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NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 6 Maths Chapter 3 are designed as per the CBSE Class 6 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 6 Chapter 3. ## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.1

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

Solution 1:
(a) 24
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
24 = 6 × 4
Stop here as 4 and 6 have occurred earlier
Hence, the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

(b) 15
15 = 1 × 15
15 = 3 × 5
15 = 5 × 3
Stop here since 3 and 5 have occurred earlier
Hence, the factors of 15 are 1, 3, 5 and 15.

(c) 21
21 = 1 × 21
21 = 3 × 7
21 = 7 × 3
Stop here as 3 and 7 have occurred earlier
Hence, the factors of 21 are 1, 3, 7 and 21.

(d) 27
27 = 1 × 27
27 = 3 × 9
27 = 9 × 3
Stop here as 3 and 9 have occurred earlier
Hence, the factors of 27 are 1, 3, 9 and 27.

(e) 12
12 = 1 × 12
12 = 2 × 6
12 = 3 × 4
12 = 4 × 3
Stop here as 3 and 4 have occurred earlier
Hence, the factors of 12 are 1, 2, 3, 4, 6 and 12.

(f) 20
20 = 1 × 20
20 = 2 × 10
20 = 4 × 5
20 = 5 × 4
Stop here as 4 and 5 have occurred earlier
Hence, the factors of 20 are 1, 2, 4, 5, 10 and 20.

(g) 18
18 = 1 × 18
18 = 2 × 9
18 = 3 × 6
18 = 6 × 3
Stop here since 3 and 6 have occurred earlier
Hence, the factors of 18 are 1, 2, 3, 6, 9 and 18.

(h) 23
23 = 1 × 23
23 = 23 × 1
Since 1 and 23 have occurred earlier
Hence, the factors of 23 are 1 and 23

(i) 36
36 = 1 × 36
36 = 2 × 18
36 = 3 × 12
36 = 4 × 9
36 = 6 × 6
Stop here as both the factors (6) are same. Thus the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18
and 36.

#### Question 2:Write first five multiples of:(a) 5(b) 8(c) 9

Solution 2:

(a) The required multiples are:
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25
Here the first five multiples of 5 are 5, 10, 15, 20 and 25.

(b) The required multiples are:
8 × 1 = 8
8 × 2 = 16
8 × 3 = 24
8 × 4 = 32
8 × 5 = 40
Here the first five multiples of 8 are 8, 16, 24, 32 and 40.

(c) The required multiples are:
9 × 1 = 9
9 × 2 = 18
9 × 3 = 27
9 × 4 = 36
9 × 5 = 45
Here, the first five multiples of 9 are 9, 18, 27, 36 and 45.

#### Question 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

Solution 3:
(i) 35 is a multiple of 7
Hence, option (b).

(ii) 15 is a factor of 30
Hence, option (d).

(iii) 16 is a multiple of 8
Hence, option (a).

(iv) 20 is a factor of 20
Hence, option (f)

(v) 25 is a factor of 50
Hence, option (e).

#### Question 4:Find all the multiples of 9 up to 100.

Solution 4:
9 × 1 = 9
9 × 2 = 18
9 × 3 = 27
9 × 4 = 36
9 × 5 = 45
9 × 6 = 54
9 × 7 = 63
9 × 8 = 72
9 × 9 = 81
9 × 10 = 90
9 × 11 = 99
So, all the multiples of 9 up to 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.2

#### Question 1:What is the sum of any two (a) Odd numbers? (b) Even numbers?

Solution 1:
(a)
The sum of any two odd numbers is even numbers.
Examples: 5 + 3 = 8
15 + 13 = 28

(b)
The sum of any two even numbers is even numbers
Examples: 2 + 8 = 10
12 + 28 = 40

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

Solution 2:
(a)
False.
The sum of three odd numbers is odd.
Example: 7 + 9 + 11 = 27 that means odd number.

(b)
True.
The sum of two odd numbers and one even numbers is even.
Example: 3 + 5 + 8 = 16 that means is even number.

(c)
True.
The product of three odd numbers is odd. Example: 3 × 7 × 9 = 189 that means is odd number.

(d)
False.
If an even number is divided by 2 so the quotient is even.
Example: 8 ÷ 2 = 4

(e)
False.
All prime numbers are not odd.
Example: 2 is a prime number but it is also an even number.

(f)
False.
Since, 1 and the number itself are factors of the number

(g)
False.
Sum of two prime numbers may also be odd number
Example: 2 + 5 = 7 that means odd number.

(h)
True.
2 is the only even prime number.

(i)
False.
Since, 2 is a prime number.

(j)
True.
The product of two even numbers is always even.
Example: 2 × 4 = 8 that means even number.

#### Question 3:The numbers 13 and 31 are prime numbers. Both these numbers have samedigits 1 and 3. Find such pairs of prime numbers up to 100.

Solution 3:
Prime numbers with same digits up to 100 are as follows:
17 and 71
37 and 73
79 and 97

#### Question 4:Write down separately the prime and composite numbers less than 20

Solution 4:
As we can see 2, 3, 5, 7, 11, 13, 17 and 19 are the prime numbers less than 20
As we can see 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18 are the composite numbers less than 20

#### Question 5:What is the greatest prime number between 1 and 10?

Solution 5:
2, 3, 5 and 7 are the prime numbers between 1 and 10. 7 is the greatest prime number among them.

Solution 6:
(a) 3 + 41 = 44
(b) 5 + 31 = 36
(c) 5 + 19 = 24
(d) 5 + 13 = 18

#### Question 7:Give three pairs of prime numbers whose difference is 2. [Remark: Twoprime numbers whose difference is 2 are called twin primes].

Solution 7:
The three pairs of prime numbers whose difference is 2 are:
3, 5
5, 7
11, 13

#### Question 8:Which of the following numbers are prime?(a) 23(b) 51(c) 37(d) 26

Solution 8:

(a) 23
1 × 23 = 23
23 × 1 = 23
As a result 23 has only two factors 1 and 23. Hence, it is a prime number.

(b) 51
1 × 51 = 51
3 × 17 = 51
As a result 51 has four factors 1, 3, 17 and 51. Hence, it is not a prime number, it is a composite number.

(c) 37
1 × 37 = 37
37 × 1 = 37
As a result 37 has two factors 1 and 37. Hence, it is a prime number.

(d) 26
1 × 26 = 26
2 × 13 = 26
As a result 26 has four factors 1, 2, 13 and 26. Hence, it is not a prime number, it is a composite number.

#### Question 9:Write seven consecutive composite numbers less than 100 so that there isno prime number between them.

Solution 9:
The seven composite numbers between 89 and 97 both which are prime numbers are 90, 91, 92, 93, 94, 95 and 96
Numbers Factors
90 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
91 1, 7, 13, 91
92 1, 2, 4, 23, 46, 92
93 1, 3, 31, 93
94 1, 2, 47, 94
95 1, 5, 19, 95
96 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

#### Question 10:Express each of the following numbers as the sum of three odd primes:(a) 21(b) 31(c) 53(d) 61

Solution 10:

(a) 3 + 5 + 13 = 21
(b) 3 + 5 + 23 = 31
(c) 13 + 17 + 23= 53
(d) 7 + 13 + 41 = 61

#### Question 11:Write five pairs of prime numbers less than 20 whose sum is divisible by 5.(Hint: 3 + 7 = 10)

Solution 11:
Here are the five pairs of prime numbers less than 20 whose sum is divisible by 5 are
2 + 3 = 5
2 + 13 = 15
3 + 17 = 20
7 + 13 = 20
19 + 11 = 30

#### Question 12:Fill in the blanks:(a) A number which has only two factors is called a ______.(b) A number which has more than two factors is called a ______.(c) 1 is neither ______ nor ______.(d) The smallest prime number is ______.(e) The smallest composite number is _____.(f) The smallest even number is ______.

Solution 12:
(a) A number which has only two factors is called a prime number.
(b) A number which has more than two factors is called a composite number.
(c) 1 is neither prime number nor composite number.
(d) The smallest prime number is 2
(e) The smallest composite number is 4
(f) The smallest even number is 2.

## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.3

#### Question 1:Using divisibility tests, determine which of the following numbers aredivisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say, yes orno): Solution 1: #### (d) 6000(e) 12159(f) 14560(g) 21084(h) 31795072(i) 1700(j) 2150

Solution 2:
(a) 572
72 are the last two digits. Since, 72 is divisible by 4. Hence, 572 is divisible by 4.
572 are the last three digits. Since, 572 is not divisible by 8. Hence, 572 is not divisible by 8.

(b) 726352
52 are the last two digits. Since, 52 is divisible by 4. Hence, 726352 is divisible by 4.
352 are the last three digits. Since 352 is divisible by 8. Hence, 726352 is divisible by 8.

(c) 5500
Since, last two digits are 00. Hence 5500 is divisible by 4.
500 are the last three digits. Since, 500 is not divisible by 8. Hence, 5500 is not divisible by 8

(d) 6000
Since, last two digits are 00. Hence 6000 is divisible by 4.
Since, last three digits are 000. Hence, 6000 is divisible by 8.

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

(f) 14560
60 are the last two digits. Since 60 is divisible by 4. Hence, 14560 is divisible by 4.
560 are the last three digits. Since, 560 is divisible by 8. Hence, 14560 is divisible by 8.

(g) 21084
84 are the last two digits. Since, 84 is divisible by 4. Hence, 21084 is divisible by 4.
084 are the last three digits. Since, 084 is not divisible by 8. Hence, 21084 is not divisible by 8.

(h) 31795072
72 are the last two digits. Since, 72 is divisible by 4. Hence, 31795072 is divisible by 4
072 are the last three digits. Since, 072 is divisible by 8. Hence, 31795072 is divisible by 8

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

(j) 2150
50 are the last two digits. Since, 50 is not divisible by 4. Hence, 2150 is not divisible by 4
150 are the last three digits. Since, 150 is not divisible by 8. Hence, 2150 is not divisible by 8

#### Question 3:Using divisibility tests, determine which of following numbers are divisibleby 6:(a) 297144(b) 1258(c) 4335(d) 61233(e) 901352(f) 438750(g) 1790184(h) 12583(i) 639210(j) 17852

Solution 3:
(a) 297144
Since, last digit of the number is 4. Hence, the number is divisible by 2
After adding all the digits of the number, we get 27 which is divisible by 3. Hence, the number is divisible by 3
As a result the number is divisible by both 2 and 3. Hence, the number is divisible by 6

(b) 1258
Since, last digit of the number is 8. Hence, the number is divisible by 2
After adding all the digits of the number, we get 16 which is not divisible by 3. Hence, the number is not divisible by 3
∴ the number is not divisible by both 2 and 3. Hence, the number is not divisible by 6

(c) 4335
Since, last digit of the number is 5 which is not divisible by 2. Hence, the number is not divisible by 2
After adding all the digits of the number, we get 15 which is divisible by 3. Hence, the number is divisible by 3
∴ the number is not divisible by both 2 and 3. Hence, the number is not divisible by 6

(d) 61233
Since, the last digit of the number is 3 which is not divisible by 2. Hence, the number is not
divisible by 2
After adding all the digits of the number, we get 15 which is divisible by 3. Hence, the number is divisible by 3                                                                                      ∴ the number is not divisible by both 2 and 3. Hence, the number is not divisible by 6

(e) 901352
Since, the last digit of the number is 2. Hence, the number is divisible by 2
After adding all the digits of the number, we get 20 which is not divisible by 3. Hence, the number is not divisible by 3
∴ the number is not divisible by both 2 and 3. Hence, the number is not divisible by 6

(f) 438750
Since, the last digit of the number is 0. Hence, the number is divisible by 2
After adding all the digits of the number, we get 27 which is divisible by 3. Hence, the number is divisible by 3
∴ the number is divisible by both 2 and 3. Hence, the number is divisible by 6

(g) 1790184
Since, the last digit of the number is 4. Hence, the number is divisible by 2
By adding all the digits of the number, we get 30 which is divisible by 3. Hence, the number is divisible by 3
∴ the number is divisible by both 2 and 3. Hence, the number is divisible by 6

(h) 12583
Since, the last digit of the number is 3. Hence, the number is not divisible by 2
By adding all the digits of the number, we get 19 which is not divisible by 3. Hence, the number is not divisible by 3
∴ the number is not divisible by both 2 and 3. Hence, the number is not divisible by 6

(i) 639210
Since, the last digit of the number is 0. Hence, the number is divisible by 2
By adding all the digits of the number, we get 21 which is divisible by 3. Hence, the number is divisible by 3
∴ the number is divisible by both 2 and 3. Hence, the number is divisible by 6

(j) 17852
Since, the last digit of the number is 2. Hence, the number is divisible by 2
By adding all the digits of the number, we get 23 which is not divisible by 3. Hence, the number is not divisible by 3
∴ the number is not divisible by both 2 and 3. Hence, the number is not divisible by 6

#### (f) 901153

Solution 4:
(a) 5445
Here sum of the digits at odd places = 5 + 4
= 9
Here sum of the digits at even places = 4 + 5
= 9
Difference = 9 – 9 = 0
As the difference between sum of digits at odd places and sum of digits at even places is 0.
Hence,
5445 is divisible by 11

(b) 10824
Here sum of digits at odd places = 4 + 8 + 1
= 13
Here sum of digits at even places = 2 + 0
= 2
Difference = 13 – 2 = 11
As the difference between sum of digits at odd places and sum of digits at even places is 11 which is divisible by 11. Hence, 10824 is divisible by 11

(c) 7138965
Here sum of digits at odd places = 5 + 9 + 3 + 7 = 24
Here sum of digits at even places = 6 + 8 + 1 = 15
Difference = 24 – 15 = 9
As the difference between sum of digits at odd places and sum of digits at even places is 9 which is not divisible by 11. Hence, 7138965 is not divisible by 11

(d) 70169308
Sum of digits at odd places = 8 + 3 + 6 + 0
= 17
Sum of digits at even places = 0 + 9 + 1 + 7
= 17
Difference = 17 – 17 = 0
Since, the difference between sum of digits at odd places and sum of digits at even places is 0. Hence,
70169308 is divisible by 11.

(e) 10000001
Sum of digits at odd places = 1
Sum of digits at even places = 1
Difference = 1 – 1 = 0
Since, the difference between sum of digits at odd places and sum of digits at even places is 0. Hence,
10000001 is divisible by 11.

(f) 901153
Here sum of digits at odd places = 3 + 1 + 0
= 4
Here sum of digits at even places = 5 + 1 + 9
= 15
Difference = 15 – 4 = 11
Since, the difference between sum of digits at odd places and sum of digits at even places is 11 which is divisible by 11. Hence, 901153 is divisible by 11.

#### Question 5:Write the smallest digit and the greatest digit in the blank space of each ofthe following numbers so that the number formed is divisible by 3:(a) __ 6724(b) 4765 __ 2

Solution 5:
(a) __ 6724
Here sum of the given digits = 19
Here sum of its digit should be divisible by 3 to make the number divisible by 3.
As 21 is the smallest multiple of 3 which comes after 19
So, smallest number = 21 – 19
= 2
Now 2 + 3 + 3 = 8
But 2 + 3 + 3 + 3 = 11
Now, if we put 8, sum of digits will be 27 which is divisible by 3.
Therefore the number will be divisible by 3
Hence, the largest number is 8.

(b) 4765 __ 2
Here sum of the given digits = 24
Here sum of its digits should be divisible by 3 to make the number divisible by 3.
Since, 24 is already divisible by 3. Hence, the smallest number that can be replaced is 0
Now, 0 + 3 = 3
3 + 3 = 6
3 + 3 + 3 = 9
3 + 3 + 3 + 3 = 12
If we put 9, sum of its digits becomes 33. As 33 is divisible by 3.
Therefore the number will be divisible by 3
Hence, the largest number is 9.

#### (a) 92 __ 389(b) 8 __ 9484

Solution 6:
(a) 92 __ 389
Let ‘a’ be placed here
Sum of its digits at odd places = 9 + 3 + 2
= 14
Sum of its digits at even places = 8 + a + 9
= 17 + a
Difference = 17 + a – 14
= 3 + a
The difference should be 0 or a multiple of 11, so the number is divisible by 11
If 3 + a = 0
a = -3
But it cannot be a negative
Taking a closest multiple of 11 which is near to 3
It is 11 which is near to 3
Now, 3 + a = 11
a = 11 – 3
a = 8
Therefore the required digit is 8.

(b) 8 __ 9484
Let ‘a’ be placed here
Here sum of its digits at odd places = 4 + 4 + a
= 8 + a
Here sum of its digits at even places = 8 + 9 + 8
= 25
Difference = 25 – (8 + a)
= 17 – a
The difference should be 0 or a multiple of 11, then the number is divisible by 11
If 17 – a = 0
a = 17 (which is not possible)
Now, take a multiple of 11.
Let’s take 11
17 – a = 11
a = 17 – 11
a = 6
Therefore the required digit is 6.

## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.4

#### Question 1:Find the common factors of:(a) 20 and 28(b) 15 and 25(c) 35 and 50(d) 56 and 120

Solution 1:
(a) 20 and 28
Here 1, 2, 4, 5, 10 and 20 are factors of 20.
Here 1, 2, 4, 7, 14 and 28 are factors of 28.
Common factors = 1, 2, 4

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

(c) 35 and 50
Here 1, 5, 7 and 35 are factors of 35
Here 1, 2, 5, 10, 25 and 30 are factors of 50
Common factors = 1, 5

(d) 56 and 120
Here 1, 2, 4, 7, 8, 14, 28 and 56 are factors of 56
Here 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120 are factors of 120
Common factors = 1, 2, 4, 8

#### Question 2:Find the common factors of:(a) 4, 8 and 12(b) 5, 15 and 25

Solution 2:
(a) 4, 8 and 12
1, 2, 4 are factors of 4
1, 2, 4, 8 are factors of 8
Here 1, 2, 3, 4, 6, 12 are factors of 12
Common factors = 1, 2, 4

(b) 5, 15 and 25
1, 5 are factors of 5

1, 3, 5, 15 are factors of 15
1, 5, 25 are factors of 25
Common factors = 1, 5

#### Question 3:Find first three common multiples of:(a) 6 and 8(b) 12 and 18

Solution 3:
(a) 6 and 18
6, 12, 18, 24, 30 are multiples of 6
8, 16, 24, 32 are multiples of 8
Three common multiples are 24, 48, and 72.

(b) 12 and 18
12, 24, 36, 48 are multiples of 12
18, 36, 54, 72 are multiples of 18
Three common factors are 36, 72, and 108

#### Question 4:Write all the numbers less than 100 which are common multiples of 3 and4.

Solution 4:
Here multiples of 3 are 3, 6, 9, 12, 15
Here multiples of 4 are 4, 8, 12, 16, 20
Common multiples are 12, 24, 36, 48, 60, 72, 84 and 96

#### Question 5:Which of the following numbers are co-prime?(a) 18 and 35(b) 15 and 37(c) 30 and 415(d) 17 and 68(e) 216 and 215(f) 81 and 16

Solution 5:
(a) 18 and 35
Here factors of 18 are 1, 2, 3, 6, 9, 18
Here factors of 35 are 1, 5, 7, 35
Common factor = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime

(b) 15 and 37
Here factors of 15 are 1, 3, 5, 15
Here factors of 37 are 1, 37
Common factors = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime

(c) 30 and 415
Here factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415 are 1, 5, 83, 415
Common factors = 1, 5
Since, their common factor is other than 1. Hence, the given two numbers are not co-prime

(d) 17 and 68
Factors of 17 are 1, 17
Factors of 68 are 1, 2, 4, 17, 34, 68
Common factors = 1, 17
Since, their common factor is other than 1. Hence, the given two numbers are not co-prime

(e) 216 and 215
Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54
Factors of 215 are 1, 5, 43, 215
Common factors = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime

(f) 81 and 16
Factors of 81 are 1, 3, 9, 27, 81
Factors of 16 are 1, 2, 4, 8, 16
Common factors = 1
Since, their common factor is 1. Hence, the given two numbers are co-prime

#### Question 6:A number is divisible by both 5 and 12. By which other number will thatnumber be always divisible?

Solution 6:
Factors of 5 are 1, 5
Factors of 12 are 1, 2, 3, 4, 6, 12
Their common factor = 1
As their common factor is 1. The given two numbers are co-prime and is also divisible by their product 60
Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

#### Question 7:A number is divisible by 12. By what other numbers will that number bedivisible?

Solution 7:
As, the number is divisible by 12. Hence, it also divisible by its factors i.e 1, 2, 3, 4, 6, 12
Therefore 1, 2, 3, 4, 6 are the numbers other than 12 by which this number is also divisible

## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.5

#### Question 1:Which of the following statements are true?(a) If a number is divisible by 3, it must be divisible by 9.(b) If a number is divisible by 9, it must be divisible by 3.(c) A number is divisible by 18, if it is divisible by both 3 and 6.(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.(e) If two numbers are co-primes, at least one of them must be prime.(f) All numbers which are divisible by 4 must also be divisible by 8.(g) All numbers which are divisible by 8 must also be divisible by 4.(h) If a number exactly divides two numbers separately, it must exactlydivide their sum.(i) If a number exactly divides the sum of two numbers, it must exactlydivide the two numbers separately

Solutions 1:
(a) False because 6 is divisible by 3 but is not divisible by 9.
(b) True, as 9 = 3 × 3. As a result, if a number is divisible by 9, it will also be divisible by 3.
(c) False. As 30 is divisible by both 3 and 6 but is not divisible by 18.
(d) True, as 9 × 10 = 90. As a result, if a number is divisible by both 9 and 10 then it is divisible by 90.
(e) False. As 15 and 32 are co-primes and also composite numbers.
(f) False, because 12 is divisible by 4 but is not divisible by 8.
(g) True, as 2 × 4 = 8. Hence, if a number is divisible by 8, it will also be divisible by 2 and
4.
(h) True, as 2 divides 4 and 8 and it also divides 12 (4 + 8 = 12).
(i) False, As 2 divides 12 but it does not divide 7 and 5.

#### (a)  (b) Solution 2:
(a) Since, 6 = 2 × 3 and 10 = 5 × 2 (b) Since, 60 = 30 × 2
30 = 10 × 3
10 = 5 × 2 #### Question 3:Which factors are not included in the prime factorization of a compositenumber?

Solution 3:
Here 1 and the number itself are not included in the prime factorization of a composite number.

#### Question 4:Write the greatest 4-digit number and express it in terms of its primefactors.

Solution 4:
The greatest four digit number is 9999
Therefore 9999 = 3 × 3 × 11 × 101 #### Question 5:Write the smallest 5-digit number and express it in the form of its primefactors.

Solution 5:
The smallest five digit number = 10000 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

#### Question 6:Find all the prime factors of 1729 and arrange them in ascending order.Now state the relation, if any; between two consecutive prime factors.

Solution 6: 1729 = 7 × 13 × 19
13 – 7 = 6
19 – 13 = 6
Hence, the difference between two consecutive prime factors is 6.

#### Question 7:The product of three consecutive numbers is always divisible by 6. Verifythis statement with the help of some examples.

Solution 7:
(i) 2 × 3 × 4 = 24 which is divisible by 6

(ii) 5 × 6 × 7 = 210 which is divisible by 6

#### Question 8:The sum of two consecutive odd numbers is divisible by 4. Verify thisstatement with the help of some examples.

Solution 8:
(i) 5 + 3 = 8 which is divisible by 4
(ii) 7 + 9 = 16 which is divisible by 4
(iii) 13 + 15 = 28 which is divisible by 4

#### Question 9:In which of the following expressions, prime factorization has been done?(a) 24 = 2 × 3 × 4(b) 56 = 7 × 2 × 2 × 2(c) 70 = 2 × 5 × 7(d) 54 = 2 × 3 × 9

Solution 9:
(a) 24 = 2 × 3 × 4
Since, 4 is composite. Hence, prime factorization has not been done.

(b) 56 = 7 × 2 × 2 × 2
Since, all the factors are prime. Hence, prime factorization has been done.

(c) 70 = 2 × 5 × 7
Since, all the factors are prime. Hence, prime factorization has been done.

(d) 54 = 2 × 3 × 9
Since, 9 is composite. Hence prime factorization has not been done.

#### Question 10:Determine if 25110 is divisible by 45. [Hint: 5 and 9 are co-prime numbers.Test the divisibility of the number by 5 and 9].

Solution 10:
45 = 5 × 9
1, 5 are factors of 5
1, 3, 9 are factors of 9
Hence, 5 and 9 are co-prime numbers
The last digit of 25110 is 0. Hence, it is divisible by 5
Sum of digits 25110
2 + 5 + 1 + 1 + 0
= 9

Since, the sum of digits of 25110 is divisible by 9. Hence, 25110 is divisible by 9
As the number is divisible by both 5 and 9
Therefore 25110 is divisible by 45

#### Question 11:18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, anumber is divisible by both 4 and 6. Can we say that the number must alsobe divisible by 4 × 6 = 24? If not, give an example to justify your answer.

Solution 11:
No, since, 12 and 36 are both divisible by 4 and 6. But 12 and 36 are not divisible by 24.

#### Question 12:I am the smallest number, having four different prime factors. Can youfind me?

Solution 12:
As it is the smallest number. Therefore it will be the product of 4 smallest prime numbers
2 × 3 × 5 × 7 = 210

## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.6

#### Question 1:Find the HCF of the following numbers:(a) 18, 48(b) 30, 42(c) 18, 60(d) 27, 63(e) 36, 84(f) 34, 102(g) 70, 105, 175(h) 91, 112, 49(i) 18, 54, 81(j) 12, 45, 75

Solution 1:
(a) 18, 48 18 = 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
HCF = 2 × 3 = 6
As a result the HCF of 18, 48 is 6.

(b) 30, 42  30 = 2 × 3 × 5
42 = 2 × 3 × 7
HCF = 2 × 3 = 6
As a result the HCF of 30, 42 is 6

(c) 18, 60 18 = 2 × 3 × 3
60 = 2 × 2 × 3 × 5
HCF = 2 × 3 = 6
As a result the HCF of 18, 60 is 6.

(d) 27, 63 27 = 3 × 3 × 3
63 = 3 × 3 × 7
HCF = 3 × 3 = 9
As a result the HCF of 27, 63 is 9.

(e) 36, 84  36 = 2 × 2 × 3 × 3
84 = 2 × 2 × 3 × 7
HCF = 2 × 2 × 3 = 12
As a result the HCF of 36, 84 is 12.

(f) 34, 102 34 = 2 × 17
102 = 2 × 3 × 17
HCF = 2 × 17 = 34
As a result the HCF of 34, 102 is 34.

(g) 70, 105, 175  70 = 2 × 5 × 7
105 = 3 × 5 × 7
175 = 5 × 5 × 7
HCF = 5 × 7 = 35
As a result the HCF of 70, 105, and 175 is 35.

(h) 91, 112, 49 91 = 7 × 13
112 = 2 × 2 × 2 × 2 × 7
49 = 7 × 7
HCF = 7
As a result the HCF of 91, 112, and 49 is 7.

(i) 18, 54, 81  18 = 2 × 3 × 3
54 = 2 × 3 × 3 × 3
81 = 3 × 3 × 3 × 3
HCF = 3 × 3 = 9
As a result the HCF of 18, 54, and 81 is 9.

(j) 12, 45, 75 12 = 2 × 2 × 3
45 = 3 × 3 × 5
75 = 3 × 5 × 5
HCF = 3
As a result the HCF of 12, 45, and 75 is 3

#### Question 2:What is the HCF of two consecutive?(a) Numbers?(b) Even numbers?(c) Odd numbers?

Solution 2:
(a) The HCF of two consecutive numbers is 1
Example: The HCF of 2 and 3 is 1

(b) The HCF of two consecutive even numbers is 2
Example: The HCF of 2 and 4 is 2
(c) The HCF of two consecutive odd numbers is 1

Example: The HCF of 3 and 5 is 1

Solution 3:

## NCERT Solutions for Class 6 Maths Chapter 3: Playing with Numbers Ex 3.7

#### Question 1:Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find themaximum value of weight which can measure the weight of the fertilizerexact number of times.

Solution 1:
Given weight of two bags of fertilizer = 75 kg and 69 kg
Maximum weight = HCF of two bags weight that means (75, 69) 75 = 3 × 5 × 5
69 = 3 × 23
HCF = 3
As a result, 3 kg is the maximum value of weight which can measure the weight of the fertilizer exact number of times.

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

Solution 2:
First boy steps measure = 63 cm
Second boy steps measure = 70 cm
Third boy steps measure = 77 cm
LCM of 63, 70, 77 is calculated as,  LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930
As a result, 6930 cm is the distance each should cover so that all can cover the distance in complete steps.

#### Question 3:The length, breadth and height of a room are 825 cm, 675 cm and 450 cmrespectively. Find the longest tape which can measure the three dimensionsof the room exactly.

Solution 3:
Given length of a room = 825 cm
Breadth of a room = 675 cm
Height of a room = 450 cm 825 = 3 × 5 × 5 × 11
675 = 3 × 3 × 3 × 5 × 5
450 = 2 × 3 × 3 × 5 × 5
HCF = 3 × 5 × 5 = 75 cm
As a result longest tape is 75 cm which can measure the three dimensions of the room exactly.

#### Determine the smallest 3-digit number which is exactly divisible by 6, 8 and12.

Solution 4:
LCM of 6, 8, 12 = smallest number LCM = 2 × 2 × 2 × 3 = 24
Here we need to find the smallest 3-digit multiple of 24
As we know that 24 × 4 = 96 and 24 × 5 = 120
Hence, 120 is the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

#### Question 5:Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Solution 5:
LCM of 8, 10 and 12 LCM = 2 × 2 × 2 × 3 × 5 = 120
So we need to find the greatest 3-digit multiple of 120
We may find 120 × 8 = 960 and 120 × 9 = 1080
As a result, 960 is the greatest 3-digit number exactly divisible by 8, 10 and 12.

#### Question 6:The traffic lights at three different road crossings change after every 48seconds, 72 seconds and 108 seconds respectively. If they changesimultaneously at 7 a.m., at what time will they change simultaneouslyagain?

Solution 6:
So LCM of 48, 72, 108 = time period after which these lights change LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432
So, lights will change together after every 432 seconds
As a result the lights will change simultaneously at 7 minutes 12 seconds.

#### Question 7:Three tankers contain 403 litres, 434 litres, and 465 litres of dieselrespectively. Find the maximum capacity of a container that can measurethe diesel of the three containers exact number of times.

Solution 7:
So HCF of 403, 434, 465 = Maximum capacity of tanker required
403 = 13 × 31
434 = 2 × 7 × 31
465 = 3 × 5 × 31
HCF = 31
As a result, a container of 31 litres can measure the diesel of the three containers exact number of times.

#### Question 8:Find the least number which when divided by 6, 15 and 18 leave remainder5 in each case.

Solution 8:
LCM of 6, 15, 18 LCM = 2 × 3 × 3 × 5 = 90
Required number = 90 + 5
= 95
Hence, 95 is the required number.

#### Question 9:Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Solution 9:
LCM of 18, 24, 32 LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
Here we need to find the smallest 4-digit multiple of 288
We find 288 × 3 = 864 and 288 × 4 = 1152
Hence, 1152 is the smallest 4-digit number which is divisible by 18, 24 and 32

#### Question 10:Find the LCM of the following numbers:(a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4Observe a common property in the obtained LCMs. Is LCM the product oftwo numbers in each case?

Solution 10:
(a) LCM of 9, 4 LCM = 2 × 2 × 3 × 3 = 36

(b) LCM of 12, 5 LCM = 2 × 2 × 3 × 5 = 60

(c) LCM of 6, 5 LCM = 2 × 3 × 5 = 30

(d) LCM of 15, 4 LCM = 2 × 2 × 3 × 5 = 60
Yes in each case the LCM of given numbers is the product of these numbers.

#### Question 11:Find the LCM of the following numbers in which one number is the factorof the other.(a) 5, 20 (b) 6, 18 (c) 12, 48 (d) 9, 45What do you observe in the results obtained?

Solution 11:
(a) 5, 20 LCM = 2 × 2 × 5 = 20

(b) 6, 18 LCM = 2 × 3 × 3 = 18

(c) 12, 48 LCM = 2 × 2 × 2 × 2 × 3 = 48

(d) 9, 45 LCM = 3 × 3 × 5 = 45
∴ Hence, in this case the LCM of given numbers is the larger number. When a number is a factor of other number then their LCM will be the larger number. 