NCERT Class 6 Maths Chapter 2: Complete Resource for Whole Numbers
Upon reaching the 6th standard, students are introduced to the NCERT Solutions for Class 6 Maths Chapter 3 syllabus, which is quite variant and includes different topics. The third chapter among the Maths syllabus, Playing with Numbers, particularly focuses on teaching students about multiples and divisors.
As the chapter progresses, the students are introduced to topics such as common factors and multiples, divisibility rules, highest common factors, lowest common factors, etc. A proper comprehension of the chapter will enable students to understand prime and composite numbers, as well as their difference.
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Chapter Name:  Chapter 3  Playing With Numbers 
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Academic Year:  202324 
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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.
5. 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.
6. 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
$17\;{\text{and }}71.$
7. 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
$4,6,8,9,10,12,14,15,16,18$
8. 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 $7.$
9. 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 \\ $
10. 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 $
11. 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.
12. 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
$90,91,92,93,94,95,96$.
13. 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
$61 = 3 + 5 + 53$
14. 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
$3 + 7 = 10$
$2 + 3 = 5$
$13 + 2 = 15$
$13 + 7 = 20$
$3 + 17 = 20$
$5 + 5 = 10$
15. 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 $2$.
Exercise3.3
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 3.6
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 \[1\]
NCERT Class 6 Maths Chapter 3 – Free PDF
With the help of NCERT solutions for Class 6 Maths Chapter 3, the student is able to prepare for their examinations with ease. Various solved examples provided here, help students understand how they are to proceed with the calculation for these sums. Also, they allow students to grasp this basic knowledge that will prove to be useful in the later years.
You can download a free PDF of Class 6 Maths Chapter 3 available on this page below. It helps you to understand the prime role that different numbers and their multiples play in solving an equation. Formulas used and applied in this chapter are primary equations that will later be useful when the student moves forward to solve more complex equations.
NCERT Class 6 Maths Chapter 3 Exercises
Exercise 3.1 – Introduction
Purpose of this introduction 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
NCERT solutions for Class 6th Maths Chapter 3 – Playing with Numbers, then moves on to the first set of exercises 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 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
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
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
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 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.
Exercise 3.8 – Highest Common Factor
HCF or highest common factors is an important part that students are required to concentrate on. The reason for that being that this section is where most examination questions are derived from.
Exercise 3.9 – Lowest Common Factor
LCF or Lowest Common Factor also tends to be preferred by teachers to be included in examinations. Calculation process of the LCF is quite simple once the students understand the equations and how they are framed.
Exercise 3.10 – Problems of HCF and LCM
This section comprises theory and practical problems for the student to solve. At the end of the chapter, students are provided with a brief revision of the entire topic.
Key Learnings from the Chapter 3 of Class 6 Maths NCERT Solutions
Chapter 3 of Class 6 Maths NCERT Solutions is mainly based on the topics such as multiples, divisors, and factors, identification of multiples and factors, how to determine whether a number is a prime or a composite number using the factors, and how to find the HCF and LCM of numbers using the factors method. NCERT Solutions for Class 6 Maths Chapter 3 consists of sums related to these concepts
Below are the key learnings from Chapter 3 of Class 6 Maths NCERT Solutions.
A prime number can be defined as a number (except 1) which has only two factors 1 and the number itself.
Composite numbers can be defined as those numbers that have more than two factors.
1 is a factor of every number and it is neither composite nor prime.
If the sum of all the factors of a number is equal to that number, then it is called a perfect number. The number 6 is a perfect number as 6 = 1 + 2 + 3 and the number 28 is also a perfect number as 1 + 2 + 4 + 7 + 14 = 28.
Every number is a factor of itself and a multiple of itself.
Every factor of a number is an exact divisor of that number.
Every factor is less than or equal to that number but every multiple of a number is greater than or equal to that number.
The number of factors of a given number is finite but the number of multiples of a given number is infinite.
If two numbers are divisible by a number, their sum and difference are also divisible by the same number.
The HCF of two or more numbers is the highest common factor of the numbers.
The Least Common Multiple (LCM) of two or more numbers is the smallest of their common multiple of the numbers.
Benefits of NCERT Class 6 Maths Chapter 3
Benefits of availing Class 6 Maths Chapter 3 solutions from Vedantu have been listed below 
The solutions have been constructed as per examination format to assist students in scoring well.
All questions and solutions are written in simple language.
Solutions to every question are precise and to the point.
All solutions are informative and effective in helping students score well.
With the help of the Class 6th Maths, Chapter 3 offered by Vedantu, students can now aim for excellence in their examinations.
Upon reaching the 6th standard, students are introduced to the NCERT Solutions for Class 6 Maths Chapter 3 syllabus, which is quite variant and includes different topics. The third chapter among the Maths syllabus, Playing with Numbers, particularly focuses on teaching students about multiples and divisors.
As the chapter progresses, the students are introduced to topics such as common factors and multiples, divisibility rules, highest common factors, lowest common factors, etc. A proper comprehension of the chapter will enable students to understand prime and composite numbers, as well as their difference.
You can also download NCERT Class 6 Maths and NCERT Class 6 Science to help you to revise complete syllabus and score more marks in your examinations.
Chapter wise NCERT Solutions for Class 6 Maths
Chapter 3  Playing with Numbers
Along with this, students can also view additional study materials provided by Vedantu, for Class 6 Maths Chapter 1 Knowing Our Numbers
Conclusion
Vedantu's NCERT Solutions for Class 6 Maths Chapter 3  Playing With Numbers offer a comprehensive and studentfriendly approach to mastering fundamental mathematical concepts. Through a wellstructured and engaging platform, students can explore the diverse aspects of number theory with ease and confidence. The solutions provide stepbystep explanations, making complex topics accessible and promoting a deeper understanding. Vedantu's commitment to quality education ensures that learners receive the necessary support to excel in their academic journey. By leveraging these solutions, students can enhance their problemsolving skills, critical thinking, and mathematical reasoning. Overall, Vedantu's NCERT Solutions empower students to develop a strong foundation in mathematics and foster a lifelong love for learning.
FAQs on NCERT Solutions for Class 6 Maths Chapter 3  Playing With Numbers
Q1. 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.
Q2. 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.
Q3. 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.
Q4. 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.
Q5. 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.
Q6. 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.
Q7. 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.