisors.
5. Why is Class 6 Maths Chapter 3 important for higher studies?
Class 6 Maths Chapter 3 'Playing With Numbers' is crucial for higher classes since it contains various key topics or concepts that are important for future classes. If you master the topics like prime and composite numbers, factors and multiples, divisibility tests, HCF, and LCM, it will strengthen your roots in Maths and will make it easy for you to apply these concepts at higher levels. For better understanding, utilise the Vedantu Mobile app. The solutions provided are free of cost.
6. How can I improve the basic concepts of Chapter 3?
Strengthening basic concepts is critical for students, and they should not overlook them at any cost. Chapter 3 of Class 6 Maths is overflowing with topics and students should pay close attention. These fundamental concepts will benefit you not only this year but also in future classes. Hence, one should focus on grasping these topics and practising them regularly. You can visit Vedantu and get more study materials for the same.
7. How many exercises are there in Chapter 3?
There is a total of 10 exercises in Class 6 Maths Chapter 3:
3.1 – Introduction
3.2 – Factors and Multiples
3.3 – Prime and Composite Numbers
3.4 – Tests for Divisibility of Numbers
3.5 – Common Factors and Common Multiples
3.6 – Some More Divisibility Rules
3.7 – Prime Factorization
3.8 – Highest Common Factor
3.9 – Lowest Common Factor
3.10 – Problems of HCF and LCM.
8. How do you find common multiples for the given numbers?
To find common multiples of two or more numbers, list out the multiples of each number and identify the ones that appear in all lists. For example, the common multiples of 3 and 4 are 12, 24, 36, etc.
9. How do you find common factors for the given numbers?
To find common factors of two or more numbers, list out all the factors of each number and identify the ones that appear in all lists. For example, the common factors of 8 and 12 are 1, 2, and 4.