Students can learn the concept of “Tests for Divisibility of Numbers” in depth by practising the NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3. Students can thus get well-versed in the various divisibility tests on numbers. These NCERT Solutions can be used to understand the problem-solving method of questions related to the divisibility tests on numbers, as they are explained in simple language to improve the conceptual knowledge of students. The problems in this exercise are based on the frequently repeated questions in the exam.
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3
Access NCERT Solutions for Class 6 Chapter 3: Playing with Numbers Exercise 3.3
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):
Solutions:
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
Solutions:
(a) 572
72 are the last two digits. Since 72 is divisible by 4, 572 is also divisible by 4.
572 are the last three digits. Since 572 is not divisible by 8, 572 is not divisible by 8.
(b) 726352
52 are the last two digits. Since 52 is divisible by 4, 726352 is divisible by 4.
352 are the last three digits. Since 352 is divisible by 8, 726352 is divisible by 8.
(c) 5500
Since the last two digits are 00, 5500 is divisible by 4.
500 are the last three digits. Since 500 is not divisible by 8, 5500 is not divisible by 8.
(d) 6000
Since the last two digits are 00, 6000 is divisible by 4.
Since the last three digits are 000, 6000 is divisible by 8.
(e) 12159
59 are the last two digits. Since 59 is not divisible by 4, 12159 is not divisible by 4.
159 are the last three digits. Since 159 is not divisible by 8, 12159 is not divisible by 8.
(f) 14560
60 are the last two digits. Since 60 is divisible by 4, 14560 is divisible by 4.
560 are the last three digits. Since 560 is divisible by 8, 14560 is divisible by 8.
(g) 21084
84 are the last two digits. Since 84 is divisible by 4, 21084 is divisible by 4.
084 are the last three digits. Since 084 is not divisible by 8, 21084 is not divisible by 8.
(h) 31795072
72 are the last two digits. Since 72 is divisible by 4, 31795072 is divisible by 4.
072 are the last three digits. Since 072 is divisible by 8, 31795072 is divisible by 8.
(i) 1700
Since the last two digits are 00, 1700 is divisible by 4.
700 are the last three digits. Since 700 is not divisible by 8, 1700 is not divisible by 8.
(j) 2150
50 are the last two digits. Since 50 is not divisible by 4, 2150 is not divisible by 4.
150 are the last three digits. Since 150 is not divisible by 8, 2150 is not divisible by 8.
3. Using divisibility tests, determine which of the 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
Solutions:
(a) 297144
Since the last digit of the number is 4, the number is divisible by 2.
By 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.
(b) 1258
Since the last digit of the number is 8, the number is divisible by 2.
By 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 the last digit of the number is 5, which is not divisible by 2, the number is not divisible by 2.
By 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, the number is not divisible by 2.
By 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, the number is divisible by 2.
By 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, the number is divisible by 2.
By 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, 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, 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, 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, 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.
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
Solutions:
(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
The difference between the sum of digits at odd places and the sum of digits at even places is 0. Hence, 5445 is divisible by 11.
(b) 10824
Sum of digits at odd places = 4 + 8 + 1
= 13
Sum of digits at even places = 2 + 0
= 2
Difference = 13 – 2 = 11
The difference between the sum of digits at odd places and the sum of digits at even places is 11, which is divisible by 11. Hence, 10824 is divisible by 11.
(c) 7138965
Sum of digits at odd places = 5 + 9 + 3 + 7 = 24
Sum of digits at even places = 6 + 8 + 1 = 15
Difference = 24 – 15 = 9
The difference between the sum of digits at odd places and the 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
The difference between the sum of digits at odd places and the 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
The difference between the sum of digits at odd places and the sum of digits at even places is 0. Hence, 10000001 is divisible by 11.
(f) 901153
Sum of digits at odd places = 3 + 1 + 0
= 4
Sum of digits at even places = 5 + 1 + 9
= 15
Difference = 15 – 4 = 11
The difference between the sum of digits at odd places and the sum of digits at even places is 11, which is divisible by 11. Hence, 901153 is divisible by 11.
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
Solutions:
(a) __ 6724
The sum of the given digits = 19
The sum of its digit should be divisible by 3 to make the number divisible by 3
Since 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, the 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
The sum of the given digits = 24
The sum of its digits should be divisible by 3 to make the number divisible by 3.
Since 24 is already divisible by 3, 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, the sum of its digits becomes 33.
Since 33 is divisible by 3, the number will be divisible by 3.
Hence, the largest number is 9.
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
Solutions:
(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, then the number is divisible by 11
If 3 + a = 0
a = -3
But it cannot be a negative
Taking the 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
Sum of its digits at odd places = 4 + 4 + a
= 8 + a
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.
Also, explore –
NCERT Solutions for Class 6 Maths Chapter 3
What is co-prime
Two integers are said to be coprime if the common number that divides both the numbers, (common factor) is 1 and only 1.
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Very nice solutions