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NCERT Solutions for Class 6 Maths Chapter 3 - Playing with Numbers

Get NCERT solutions for all exercises of class 6 Maths chapter 3. In this chapter students will understand the concepts and logics behind calculating the factors and multiples of different numbers.

Jul 10, 2019 11:43 IST
NCERT Solutions for Class 6 Maths Chapter 3

NCERT solutions for class 6 Maths chapter 3: Playing with Numbers are provided here for free download. Here you will find the easy and accurate solutions for all the exercises in class 6 Maths NCERT chapter 3.

NCERT Solutions for Class 6 Maths

Chapter 6: Playing With Numbers

Exercise 3.1

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.

(a) 24

24 = 1 x 24

24 = 2 x 12

24 = 3 x 8

24 = 4 x 6

Therefore, all the factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24.

(b) 15

15 = 1 x 15

15 = 3 x 5

Therefore, all the factors of 15 are: 1, 3, 5 and 15.

(c) 21

21 = 1 x 21

21 = 3 x 7

Therefore, all the factors of 21 are: 1, 3, 7 and 21.

(d) 27

27 = 1 x 27

27 = 3 x 9

Therefore, all the factors of 27 are: 1, 3, 9 and 27.

(e) 12

12 = 1 x 12

12 = 2 x 6

12 = 3 x 4

Therefore, all the factors of 12 are: 1, 2, 3, 4, 6 and 12.

(f) Factors of 20 are:

20 = 1 x 20

20 = 2 x 10

20 = 4 x 5

Therefore, all the factors of 20 are: 1, 2, 4, 5, 10 and 20.

(g) Factors of 18 are:

18 = 1 x 18

18 = 2 x 9

18 = 3 x 6

Therefore, all the factors of 18 are: 1, 2, 3, 6, 9 and 18.

(h) Factors of 23 are:

23 = 1 x 23

Therefore, all the factors of prime number 23 are: 1 and 23.

(i) 36

36 = 1 x 36

36 = 2 x 18

36 = 3 x 12

36 = 4 x 9

36 = 6 x 6

Therefore, all the factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18 and 36.

2. Write first five multiples of:

(a) 5

(b) 8

(c) 9

Solution.

(a) 5

5 x 1 = 5

5 x 2 = 10

5 x 3 = 15

5 x 4 = 20

5 x 5 = 25

Therefore, five multiples of 5 are: 5, 10, 15, 20 and 25.

(b) First five multiples of 8 are:

8 x 1 = 8;

8 x 2 = 16;

8 x 3 = 24;

8×4 = 32;

8 x 5 = 40

Therefore, five multiples of 8 are: 8, 16, 24, 32 and 40.

(c) First five multiples of 9 are:

9 x 1 = 9;

9 x 2 = 18;

9 x 3 = 27;

9 x 4 = 36;

9 x 5 = 45

Therefore, five multiples of 9 are: 9, 18, 27, 36 and 45.

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.

(i) 35             (b) Multiple of 7 [∵ 7 x 5 = 35]

(ii) 15            (d) Factor of 30  [∵ 15 x 2 = 30]

(iii) 16           (a) Multiple of 8 [∵ 8 x 2 = 16]

(iv) 20           (f) Factor of 20   [∵ 20 x 1 = 20]

(v) 25           (e) Factor of 50  [∵ 25 x 2 = 50]

4. Find all the multiples of 9 upto 100.

Solution.

9 x 1 = 9

9 x 2 = 18

9 x 3 = 27

9 x 4 = 36

9 x 5 = 45

9 x 6 = 54

9 x 7 = 63

9 x 8 = 72;

9 x 9 = 81

9 x 10 = 90

9 x 11 = 99

Therefore, all the multiples of 9 upto 100 are:

9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99

Exercise 3.2

1. What is the sum of any two:

(a) Odd numbers?

(b) Even numbers?

Solution.

(a) The sum of any two odd numbers is always even.

For Example: 3 + 5 = 8

                        25 + 17 = 42

(b) The sum of any two even numbers is always even.

For Example:  8 + 6 = 14

                          20 + 16 = 36

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 only the even prime number.

(i) All even numbers are composite numbers.

(j) The product of any two even numbers is always even.

Solution.

(а) False; sum of the three odd numbers is always odd.

For example: 7 + 9 + 11 = 27

(b) True; sum of two odd numbers and one even number is always even

For example:  7 + 9 + 10 = 26

(c) True; product of three odd numbers is always odd.

For example: 7 x 9 x 11 = 693

(d) False; If an even number is divided by 2, then the quotient can either be an odd or an even number.

 For example:  6 ÷ 2 = 3 (odd)

And                   4 ÷ 2 = 2 (even)

(e) False; 2 is a prime number but it is even.

(f) False; 1 and the number itself are the factors of each prime number.

For example: 3 = 3 x 1

Thus, 1 and 3 are the factors of 3.

(g) False; sum of two prime numbers can either be an odd or an even.

For example: 7 + 2 = 9 (odd)

And                 7 + 5 = 12 (even)

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

(i) False; 2 is the smallest even number which not composite. It is prime number.

(j) True; product of any two even numbers is always even.

3. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.

Solution.

Pairs of prime numbers having same digits are:

17 and 71

37 and 73

79 and 97

4. Write down separately the prime and composite numbers less than 20.

Solution.

Prime numbers less than 20 are:

2, 3, 5, 7, 11, 13, 17 and 19

Composite numbers less than 20 are:

4, 6, 8, 9, 10, 12, 14, 15, 16 and 18

5. What is the greatest prime number between 1 and 10?

Solution.

Prime numbers between 1 and 10 are: 2, 3, 5, 7

Thus the greatest prime number between 1 and 10 is 7.

6. Express the following as the sum of two odd primes.

(a) 44

(b) 36

(c) 24

(d) 18

Solution.

(a) 44 = 31 +13

(b) 36 = 31 + 5

(c) 24 = 19 + 5

(d) 18 = 13 + 5

7. Give three pairs of prime numbers whose difference is 2.

[Remark: Two prime numbers whose difference is 2 are called twin primes]

Solution.

Three pairs of prime numbers whose difference is 2 are:

3 and 5

5 and 7

11 and 13

8. Which of the following numbers are prime?

(a) 23

(b) 51

(c) 37

(d) 26

Solution.

We know that numbers which have only two factors (1 and the number itself) are prime numbers and the other numbers which have more than two factors are called composite numbers.

(a) 23

23 = 1 x 23

Therefore, it is a prime number.

(b) 51

51 = 1 x 51

51 = 17 x 3

Therefore, it is not a prime number.

(c) 37

37 = 1 x 37

Therefore, it is a prime number.

(d) 26

26 = 1 x 26

26 = 13 x 2

Therefore, it is not a prime number.

9. Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

Solution.

There are seven composite numbers between 89 and 97 such that there is no prime number between them. These seven composite numbers are:

90, 91, 92, 93, 94, 95 and 96

10. Express each of the following numbers as the sum of three odd primes.

(a) 21

(b) 31

(c) 53

(d) 61

Solution.

(a) 21 = 3 + 5 + 13

(b) 31 = 5 + 7 + 19

(c) 53 = 13 + 17 + 23

(d) 61 = 11 + 19 + 31

11. Write five pairs of prime numbers less than 20 whose sum is divisible by 5.

(Hint: 3 + 7 = 10)

Solution.

Required pairs of prime numbers below 20 whose sum is divisible by 5 are:

2 and 3; 2 + 3 = 5

2 and 13; 2 + 13 = 15

3 and 17; 3 + 17 = 20

7 and 13; 7 + 13 = 20

9 and 11; 9 + 11 = 20

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:

(a) prime number

(b) composite number

(c) a prime number, a composite number

(d) 2

(e) 4

(f) 2

 

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)

Number

Divisible by

 

2

3

4

5

6

8

9

10

11

128

990

1586

275

6686

639210

429714

2856

3060

406839

Yes

…..

…..

…..

…..

…..

…..

…..

…..

…..

No

…..

…..

…..

…..

…..

…..

…..

…..

…..

Yes

…..

…..

…..

…..

…..

…..

…..

…..

…..

No

…..

…..

…..

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…..

…..

…..

…..

…..

No

…..

…..

…..

…..

…..

…..

…..

…..

…..

Yes

…..

…..

…..

…..

…..

…..

…..

…..

…..

No

…..

…..

…..

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…..

…..

…..

…..

 No

…..

…..

…..

…..

…..

…..

…..

…..

…..

No

…..

…..

…..

…..

…..

…..

…..

…..

…..

Solution.

Number

Divisible by

 

2

3

4

5

6

8

9

10

11

128

990

1586

275

6686

639210

429714

2856

3060

406839

Yes

Yes

Yes

No

Yes

Yes

Yes

Yes

Yes

No

No

Yes

No

No

No

Yes

Yes

Yes

Yes

Yes

Yes

No

No

No

No

No

No

Yes

Yes

No

No

Yes

No

Yes

No

Yes

No

No

Yes

No

No

Yes

No

No

No

Yes

Yes

Yes

Yes

No

Yes

No

No

No

No

No

No

Yes

No

No

No

Yes

No

No

No

No

Yes

No

Yes

No

 No

Yes

No

No

No

Yes

No

No

Yes

No

No

Yes

No

Yes

No

Yes

No

No

No

No

2. Using divisibility tests, determine which of following numbers are divisible by 4; by 8.

(a) 572

(b) 726352

(c) 5500

(d) 6000

(e) 12159

(f) 14560

(g) 21084

(j) 2150

(h) 31795072

(i) 1700

Solution.

We know that a number is divisible by 4 if its last two digits are divisible by 4.

And a number is divisible by 8 if its last three digits are divisible by 8.

(a) 572

To check divisibility by 4

Here, last two digits form 72 which is divisible by 4 so the number 572 is also divisible by 4.

To check divisibility by 8

Here, last three digits form 572 which is not divisible by 8 so the numbre 572 is not divisible by 8.

(b) 726352

To check divisibility by 4

Here, last two digits form 52 which is divisible by 4. So the number 726352 is also divisible by 4.

To check divisibility by 8

Here, last three digits form 352 which is divisible 8 so the number 726352 is also divisible by 8.

(c) 5500

To check divisibility by 4

Here, last two digits form 00 which is divisible. Hence, 5500 is divisible by 4.

To check divisibility by 8

Here, last three digits form 500 which is not divisible by 8 so the number 5500 is not divisible by 8.

(d) 6000

To check divisibility by 4

Here, the last two digits are 00. Hence, 6000 is divisible by 4.

To check divisibility by 8

Here, the last three digits are 000. Hence, 6000 is divisible by 8.

(e) 12159

To check divisibility by 4

Here, last two digits form 59 which is not divisible by 4 so the number 12159 is not divisible by 4.

To check divisibility by 8

Here, last three digits form 159 which is not divisible by 8 so the number 12159 is not divisible by 8.

(f) 14560

To check divisibility by 4

Here, last two digits form 60 which is divisible by 4 so the number 60 is also divisible by 4.

To check divisibility by 8

Here, last three digits form 560 which is divisible by 8 so the number 14560 is divisible by 8.

(g) 21084

To check divisibility by 4

Here, last two digits form 84 which is divisible by 4 so the number 21084 is also divisible by 4.

To check divisibility by 8

Here, last three digits form 084 which is not divisible by 8 so 21084 is not divisible by 8.

(h) 31795072

To check divisibility by 4

Here, last two digits form 72 which is divisible by 4 so the number is also divisible by 4.

To check divisibility by 8

Here, last three digits form 072 which us divisible by 8 so the number 31795072 is also divisible by 8.

(i) 1700

To check divisibility by 4

Here, last two digits form 0 which is divisible by 4 so the number 1700 is also divisible by 4.

To check divisibility by 8

Here, last three digits form  700 which is not divisible by 8 so the number 1700 is also not divisible by 8.

(j) 2150

To check divisibility by 4

Here, last two digits form 50 which is not divisible by 4 so the number 2150 is also not divisible by 4.

To check divisibility by 8

Here, last three digits form 150 which is not divisible by 8 so the number 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

Solution.

We know that a number is divisible by 6 if it is divisible by both 2 and 3.

Now a number is divisible by 2 if the digit at its ones place is an even number.

And a number is divisible by three if the sum of its all digits is divisible by three.

(a) 297144

Here the digit at ones place is 4 which is even so 297144  is divisible by 2.

The sum of all the digits of 297144 = 2 + 9 + 7 + 1 + 4 + 4 = 27 which is divisible by 3.

Hence, the given number 297144 is divisible by 6.

(b) 1258

Here the digit at ones place is 8 which is even so 1258 is divisible by 2.

The sum of all digits of 1258 = l + 2 + 5 + 8 = 16 which is not divisible by 3.

Since the given number 1258 is not divisible by both 2 and 3 therefore, it is not divisible by 6.

(c) 4335

Here the digit at ones place is 5 which is not even so 4335 it is not divisible by 2.

Hence, 4335 it is also not divisible by 6.

(d) 61233

Here, the digit at ones place is 3 which is not even so the number 61233 is not divisible by 2.

Hence, the number 61233 is not divisible by 6.

(e) 901352

Here, the digit at ones place is 2 which is even so the number 901352 is divisible by 2.

The sum of all the digits of 901352 = 9 + 0 + 1 + 3 + 5 + 2 = 20 which is not divisible by 3.

Since, the number 901352 is not divisible by both 2 and 3 so it is not divisible by 6.

(f) 438750

Here, the digit at ones place of the given number is 0 so the number 438750 is divisible by 2.

The sum of all the digits of 438750 = 4 + 3 + 8 + 7 + 5 + 0 = 27 which is divisible by 3.

Hence, the given number is divisible by 6.

(g) 1790184

Here, the digit at ones place is 4 which is even so the number 1790184 is divisible by 2.

The sum of all the digits of 1790184 = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 which is divisible by 3.

Hence, the given number is divisible by 6.

(h) 12583

Here, the digit at ones place is 3 which is odd so the number 12583 is not divisible by 2.

Hence, the number 12583 is not divisible by 6.

(i) 639210

Here, the digit at ones place is 0 so the number 639210 is divisible by 2.

The sum of all the digits of 639210 = 6 + 3 + 9 + 2 + 1 + 0 = 21 which is divisible by 3.

Hence, the given number is divisible by 6.

(j) 17852

The digit at ones place of the given number is 2 which is even hence 17852 is divisible by 2.

The sum of all the digits of 17852 = 1 + 7 + 8 + 5 + 2 = 23 which is not divisible by 3.

Since, the number 17852 is not divisible by both 2 and 3 so it 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.

Solution.

We know that a number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number is either 0 or divisible by 11.

(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

Hence, the given number 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 which is divisible by 11.

Hence, the given number 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 which is not divisible by 11.

Hence, the given number 7138965 is not divisible by 11.

(d) 70169308

Sum of all the digits at odd places = 8 + 3 + 6 + 0 = 17

Sum of all the digits at even places = 0 + 9 + 1 + 7 = 17

Difference = 17-17 = 0

Hence, the given number 70169308 is divisible by 11.

(e) 10000001

Sum of all the digits at odd places = 1 + 0 + 0 + 0 = 1

Sum of all the digits at even places = 0 + 0 + 0 + 1 = 1

Difference = 1 – 1 = 0

Hence, the given number 10000001 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

Solution.

We know that number is divisible by 3 if the sum of all its digits is also divisible by 3.

(a) ___ 6724

Sum of the given digits = 4 + 2 + 7 + 6 = 19

After 19 the next number which is multiple of 3 is 21.

So, the smallest digit to be placed is blank space = 21 – 19 = 2

Now, the greatest digit to be placed in blank space = 8

Such that the sum = 19 + 8 = 27 is divisible by 3.

Thus, the smallest digit = 3

And the greatest digit = 8

(b) 4765 ____ 2.

Sum of the given digits = 2 + 5 + 6 + 7 + 4 = 24 which is divisible 3.

So, the smallest digits to be placed in blank space = 0

Such that, sum = 24 + 0 = 24 is divisible by 3.

Now, the greatest digit to be placed in blank space = 9.

Such that the sum = 24 + 9 = 33 which is divisible by 3.

Thus, the smallest digit = 0

And the greatest digit = 9

6. Write a digit in the blank space of each of the following numbers so that the numbers formed is divisible by 11.

(a) 92 ___ 389

(b) 8 ___ 9484

Solution.

We know that a number is divisible by 11 if the difference between the sum of the digits at odd places (from the right) and the sum of the digits at even places (from the right) of the number is either 0 or divisible by 11.

(a) 92 ___ 389

Let p be the digit to be placed in blank space.

Then sum of the digits at odd places = 9 + 3 + 2 = 14

Sum of the digits at even places = 8 + p + 9 = 17 + a

Difference = 17 + p – 14 = a + 3

As the difference should be either 0 or divisible by 11

If p + 3 = 0

Then p = –3 but negative value cannot be placed in blank space.

So, p + 3 = 11

∴ p = 11 – 3 = 8

So, the required digit = 8

(b) 8 ___ 9484

Let p be the digit to be placed in blank space.

Sum of the digits at odd places = 4 + 4 + p = 8 + p

Sum of the digits at even places = 8 + 9 + 8 = 25

∴ Difference = 25 – (8 + p)

                       = 25 – 8 – p

                       = 17 – p

As the difference should be either 0 or divisible by 11

So, either 17 – p = 0

Then p = 17 which is not possible.

So we take 17 – p = 11

∴ p = 17 – 11 = 6

So, the missing digit = 6

Exercise 3.4

1. Find the common factors of:

(a) 20 and 28

(b) 15 and 25

(c) 35 and 50

(d) 56 and 120

Solution.

(a) Factors of 20 = 1, 2, 4, 5, 10, 20

Factors of 28 = 1, 2, 4, 7, 28

Hence, the common factors of 20 and 28 = 1, 2, 4

(b) Factors of 15 = 1, 3, 5, 15

Factors of 25 = 1, 5, 25

Hence, the common factors of 15 and 25 = 1, 5

(c) Factors of 35 = 1, 5, 7, 35

Factors of 50 = 1, 2, 5, 10, 25, 50

Hence, the common factors of 35 and 50 = 1, 5

(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Hence, the common factors of 56 and 120 = 1, 2, 4, 8

2. Find the common factors of:

(a) 4, 8 and 12

(b) 5, 15 and 25

(6) 8 9484

Solution.

(a) Factors of 4 = 1, 2, 4

Factors of 8 = 1, 2, 4, 8

Factors of 12 = 1, 2, 3, 4, 6, 12

Hence, the common factors of 4, 8 and 12 = 1, 2, 4

(b) Factors of 5 = 1, 5

Factors of 15 = 1, 3, 5, 15

Factors of 25 = 1, 5, 25

Hence, the common factors of 5, 15 and 25 = 1, 5

3. Find first three common multiples of:

(a) 6 and 8

(b) 12 and 18

Solution.

(a) 6 and 8

Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72

Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72

So the first three common multiples of 6 and 8 = 24, 48, 72

(b) 12 and 18

Multiples of 12 = 12, 24, 36, 48, 50, 72, 84, 96, 108

Multiples of 18 = 18, 36, 54, 72, 90, 108

So the first three common multiples of 12 and 18 = 36, 72, 108

4. Write all the numbers less than 100 which are common multiples of 3 and 4.

Solution.

Multiples of 3 less than 100 are = 3, 6, 9, 12, 15, 18, 21, 24, 27,……, 96, 99

Multiples of 4 less than 100 are = 4, 8, 13, 16, 20, 24, 28, 32, ………, 96

Hence, the common multiples of 3 and 4 less than 100 are: 12, 24, 36, 48, 60, 72, 84 and 96.

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.

We know that any two numbers are said to be co-prime if they have only 1 as their common factor.

(a) 18 and 35

Factors of 18 are 1, 2, 3, 6, 9, 18

Factors of 35 are 1, 5, 7, 35

Since, the common factor of 18 and 35 is only 1

Hence, they are co-prime.

(b) 15 and 37

Factors of 15 are 1, 3, 5, 15

Factors of 37 are 1, 37

Since, the common factor of 15 and 37 is only 1.

Hence, they are co-prime.

(c) 30 and 415

Factors of 30 are 1, 2, 3, 5, 6, 15, 30

Factors of 415 are 1, 5, 83

Since, the common factor of 30 and 415 are 1 and 5

Hence, they are not co-prime.

(d) 17 and 68

Factors of 17 are 1, 17

Factors of 68 are 1, 2, 4, 17, 34, 68

Since, the common factor of 17 and 68 are 1 and 17

Hence, they are not co-prime.

(e) 216 and 215

Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 54, 72, 108, 216

Factors of 215 are 1, 5, 43

Since the common factor of 216 and 215 is only 1

Hence, they 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

Since the common factor of 81 and 16 is only 1

Hence, they are co-prime.

6. A number is divisible by both 5 and 12. By which other will that number be always divisible?

Solution.

Factors of 5 = 1, 5

Factors of 12 = 1, 2, 3, 4, 6, 12

Thus 5 and 12 are co-prime numbers so the number will be divisible by the product 5 × 12 = 60.

So, the given number will always be divisible by 60.

7. A number is divisible by 12. By what other will that number be divisible?

Solution.

If number is divisible by another number, then it is also divisible by each of the factors of that number.

Here, factors of 12 are 1, 2, 3, 4, 6, 12

Hence the number which is divisible by 12 will also be divisible by 1, 2, 3, 4, 6 and 12.

Exercise 3.5

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 exactly divide their sum.

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

Solution.

(a) False; 21 is divisible by 3 but not by 9.

(b) True; Since 9 = 3 × 3, which means 3 is a factor of 9.

And we know that if number is divisible by another number, then it is also divisible by each of the factors of that number.

So, if a number is divisible by 9, it must be divisible by 3.

(c) False; Because 24 is divisible by both 3 and 6 but not by 30.

(d) True; Since 9 × 10 = 90

So, if a number is divisible by 9 and 10 both, then it must be divisible by 90.

(e) False; since 18 and 35 are co-primes (they have only 1 as their common multiple) but both of them are composites.

(f) False; 12 is divisible by 4 but not by 8.

(g) True; As 4 is a factor of 8 so all the numbers which are divisible by 8 will also be divisible by 4.

(h) True; Since 3 divides 6 and 9, it also divides 6 + 9 = 15.

(i) False; Since 3 divides 15 but it does not divide 8 and 7.

2.Here are two different factor trees for 60. Write the missing numbers.

(a)

(b)

Solution.

(a) Since 6 = 2 ×3 and 10 = 5 ×2

Missing numbers in the factor tree of 60 can be filled as follows:

(b) Since 60 = 30 ×2, 30 = 10 ×3 and 10 = 5×2

Thus, missing numbers in the factor tree of 60 can be filled as follows:

3.Which factors are not included in the prime factorisation of a composite number?

Solution.

1 and the number itself are not included in the prime factorisation of a composite number.

4.Write the greatest 4-digit number and express it in terms of its prime factors.

Solution.

The greatest 4-digit number = 9999

Its prime factors are calculated as follows:

Hence, 9999 can be expressed in terms of its prime factors as:

9999 = 3 x 3 x 11 x 101.

5.Write the smallest 5-digit number and express it in the form of its prime factors.

Solution.

The smallest 5-digit number = 10000

Its prime factors are calculated as follows:

Hence, 10000 can be expressed in terms of its prime factors as:

10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5.

6. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relations, if any, between the two consecutive prime factors.

Solution.

Prime factors of 1729 are calculated as follows:

Hence, the prime factors of 1729 = 7 x 13 x 19.

Here, 13 – 7 = 6 and 19 – 13 = 6

We can see that the difference between two consecutive prime factors is 6.

7.The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Solution.

(i) 5, 6, 7

5 × 6 × 7 = 210 which is divisible by 6.

(ii) 8, 9, 10

8 × 9 × 10 = 720 which is divisible by 6.

(iii) 18, 19, 20

18 × 19 × 20 = 6840 which is divisibke by 6.

8.The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Solution.

(i) 7 + 9 = 16 which is divisible by 4.

(ii) 59 + 61 = 120 which is divisible by 4.

(iii) 83 + 85 = 168 which is divisible by 4.

9.In which of the following expressions, prime factorisation has been done?

(a) 24 = 2 x 3 x 4

(b) 56 = 7 x 2 x 2 x 2

(c) 70 = 2 x 5 x 7

(d) 54 = 2 x 3 x 9.

Solution.

(a) 24 = 2 x 3 x 4

Here, 4 is not a prime number.

Therefore, 24 = 2 x 3 x 4 is not a prime factorisation.

(b) 56 = 7 x 2 x 2 x 2

Here, all factors are prime numbers

Therefore, 56 = 7 x 2 x 2 x 2 is a prime factorisation.

(c) 70 = 2 x 5 x 7

Here, all factors are prime numbers.

Therefore, 70 = 2 x 5 x 7 is a prime factorisation.

(d) 54 = 2 x 3 x 9

Here, 9 is not a prime number.

Therefore, 54 = 2 x 3 x 9 is not a prime factorisation.

10.Determine if 25110 is divisible by 45.

Solution.

45 = 5 x 9

Now factors of 5 = 1, 5

And factors of 9 = 1, 3, 9

Since, 5 and 9 have only 1 as common factor so they are co-prime numbers.

Now we will check the divisibility of 25110 by 5 and 9.

To check divisibility by 5:

As the ones place of the number 25110 is 0. So, it is divisible by 5.

To check divisibility by 9:

Sum of all the digits of number 25110 = 2 + 5 + 1 + 1 + 0 = 9 which is divisible by 9.

So, the number 25110is divisible by both 5 and 9, so it is also divisible by 45.

11. 18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24? If not, give an example to justify your answer.

Solution.

Since, the numbers 4 and 6 are not co-prime, so it is not necessary that a number which is divisible by both 4 and 6, must also be divisible by their product 4 x 6 = 24.

For example: 60 is divisible by both 4 and 6 but not by 24.

12. I am the smallest number, having four different prime factors. Can you find me?

Solution.

Four smallest prime numbers are 2, 3, 5 and 7.

Hence, the required number = 2 x 3 x 5 x 7 = 210

Exercise 3.6

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.

(a) 18, 48

Thus, prime factorisations of 18 and 48 are given as:

18 = 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

Common factors are 2 and 3

Hence, the HCF of 18 and 48 = 2 x 3 = 6

(b) 30, 42

Thus, prime factorisations of 30 and 42are given as:

30 = 2 × 3 × 5

42 = 2 × 3 × 7

Common factors are 2 and 3

Hence, the HCF of 30 and 42 = 2 x 3 = 6

(c) 18, 60

Prime factorisations of 18 and 60 are given as:

18 = 2 × 3 × 3

60= 2 × 2 × 3 × 5

Common factors are 2 and 3

Hence, the HCF of 18 and 60 = 2 x 3 = 6

(d) 27, 63

Prime factorisations of 27 and 63 are given as:

27 = 3 × 3 × 3

63 = 3 × 3× 7

Common factors are 3 and 3.

Hence, the HCF of 27 and 63 = 3 x 3 = 9

(e) 36, 84

Prime factorisations of 36 and 84 are given as:

36 = 2 × 2 ×3 × 3

84 = 2 × 2 ×3 × 7

Common factors are 2, 2 and 3.

Hence, the HCFof 36 and 84 = 2 x 2 x 3 = 12

(f) 34, 102

Prime factorisations of 34 and 102 are given as:

34 = 2 ×17

102 = 2 × 3 ×17

Common factors are 2 and 17.

Hence, the HCF of 34 and 102 = 2 x 17 = 34

(g) 70, 105, 175

Prime factorisations of 70, 105 and 175 are given as:

70 = 2 × 5 ×7

105 = 3 × 5 × 7

175 = 5× 5 × 7

Here, common factors are 5 and 7.

Hence, the HCF of 70, 105 and 175 = 5 x 7 = 35

(h) 91, 112, 49

Prime factorisations of 91, 112 and 49 are given as:

91 = 7 × 13

112 = 2 × 2 × 2 × 2 ×7

49 = 7× 7

Here, common factor is 7.

Hence, the HCFof 91, 112 and 49 = 7

(i) 18, 54, 81

Prime factorisations of 18, 54 and 81 are given as:

18 = 2 × 3×3

54 = 2 ×3× 3× 3

81 = 3 × 3 × 3× 3

Here, common factors are 3 and 3.

Hence, the HCFof 18, 54 and 81 = 3 x 3 = 9.

(j) 12, 45, 75

Prime factorisations of 12, 45 and 75 are given as:

12 = 2 ×2× 3

45 = 3 × 3 ×5

75 = 3 ×5×5

Here, common factor is 3.

Hence, the HCF of 12, 45 and 75 = 3

2.What is the HCF of two consecutive

(a) numbers?

(b) even numbers?

(c) odd numbers?

Solution.

(a) Since the common factor of two consecutive numbers is 1

Hence, the HCF of two consecutive numbers = 1

(b) Since, the common factors of two consecutive even numbers are 1 and 2

Hence, the HCF two consecutive even numbers = 1 x 2 = 2

(c) Since, the common factor of two consecutive odd numbers is 1

Hence, the HCF of two consecutive odd numbers = 1

3.HCF of co-prime numbers 4 and 15 was found as follows by factorisation:

4 = 2 x 2 and 15 = 3 x 15since there is no common prime factor, so HCF of 4 and 15 is 0.

Is the answer correct? If not, what is the correct HCF?

Solution.

No, answer is not correct.

1 is the prime factor of co-prime numbers.

Hence, the correct HCF of 4 and 15 is 1.

Exercise 3.7

1.Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Solution.

Maximum value of weight = HCF of 75 and69

75 = 3 × 5 × 5

69 = 3 × 23

Here, the common factor is 3.

∴ HCF of 75 and 69 = 3

Hence, maximum value of weight which can measure the weight of the fertiliser exact number of times = 3 kg.

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

Solution.

Minimum distance = LCM of 63, 70, 77

∴ LCM of 63, 70 and 77 = 2 x 3 x 3 x 5 x 7 x 11 = 6930

Hence, the minimum distance each boy should cover so that all can cover the distance in complete steps = 6930 cm

3.The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Solution.

The longest tape = HCF of 825, 675 and 450

825 = 3 x 5 x 5 x 11

675 = 3 x 3 x 3 x 5 x 5

450 = 2 x 3 x 3 x 5 x 5

Here, common factors are 3, 5, 5

∴ HCF of 825, 675 and 450 = 3 x 5 x 5 = 75

Hence, the required longest tape which can measure the three dimensions of the room exactly = 75 cm.

4.Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Solution.

Smallest number which is exactly divisible by 6, 8 and 12 = LCM of 6, 8 and 12

∴ LCM of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24

Since, all the multiples of 24 will also be divisible by 6, 8 and 12.

But we need to find the smallest three digit number which is divisible by 6, 8 and 12.

Here 24 × 4 = 96 and 24 × 5 = 120

Thus, 120 is the required smallest three digit number which is exactly divisible by 6, 8 and 12.

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

Solution.

We first find the LCM of 8, 10 and 12

∴ LCM of 8, 10 and 12 = 2 x 2 x 2 x 3 x 5 = 120

We need to find the greatest 3-digit number exactly divisible by 8, 10 and 12.

Here, 120x 8 = 960 and 120x 9 = 1080

Hence, 960 is the required greatest three digit number.

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

Solution.

Time at which the lights will change simultaneously = LCM of 48, 72 and 108

∴ LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3= 432

Hence, the required time= 432 seconds = 7 minutes 12 seconds

i.e., the lights will change simultaneously at 7 minutes 12 seconds past 7 a.m.

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Solution.

Maximum capacity of the container = HCF of 403, 434 and 465.

403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

Here, common factors is = 31.

∴ HCF of 403, 434 and 465 = 31

Hence, the maximum capacity of the required container = 31 litres.

8.Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Solution.

Least number which is exactly divisible by 6, 15 and 18 = LCM of 6, 15 and 18

∴ LCM of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90

So, 90 is the least number exactly divisible by 6, 15 and 18.

To get a remainder 5, the least number will be 90 + 5 = 95.

Hence, the required number is 95.

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

Solution.

The smallest number divisible by 18, 24 and 32 = LCM of 18, 24 and 32

∴ LCM of 18, 24 and 32= 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288

Since, 288 is the smallest number which is exactly divisible by 18, 24 and 32 but it is not a 4-digit number.

Now 288 × 3 = 864 and 288 × 4 = 1152

So, the smallest 4-digit number which is divisible by 18, 24 and 32 =1152

10.Find the LCM of the following numbers:

(a) 9 and 4

(b) 12 and 5

(c) 6 and 5

(d) 15 and 4

Observe a common property in the obtained  LCMs. Is LCM the product of two numbers in each case?

Solution.

(a) 9 and 4

∴ LCM of 9 and 4= 2 x 2 x 3 x 5 = 36

Also 9 x 4 = 36

Thus, LCM of 9 and 4 = Product of 9 and 4.

(b) 12 and 5

∴ LCM of 12 and 5 = 2 x 2 x 3 x 5 = 60

Also 12 x 5 = 60

Thus, LCM of 12 and 5 = Product of 12 and 5

(c) 6 and 5

∴ LCM of 6 and 5= 2 x 3 x 5 = 30.

Also 6 x 5 = 30.

Thus, LCM of 6 and 5 = Product of 6 and 5

(d) 15 and 4

∴ LCM of 15 and 4 = 2 x 2 x 3 x 5 = 60

Also 15 x 4 = 60

Thus, LCM of 15 and 4 = Product of 15 and 4

Here, in all of the questions given above the LCM of numbers comes to be equal to their product. Actually, when two numbers are co-prime their LCM is equal to their product.

11.Find the LCM of the following numbers in which one number is the factor of the other.

(a) 5, 20

(b) 6, 18

(c) 12, 48

(d) 9, 45

What do you observe in the results obtained?

Solution.

(a) 5 and 20

∴ LCM of 5 and 20 = 2 x 2 x 5 = 20

Hence, the LCM of 5 and 20 = 20

(b) 6 and 18

∴ LCM of 6 and 18 = 2 x 3 x 3 = 18

Hence, the LCM of 6 and 18 = 18

(c) 12 and 48

LCM of 12 and 48 = 2 x 2 x 2 x 2 x 3 = 48

Hence, the LCM of 12 and 48 = 48

(d) 9 and 45

∴ LCM of 9 and 45= 3 x 3 x 5 = 45

Hence, the LCM of 9 and 45 = 45

Here is all the questions given above, we observe that the LCM of the two numbers, with one number being a factor of the other, is the greater number.

Check chapter-wise NCERT Solutions for Class 6 Maths from the links given below:

NCERT Solutions for Class 6 Maths Chapter 1 - Knowing Our Numbers

NCERT Solutions for Class 6 Maths Chapter 2 - Whole Numbers

NCERT solutions for other chapters will be provided here very soon. Check here for the detailed and appropriate solutions.