NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.3

Students 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 about 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 conceptual knowledge among students.The problems in this exercise are based on the frequently repeated questions in the exam.

NCERT Solutions for Class 6 Chapter 3: Playing with Numbers Exercise 3.3 Download PDF

 

ncert solutions class 6 maths chapter 3 ex 3.3
ncert solutions class 6 maths chapter 3 ex 3.3
ncert solutions class 6 maths chapter 3 ex 3.3
ncert solutions class 6 maths chapter 3 ex 3.3
ncert solutions class 6 maths chapter 3 ex 3.3
ncert solutions class 6 maths chapter 3 ex 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):

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.3 - 1

Solutions:

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.3 - 2

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. Hence, 572 is also 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

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

Solutions:

(a) 297144

Since, last digit of the number is 4. Hence, 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, last digit of the number is 8. Hence, 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, last digit of the number is 5 which is not divisible by 2. Hence, 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. Hence, 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. Hence, 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. Hence, 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. 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

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

Since, 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

Sum of digits at odd places = 4 + 8 + 1

= 13

Sum of digits at even places = 2 + 0

= 2

Difference = 13 – 2 = 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, 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

Since, 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

Sum of digits at odd places = 3 + 1 + 0

= 4

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

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

Sum of the given digits = 19

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, 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

Sum of the given digits = 24

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. Since, 33 is divisible by 3.

Therefore 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 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

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

2 Comments

  1. What is co-prime

    1. 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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