Monday, May 25, 2020

NCERT Solutions for Class 6 Mathematics Chapter 3 Playing with Numbers Exercise 3.3

Exercise 3.3



Introduction


Introduction - Ex 3.3 - "Playing with Numbers" Chapter 3 - Class 6th Maths




Q 1 - Ex 3.3 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3

Page No 57:

Question 1:

Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say, yes or no):
Number
Divisible by
2
3
4
5
6
8
9
10
11
128
Yes
No
Yes
No
No
Yes
No
No
No
990
1586
275
6686
639210
429714
2856
3060
406839

Answer:

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


Q 2 - Ex 3.3 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3

Question 2:

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

Answer:

(a) 572
The last two digits are 72. Since 72 is divisible by 4, the given number
is also divisible by 4.
The last three digits are 572. Since 572 is not divisible by 8, the given number is also not divisible by 8.
(b) 726352
The last two digits are 52. As 52 is divisible by 4, the given number is also divisible by 4.
The last three digits are 352. Since 352 is divisible by 8, the given number is also divisible by 8.
(c) 5500
Since last two digits are 00, it is divisible by 4.
The last 3 digits are 500. Since 500 is not divisible by 8, the given number is also not divisible by 8.
(d) 6000
Since the last 2 digits are 00, the given number is divisible by 4.
Since the last 3 digits are 000, the given number is divisible by 8.
(e) 12159
The last 2 digits are 59. Since 59 is not divisible by 4, the given number is also not divisible by 4.
The last 3 digits are 159. Since 159 is not divisible by 8, the given number is not divisible by 8.
(f) 14560
The last two digits are 60. Since 60 is divisible by 4, the given number is divisible by 4.
The last 3 digits are 560. Since 560 is divisible by 8, the given number is divisible by 8.
(g) 21084
The last two digits are 84. Since 84 is divisible by 4, the given number is divisible by 4.
The last three digits are 084. Since 084 is not divisible by 8, the given number is not divisible by 8.
(h) 31795072
The last two digits are 72. Since 72 is divisible by 4, the given number is divisible by 4.
The last three digits are 072. Since 072 is divisible by 8, the given number is divisible by 8.
(i) 1700
The last two digits are 00. Since 00 is divisible by 4, the given number is divisible by 4.
The last three digits are 700. Since 700 is not divisible by 8, the given number is not divisible by 8.
(j) 2150
The last two digits are 50. Since 50 is not divisible by 4, the given number is not divisible by 4.
The last three digits are 150. Since 150 is not divisible by 8, the given number is not divisible by 8.


Q 3 - Ex 3.3 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3

Question 3:

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

Answer:

(a) 297144
Since the last digit of the number is 4, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3.
As the number is divisible by both 2 and 3, it is divisible by 6.
(b) 1258
Since the last digit of the number is 8, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 16. Since 16 is not divisible by 3, the given number is also not divisible by 3.
As the number is not divisible by both 2 and 3, it is not divisible by 6.
(c) 4335
The last digit of the number is 5, which is not divisible by 2. Therefore, the given number is also not divisible by 2.
On adding all the digits of the number, the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3.
As the number is not divisible by both 2 and 3, it is not divisible by 6.
(d) 61233
The last digit of the number is 3, which is not divisible by 2. Therefore, the given number is also not divisible by 2.
On adding all the digits of the number, the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3.
As the number is not divisible by both 2 and 3, it is not divisible by 6.
(e) 901352
Since the last digit of the number is 2, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 20. Since 20 is not divisible by 3, the given number is also not divisible by 3.
As the number is not divisible by both 2 and 3, it is not divisible by 6.
(f) 438750
Since the last digit of the number is 0, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3.
As the number is divisible by both 2 and 3, it is divisible by 6.
(g) 1790184
Since the last digit of the number is 4, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 30. Since 30 is divisible by 3, the given number is also divisible by 3.
As the number is divisible by both 2 and 3, it is divisible by 6.
(h) 12583
Since the last digit of the number is 3, it is not divisible by 2.
On adding all the digits of the number, the sum obtained is 19. Since 19 is not divisible by 3, the given number is also not divisible by 3.
As the number is not divisible by both 2 and 3, it is not divisible by 6.
(i) 639210
Since the last digit of the number is 0, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 21. Since 21 is divisible by 3, the given number is also divisible by 3.
As the number is divisible by both 2 and 3, it is divisible by 6.
(j) 17852
Since the last digit of the number is 2, it is divisible by 2.
On adding all the digits of the number, the sum obtained is 23. Since 23 is not divisible by 3, the given number is also not divisible by 3.
As the number is not divisible by both 2 and 3, it is not divisible by 6.


Q 4 - Ex 3.3 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3

Question 4:

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

Answer:

(a) 5445
Sum of the digits at odd places = 5 + 4 = 9
Sum of the digits at even places = 4 + 5 = 9
Difference = 9 − 9 = 0
As the difference between the sum of the digits at odd places and the sum of the digits at even places is 0, therefore, 5445 is divisible by 11.
(b) 10824
Sum of the digits at odd places = 4 + 8 + 1 = 13
Sum of the digits at even places = 2 + 0 = 2
Difference = 13 − 2 = 11
The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 10824 is divisible by 11.
(c) 7138965
Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24
Sum of the digits at even places = 6 + 8 + 1 = 15
Difference = 24 − 15 = 9
The difference between the sum of the digits at odd places and the sum of digits at even places is 9, which is not divisible by 11. Therefore, 7138965 is not divisible by 11.
(d) 70169308
Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17
Sum of the digits at even places = 0 + 9 + 1 + 7 = 17
Difference = 17 − 17 = 0
As the difference between the sum of the digits at odd places and the sum of the digits at even places is 0, therefore, 70169308 is divisible by 11.
(e) 10000001
Sum of the digits at odd places = 1
Sum of the digits at even places = 1
Difference = 1 − 1 = 0
As the difference between the sum of the digits at odd places and the sum of the digits at even places is 0, therefore, 10000001 is divisible by 11.
(f) 901153
Sum of the digits at odd places = 3 + 1 + 0 = 4
Sum of the digits at even places = 5 + 1 + 9 = 15
Difference = 15 − 4 = 11
The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 901153 is divisible by 11.



Q 5 - Ex 3.3 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3

Question 5:

Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3:
(a) ___6724 (b) 4765 ___2

Answer:

(a) _6724
Sum of the remaining digits = 19
To make the number divisible by 3, the sum of its digits should be divisible by 3.
The smallest multiple of 3 which comes after 19 is 21.
Therefore, smallest number = 21 − 19 = 2
Now, 2 + 3 + 3 = 8
However, 2 + 3 + 3 + 3 = 11
If we put 8, then the sum of the digits will be 27 and as 27 is divisible by 3, the number will also be divisible by 3.
Therefore, the largest number is 8.
(b) 4765_2
Sum of the remaining digits = 24
To make the number divisible by 3, the sum of its digits should be divisible by 3. As 24 is already divisible by 3, the smallest number that can be placed here is 0.
Now, 0 + 3 = 3
3 + 3 = 6
3 + 3 + 3 = 9
However, 3 + 3 + 3 + 3 = 12
If we put 9, then the sum of the digits will be 33 and as 33 is divisible by 3, the number will also be divisible by 3.
Therefore, the largest number is 9.


Q 6 - Ex 3.3 - Playing With Numbers - NCERT Maths Class 6th - Chapter 3

Page No 58:

Question 6:

Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:
(a) 92 ___ 389 (b) 8 ___9484

Answer:

(a) 92_389
Let a be placed in the blank.
Sum of the digits at odd places = 9 + 3 + 2 = 14
Sum of the digits at even places = 8 + + 9 = 17 + a
Difference = 17 + a − 14 = 3 + a
For a number to be divisible by 11, this difference should be zero or a multiple of 11.
If 3 + a = 0, then
a = − 3
However, it cannot be negative.
A closest multiple of 11, which is near to 3, has to be taken. It is 11 itself.
3 + a = 11
a = 8
Therefore, the required digit is 8.
(b) 8_9484
Let a be placed in the blank.
Sum of the digits at odd places = 4 + 4 + a = 8 + a
Sum of the digits at even places = 8 + 9 + 8 = 25
Difference = 25 − (8 + a)
= 17 − a
For a number to be divisible by 11, this difference should be zero or a multiple of 11.
If 17 − a = 0, then
a = 17
This is not possible.
A multiple of 11 has to be taken. Taking 11, we obtain
17 − a = 11
a = 6
Therefore, the required digit is 6.

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