Selina Solutions Concise Mathematics Class 6 Chapter 9 Playing With Numbers provides students a step by step method of solving problems, along with illustrations. Students gain a better idea of concepts covered, by solving tricky questions of the textbook thoroughly. Experts at BYJU’S have solved the questions keeping in mind the understanding capacity of students. Students boost their problem solving skills by referring to these Solutions, provided in simple language.
Some of the concepts discussed in this chapter are interchanging the digits of a number, the generalized form of numbers, tests of divisibility, etc. in a precise and lucid manner. By practising Selina Solutions, students not only clear their doubts but also secure high marks in the annual examinations. Students are advised to download the PDF of Selina Solutions Concise Mathematics Class 6 Chapter 9 Playing With Numbers, from the links given here
Selina Solutions Concise Mathematics Class 6 Chapter 9: Playing With Numbers Download PDF
Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 9: Playing With Numbers
Exercise 9(A) Solutions
Exercise 9(B) Solutions
Exercise 9(C) Solutions
Access Selina Solutions Concise Mathematics Class 6 Chapter 9: Playing With Numbers
Exercise 9(A)
(Using BODMOS)
1. 19 – (1 + 5) – 3
Solution:
Given
19 – (1 + 5) – 3
On further calculation, we get
= 19 – 6 – 3
= 19 – 9
= 10
2. 30 × 6 ÷ (5 – 2)
Solution:
Given
30 × 6 ÷ (5 – 2)
On further calculation, we get
= 30 × 6 ÷ 3
= 30 × 2
= 60
3. 28 – (3 × 8) ÷ 6
Solution:
Given
28 – (3 × 8) ÷ 6
Calculating further, we get
= 28 – 24 ÷ 6
= 28 – 4
= 24
4. 9 – [(4 – 3) + 2 × 5]
Solution:
Given
9 – [(4 – 3) + 2 × 5]
We get
= 9 – [1 + 10]
= 9 – 11
= – 2
5. [18 – (15 ÷ 5) + 6]
Solution:
Given
[18 – (15 ÷ 5) + 6]On further calculation, we get
= [18 – 3 + 6]
= 18 + 3
= 21
6. [(4 × 2) – (4 ÷ 2)] + 8
Solution:
Given
[(4 × 2) – (4 ÷ 2)] + 8On further calculation, we get
= [8 – 2] + 8
= 6 + 8
= 14
7. 48 + 96 ÷ 24 – 6 × 18
Solution:
Given
48 + 96 ÷ 24 – 6 × 18
We get
= 48 + 4 – 6 × 18
= 48 + 4 – 108
= 52 – 108
= – 56
8. 22 – [3 – {8 – (4 + 6)}]
Solution:
Given
22 – [3 – {8 – (4 + 6)}]
On calculating further, we get
= 22 – [3 – {8 – 10}]
= 22 – [3 + 2]
= 22 – 5
= 17
9. 34 – [29 – {30 + 66 ÷ (24 –
Solution:
Given
34 – [29 – {30 + 66 ÷ (24 –
On further calculation, we get
= 34 – [29 – {30 + 66 ÷ (24 – 2)}]
= 34 – [29 – {30 + 66 ÷ 22}]
= 34 – [29 – {30 + 3}]
= 34 – [29 – 33]
= 34 – [-4]
= 34 + 4
= 38
10. 60 – {16 ÷ (4 × 6 – 8)}
Solution:
Given
60 – {16 ÷ (4 × 6 – 8)}
On further calculation, we get
= 60 – {16 ÷ (24 – 8)}
= 60 – {16 ÷ 16}
= 60 – 1
= 59
11. 25 – [12 – {5 + 18 ÷ (4 –
Solution:
Given
25 – [12 – {5 + 18 ÷ (4 –
By calculating further, we get
= 25 – [12 – {5 + 18 ÷ (4 – 2)}]
= 25 – [12 – {5 + 18 ÷ 2}]
= 25 – [12 – {5 + 9}]
= 25 – [12 – 14]
= 25 – [-2]
= 25 + 2
= 27
12. 15 – [16 – {12 + 21 ÷ (9 – 2)}]
Solution:
Given
15 – [16 – {12 + 21 ÷ (9 – 2)}]
On further calculation, we get
= 15 – [16 – {12 + 21 ÷ 7}]
= 15 – [16 – {12 + 3}]
= 15 – [16 – 15]
= 15 – 1
= 14
Exercise 9(B)
1. Fill in the blanks:
(i) On dividing 9 by 7, quotient = …………….. and remainder = …………….
(ii) On dividing 18 by 6, quotient = ………….. and remainder = …………
(iii) Factor of a number is ………….. of ………..
(iv) Every number is a factor of ………….
(v) Every number is a multiple of …………
(vi) ………. is factor of every number.
(vii) For every number, its factors are ………. and its multiples are ……….
(viii) x is a factor of y, then y is a …….. of x.
Solution:
(i) On dividing 9 by 7, quotient = 1 and remainder = 2
(ii) On dividing 18 by 6, quotient = 3 and remainder = 0
(iii) Factor of a number is an exact division of the number
(iv) Every number is a factor of itself
(v) Every number is a multiple of itself
(vi) One is factor of every number
(vii) For every number, its factors are finite and its multiples are infinite
(viii) x is a factor of y, then y is a multiple of x
2. Write all the factors of:
(i) 16
(ii) 21
(iii) 39
(iv) 48
(v) 64
(vi) 98
Solution:
(i) 16
Solution:
All factors of 16 are:
1, 2, 4, 8, 16
(ii) 21
Solution:
All factors of 21 are:
1, 3, 7, 21
(iii) 39
Solution:
All factors of 39 are:
1, 3, 13, 39
(iv) 48
Solution:
All factors of 48 are:
1, 2, 3, 4,6, 8, 12, 16, 24, 48
(v) 64
Solution:
All factors of 64 are:
1, 2, 4, 8, 16, 32, 64
(vi) 98
Solution:
All factors of 98 are:
1, 2, 7, 14, 49, 98
3. Write the first six multiples of:
(i) 4
(ii) 9
(iii) 11
(iv) 15
(v) 18
(vi) 16
Solution:
(i) 4
Following are the first six multiples of 4
1 × 4 = 4
2 × 4 = 8
3 × 4 = 12
4 × 4 = 16
5 × 4 = 20
6 × 4 = 24
Hence, multiples of 4 are 4, 8, 12, 16, 20 and 24
(ii) 9
Following are the first six multiples of 9
1 × 9 = 9
2 × 9 = 18
3 × 9 = 27
4 × 9 = 36
5 × 9 = 45
6 × 9 = 54
Hence, multiples of 9 are 9, 18, 27, 36, 45 and 54
(iii) 11
Following are the first six multiples of 11
1 × 11 = 11
2 × 11 = 22
3 × 11 = 33
4 × 11 = 44
5 × 11 = 55
6 × 11 = 66
Hence, multiples of 11 are 11, 22, 33, 44, 55 and 66
(iv) 15
Following are the first six multiples of 15
1 × 15 = 15
2 × 15 = 30
3 × 15 = 45
4 × 15 = 60
5 × 15 = 75
6 × 15 = 90
Hence, multiples of 15 are 15, 30, 45, 60, 75 and 90
(v) 18
Following are the first six multiples of 18
1 × 18 = 18
2 × 18 = 36
3 × 18 = 54
4 × 18 = 72
5 × 18 = 90
6 × 18 = 108
Hence, multiples of 18 are 18, 36, 54, 72, 90 and 108
(vi) 16
Following are the first six multiples of 16
1 × 16 = 16
2 × 16 = 32
3 × 16 = 48
4 × 16 = 64
5 × 16 = 80
6 × 16 = 96
Hence, multiples of 16 are 16, 32, 48, 64, 80 and 96
4. The product of two numbers is 36 and their sum is 13. Find the numbers.
Solution:
36 can be written as
1 × 36 = 36
2 × 18 = 36
3 × 12 = 36
4 × 9 = 36
6 × 6 = 36
Here, the sum of 4 and 9 is 13
Hence, 4 and 9 are the two numbers
5. The product of two numbers is 48 and their sum is 16. Find the numbers.
Solution:
48 can be written as
1 × 48 = 48
2 × 24 = 48
3 × 16 = 48
4 × 12 = 48
6 × 8 = 48
Here, the sum of 4 and 12 is 16
Hence, 4 and 12 are the two numbers
6. Write two numbers which differ by 3 and whose product is 54.
Solution:
54 can be written as
1 × 54 = 54
2 × 27 = 54
3 × 18 = 54
6 × 9 = 54
Here, the difference between 6 and 9 is 3
Hence, 6 and 9 are the two numbers
7. Without making any actual division show that 7007 is divisible by 7.
Solution:
Given
7007
This can be written as
= 7000 + 7
= 7 × (1000 + 1)
= 7 × 1001
Clearly, 7007 is divisible by 7
8. Without making any actual division show that 2300023 is divisible by 23
Solution:
Given
2300023
This can be written as
= 2300000 + 23
= 23 × (100000 + 1)
= 23 × 100001
Clearly, 2300023 is divisible by 23
9. Without making any actual division, show that each of the following numbers is divisible by 11
(i) 11011
(ii) 110011
(iii) 11000011
Solution:
(i) 11011
This can be written as
= 11000 + 11
= 11 × (1000 + 1)
= 11 × 1001
Clearly, 11011 is divisible by 11
(ii) 110011
This can be written as
= 110000 + 11
= 11 × (10000 + 1)
= 11 × 10001
Clearly, 110011 is divisible by 11
(iii) 11000011
This can be written as
= 11000000 + 11
= 11 × (1000000 + 1)
= 11 × 1000001
Clearly, 11000011 is divisible by 11
10. Without actual division, show that each of the following numbers is divisible by 8
(i) 1608
(ii) 56008
(iii) 240008
Solution:
(i) 1608
This can be written as
= 1600 + 8
= 8 × (200 + 1)
= 8 × 201
Clearly, 1608 is divisible by 8
(ii) 56008
This can be written as
= 56000 + 8
= 8 × (7000 + 1)
= 8 × 7001
Clearly, 56008 is divisible by 8
(iii) 240008
This can be written as
= 240000 + 8
= 8 × (30000 + 1)
= 8 × 30001
Clearly, 240008 is divisible by 8
Exercise 9(C)
1. Find which of the following numbers are divisible by 2:
(i) 352
(ii) 523
(iii) 496
(iv) 649
Solution:
(i) 352
The given number = 352
Digit at unit’s place = 2
Hence, the number is divisible by 2
(ii) 523
The given number = 523
Digit at unit’s place = 3
Hence, the number is not divisible by 2
(iii) 496
The given number = 496
Digit at unit’s place = 6
Hence, the number is divisible by 2
(iv) 649
The given number = 649
Digit at unit’s place = 9
Hence, the number is not divisible by 2
2. Find which of the following number are divisible by 4:
(i) 222
(ii) 532
(iii) 678
(iv) 9232
Solution:
(i) 222
The given number = 222
The number formed by ten’s and unit digit is 22, which is not divisible by 4.
Hence, the number is not divisible by 4
(ii) 532
The given number = 532
The number formed by ten’s and unit digit is 32, which is divisible by 4.
Hence, the number is divisible by 4
(iii) 678
The given number = 678
The number formed by ten’s and unit digit is 78, which is not divisible by 4
Hence, the number is not divisible by 4
(iv) 9232
The given number = 9232
The number formed by ten’s and unit digit is 32, which is divisible by 4
Hence, the number is divisible by 4
3. Find which of the following numbers are divisible by 8:
(i) 324
(ii) 2536
(iii) 92760
(iv) 444320
Solution:
(i) 324
The given number = 324
The number formed by hundred’s, ten’s and unit digit is 324, which is not divisible by 8
Hence, 324 is not divisible by 8
(ii) 2536
The given number = 2536
The number formed by hundred’s, ten’s and unit digit is 536, which is divisible by 8
Hence, 2536 is divisible by 8
(iii) 92760
The given number = 92760
The number formed by hundred’s, ten’s and unit digit is 760, which is divisible by 8
Hence, 92760 is divisible by 8
(iv) 444320
The given number = 444320
The number formed by hundred’s, ten’s and unit digit is 320, which is divisible by 8
Hence, 444320 is divisible by 8
4. Find which of the following numbers are divisible by 3:
(i) 221
(ii) 543
(iii) 28492
(iv) 92349
Solution:
(i) 221
The given number = 221
For a number to be divisible by 3, sum of digits must be divisible by 3
Sum of digits = 2 + 2 + 1 = 5
Since 5 is not divisible by 3
Hence, 221 is not divisible by 3
(ii) 543
The given number = 543
For a number to be divisible by 3, sum of digits must be divisible by 3
Sum of digits = 5 + 4 + 3 = 12
Since 12 is divisible by 3
Hence, 543 is divisible by 3
(iii) 28492
The given number = 28492
For a number to be divisible by 3, sum of digits must be divisible by 3
Sum of digits = 2 + 8 + 4 + 9 + 2 = 25
Since 25 is not divisible by 3
Hence, 28492 is not divisible by 3
(iv) 92349
The given number = 92349
For a number to be divisible by 3, sum of digits must be divisible by 3
Sum of digits = 9 + 2 + 3 + 4 + 9 = 27
Since 27 is divisible by 3
Hence, 92349 is divisible by 3
5. Find which of the following numbers are divisible by 9:
(i) 1332
(ii) 53247
(iii) 4968
(iv) 200314
Solution:
(i) 1332
The given number = 1332
For a number to be divisible by 9, sum of digits must be divisible by 9
Sum of digits = 1 + 3 + 3 + 2 = 9
Since 9 is divisible by 9
Hence, 1332 is divisible by 9
(ii) 53247
The given number = 53247
For a number to be divisible by 9, sum of digits must be divisible by 9
Sum of digits = 5 + 3 + 2 + 4 + 7 = 21
Since 21 is not divisible by 9
Hence, 53247 is not divisible by 9
(iii) 4968
The given number = 4968
For a number to be divisible by 9, sum of digits must be divisible by 9
Sum of digits = 4 + 9 + 6 + 8 = 27
Since 27 is divisible by 9
Hence, 4968 is divisible by 9
(iv) 200314
The given number = 200314
For a number to be divisible by 9, sum of digits must be divisible by 9
Sum of digits = 2 + 0 + 0 + 3 + 1 + 4 = 10
Since 10 is not divisible by 9
Hence, 200314 is not divisible by 9
6. Find which of the following number are divisible by 6:
(i) 324
(ii) 2010
(iii) 33278
(iv) 15505
Solution:
A number which is divisible by either 2 and 3 or both then the given number is divisible by 6
(i) 324
The given number = 324
Sum of digits = 3 + 2 + 4 = 9
which is divisible by 3
Therefore, 324 is divisible by 6
(ii) 2010
The given number = 2010
Sum of digits = 2 + 0 + 1 + 0 = 3
which is divisible by 3
Therefore, 2010 is divisible by 6
(iii) 33278
The given number = 33278
Sum of digits = 3 + 3 + 2 + 7 + 8 = 23
Unit digit is 3, which is odd
Therefore, 33278 is not divisible by 6
(iv) 15505
The given number = 15505
Sum of digits = 1 + 5 + 5 + 0 + 5 = 16
which is divisible by 2
Therefore, 15505 is divisible by 6
7. Find which of the following numbers are divisible by 5:
(i) 5080
(ii) 66666
(iii) 755
(iv) 9207
Solution:
(i) 5080
The given number = 5080
For a number to be divisible by 5, unit’s digit must be 0 or 5
Here, unit digit is 0
Therefore, 5080 is divisible by 5
(ii) 66666
The given number = 66666
For a number to be divisible by 5, unit’s digit must be 0 or 5
Here, unit digit is 6
Therefore, 66666 is not divisible by 5
(iii) 755
The given number = 755
For a number to be divisible by 5, unit’s digit must be 0 or 5
Here, unit digit is 5
Therefore, 755 is divisible by 5
(iv) 9207
The given number = 9207
For a number to be divisible by 5, unit’s digit must be 0 or 5
Here, unit digit is 7
Therefore, 9207 is not divisible by 5
8. Find which of the following numbers are divisible by 10:
(i) 9990
(ii) 0
(iii) 847
(iv) 8976
Solution:
(i) 9990
The given number = 9990
For a number to be divisible by 10, unit’s digit must be 0
Here, unit digit is 0
Therefore, 9990 is divisible by 10
(ii) 0
The given number = 0
For a number to be divisible by 10, unit’s digit must be 0
Here, unit digit is 0
Therefore, 0 is divisible by 10
(iii) 847
The given number = 847
For a number to be divisible by 10, unit’s digit must be 0
Here, unit digit is 7
Therefore, 847 is not divisible by 10
(iv) 8976
The given number = 8976
For a number to be divisible by 10, unit’s digit must be 0
Here, unit digit is 6
Therefore, 8976 is not divisible by 10
9. Find which of the following numbers are divisible by 11:
(i) 5918
(ii) 68,717
(iii) 3882
(iv) 10857
Solution:
(i) 5918
The given number = 5918
If the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11 then the number is divisible by 11
Sum of digits at odd places = 5 + 1 = 6
Sum of digits at even places = 9 + 8 = 17
Difference = 17 – 6 = 11
Here, the difference is 11 which is divisible by 11
Hence, the number is divisible by 11
(ii) 68717
The given number = 68717
If the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11 then the number is divisible by 11
Sum of digits at odd places = 6 + 7 + 7 = 20
Sum of digits at even places = 8 + 1 = 9
Difference = 20 – 9 = 11
Here, difference is 11 which is divisible by 11
Hence, the number is divisible by 11
(iii) 3882
The given number = 3882
If the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11 then the number is divisible by 11
Sum of digits at odd places = 3 + 8 = 11
Sum of digits at even places = 8 + 2 = 10
Difference = 11 – 10 = 1
Here, difference is 1 which is not divisible by 11
Hence, the number is not divisible by 11
(iv) 10857
The given number = 10857
If the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11 then the number is divisible by 11
Sum of digits at odd places = 1 + 8 + 7 = 16
Sum of digits at even places = 0 + 5 = 5
Difference = 16 – 5 = 11
Here, difference is 11which is divisible by 11
Hence, the number is divisible by 11
10. Find which of the following numbers are divisible by 15:
(i) 960
(ii) 8295
(iii) 10243
(iv) 5013
Solution:
(i) 960
The given number = 960
For a number to be divisible by 15, it should be divisible by both 3 and 5
Sum of digits = 9 + 6 + 0 = 15
Since 15 is divisible by 3
So, the number is divisible by 3
Here, unit digit is 0, so it is divisible by 5
Hence, the number is divisible by 15
(ii) 8295
The given number = 8295
For a number to be divisible by 15 it should be divisible by both 3 and 5
Sum of digits = 8 + 2 + 9 + 5 = 24
Since 24 is divisible by 3
So, the number is divisible by 3
Here, unit digit is 5, so it is divisible by 5
Hence, the number is divisible by 15
(iii) 10243
The given number = 10243
For a number to be divisible by 15 it should be divisible by both 3 and 5
Sum of digits = 1 + 0 + 2 + 4 + 3 = 10
Since 10 is not divisible by 3
So, the number is not divisible by 3
Here, unit digit is 3, so it is not divisible by 5
Hence, the number is not divisible by 15
(iv) 5013
The given number = 5013
For a number to be divisible by 15 it should be divisible by both 3 and 5
Sum of digits = 5 + 0 + 1 + 3 = 9
Since 9 is divisible by 3
So, the number is divisible by 3
Here, unit digit is 3, so it is not divisible by 5
Hence, the number is not divisible by 15
11. In each of the following numbers, replace M by the smallest number to make resulting number divisible by 3:
(i) 64 M 3
(ii) 46 M 46
(iii) 27 M 53
Solution:
(i) 64 M 3
The given number = 64 M 3
For a number to be divisible by 3 sum of digits must be divisible by 3
Sum of digits = 6 + 4 + 3 = 13
The number which is divisible by 3 next to 13 is 15
Required smallest number = 15 – 13 = 2
Hence, value of M is 2
(ii) 46 M 46
The given number = 46 M 46
For a number to be divisible by 3 sum of digits must be divisible by 3
Sum of digits = 4 + 6 + 4 + 6 = 20
The number which is divisible by 3 next to 20 is 21
Required smallest number = 21 – 20 = 1
Hence, the value of M is 1
(iii) 27 M 53
The given number = 27 M 53
For a number to be divisible by 3 sum of digits must be divisible by 3
Sum of digits = 2 + 7 + 5 + 3 = 17
The number which is divisible by 3 next to 17 is 18
Required smallest number = 18 – 17 = 1
Hence, the value of M is 1
12. In each of the following numbers replace M by the smallest number to make resulting number divisible by 9
(i) 76 M 91
(ii) 77548 M
(iii) 627 M 9
Solution:
(i) 76 M 91
The given number is 76 M 91
For a number to be divisible by 9 sum of digits must be divisible by 9
Sum of digits = 7 + 6 + 9 + 1 = 23
The number which is divisible by 9 next to 23 is 27
Required smallest number = 27 – 23 = 4
Hence, the value of M is 4
(ii) 77548 M
The given number = 77548 M
For a number to be divisible by 9 sum of digits must be divisible by 9
Sum of digits = 7 + 7 + 5 + 4 + 8 = 31
The number which is divisible by 9 next to 31 is 36
Required smallest number = 36 – 31 = 5
Hence, the value of M is 5
(iii) 627 M 9
The given number = 627 M 9
For a number to be divisible by 9 sum of digits must be divisible by 9
Sum of digits = 6 + 2 + 7 + 9 = 24
The number which is divisible by 9 next to 24 is 27
Required smallest number = 27 – 24 = 3
Hence, the value of M is 3
13. In each of the following numbers, replace M by the smallest number to make resulting number divisible by 11
(i) 39 M 2
(ii) 3 M 422
(iii) 70975 M
(iv) 14 M 75
Solution:
(i) 39 M 2
The given number = 39 M 2
A number is divisible by 11, if the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11
Sum of its digits in odd places = 3 + M
Sum of its digits in even places = 9 + 2 = 11
Difference:
11 – (3 + M) = 0
11 – 3 – M = 0
8 – M = 0
M = 8
Hence, the value of M is 8
(ii) 3 M 422
The given number = 3 M 422
A number is divisible by 11, if the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11
Sum of its digits in odd places = 3 + 4 + 2 = 9
Sum of its digits in even places = M + 2 = 11
Difference:
9 – (M + 2) = 0
9 – M – 2 = 0
9 – 2 = M
M = 7
Hence, the value of M is 7
(iii) 70975 M
The given number is 70975 M
A number is divisible by 11, if the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11
Sum of its digits in odd places = 7 + 9 + 5 = 21
Sum of its digits in even places = 0 + 7 + M = 7 + M
Difference:
21 – (7 + M) = 0
21 = 7 + M
M = 14
Hence, the value of M is 14
(iv) 14 M 75
The given number is 14 M 75
A number is divisible by 11, if the difference of sum of its digit in odd places from left side and sum of digits in even places from left side is divisible by 11
Sum of its digits in odd places = 1 + M + 5
= 6 + M
Sum of its digits in even places = 4 + 7
= 11
Difference:
11 – (6 + M) = 0
11 = 6 + M
M = 11 – 6
M = 5
Hence, the value of M is 5
14. State, true or false:
(i) If a number is divisible by 4. It is divisible by 8
(ii) If a number is a factor 16 and 24, it is a factor of 48
(iii) If a number is divisible by 18, it is divisible by 3 and 6
(iv) If a divide b and c completely, then a divides (i) a + b (ii) a – b also completely.
Solution:
(i) False
The number is divisible by 4, if tens and unit digit is divisible by 4
The number is divisible by 8, if hundreds, tens and unit digit is divisible by 8
(ii) True
As 16 and 24 are factors of 48
(iii) True
The product of 3 and 6 is 18, so if a number is divisible by 18, it is divisible by 3 and 6
(iv) True
If a divides b and c completely, then a divides a + b and a – b completely
Hence, if a number is a factor of each of the two numbers, then it is a factor of their sum also
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