# Ncert solutions for class 6th maths,playing with numbers,factors, chapter 3

1. A factor of a number is an exact divisor of that number.

2. 1 is a factor of every number.

3. Every number is a factor of itself.

4. Every factor of a number is an exact divisor of that number.

5. Every factor is less than or equal to the given number.

6. The number of factors of a given number are finite.

7. Every multiple of a number is greater than or equal to so that number.

8. The number of a given number is infinite.

9. Every number is a multiple of itself .

10. A number for which sum of all its factors is equal to twice the number is called a perfect number.

11. The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers.

12. Numbers having more than two factors are called composite numbers.

13. 2 is the smallest prime number which is even.

14. Every prime number except 2 is odd.

You can see the solutions of some chapters for 6th class maths.

NCERT solutions for class 6th chapter 1

NCERT solutions for class chapter 2

ncert solutions for class 7 th maths some chapters.

Maths solutions class 10 real numbers

## Ncert solutions for class 6th maths chapter 3, playing with numbers solutions, factors

EXERCISE 1.1

Problem 1

1. Write all the factors of the following numbers:

a. 24, b. 15, c. 21, d. 27

e. 12, f. 20, d. 18, h. 23, i. 36

Solutions :

a. 24

24 = 1 × 2

24 = 2 × 12

24 = 3 × 8

24 = 4 × 6

24 = 6 × 4

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

b. 15

15 = 1 × 15

15 = 3 × 5

15 = 5 × 3

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

c. 21

21 = 1 × 21

21 = 3 × 7

21 = 7 × 3

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

d. 27

27 = 1 × 27

27 = 3 × 9

27 = 9 × 3

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

e. 12

12 = 1 × 12

12 = 2 × 6

12 = 3 × 4

12 = 4 × 3

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

f. 20

20 = 1 × 20

20 = 2 × 10

20 = 4 × 5

20 = 5 × 4

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

g. 18

18 = 1 × 18

18 = 2 × 9

18 = 3 × 6

18 = 6 × 3

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

h. 23

23 = 1 × 23

Therefore, the factors of 23 are 1 and 23.

i. 36

36 = 1 × 36

36 = 2 × 18

36 = 3 × 12

36 = 4 × 9

36 = 6 × 6

36 = 9 × 4

Therefore, 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

Solutions :

a. 5

5 × 1 = 5

5 × 2 = 10

5 × 3 = 15

5 × 4 =20

5 × 5 = 25

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

b. 8

8 × 1 = 8

8 × 2 = 16

8 × 3 = 24

8 × 4 = 32

8 × 5 = 40

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

c. 9

9 × 1 = 9

9 × 2 = 18

9 × 3 = 27

9 × 4 = 36

9 × 5 = 45

Therefore, the first 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. 3 a. Multiple of 8

ii. 1 b. Multiple of 7

iii. 16 c. Multiple of 70

iv. 2 d. Multiple of 30

v. 25 e. Factor of 50

f. Factor of 20

Solutions:

i) – b)

ii) d)

iii- a)

iv) – f)

v) – e)

4. Find all the multiples of 9 up to 100.

Solution :

The multiples of 9 up to 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

# Ncert solutions for class 6th maths chapter 3, playing with numbers solutions, factors solutions

EXERCISE 3.2

Problem 1

1. What is the sum of any two a. Odd numbers ? b. Even numbers ?

Solutions :

a. The sum of any two odd numbers is an even number.

Example : 3 + 7 = 10, 5 + 9 = 14

b. The sum of any two even numbers is an even number.

Example : 4 + 6 = 10, 8 + 4 = 12

Problem 2
2. State whether the following statements are True or False :

a. The sum of three odd numbers is even.

Solution : False

b. The sum of two odd numbers and one even number is even.

Solution : True

c. The product of three odd numbers is odd.

Solution : True

d. If an even number is divided by 2, the quotient is always odd.

Solution : False

e. All prime numbers are odd.

Solution : False

f. Prime numbers do not have any factors.

Solution : False

g. Sum of two prime numbers is always even.

Solution : False

h. 2 is the only even prime number.

Solution : True

i. All even numbers are composite numbers.

Solution : False

j. The product of two even numbers is always even.

Solution : True

Problem 3

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

Solution:

17 and 71

37 and 73

79 and 97

Problem 4

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

Solution :

Prime numbers : 2, 3, 5, 7, 11, 13, 17, 19.

Composite numbers : 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.

Problem 5

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

Solution :

The greatest prime number between 1 and 10 is 7.

Problem 6

6. Express the following as the sum of two odd primes.
a. 44. b. 36, c. 24, d. 18

Solution :

a. 44

3 + 41 = 44

b. 36

7 + 29 = 36

c. 24

5 + 19 =24

d. 18

5 + 13 = 18

Problem 7

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

Solution :

Twin prime pairs :

3 and 5

5 and 7

9 and 11

Problem 8

8. Which of the following numbers are prime ?

a. 23, 51, c. 37, d. 26

Solution :

a. 23 and c. 37 are prime numbers.

Problem 9

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

Solution :

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

Problem 10

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

a. 21, b. 31, c. 53, d.

Solution :

a. 21 = 3 + 7 + 11

b. 31 = 3 + 11 + 17

c. 53 = 11 + 13 + 29

d. 61 = 13 + 19 + 29

Problem 11

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

Solution :

2 + 3 = 5

3 + 7= 10

5 + 5 = 10

2 + 13 = 15

7 + 13 = 20

Problem 12

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 _ .

Answer : prime number nor composite number.

d. The smallest prime number is _ .

e. The smallest composite number is _ .

f. The smallest even number is _ .

## Cbse solutions for class 6th maths,chapter 3,playing with numbers.

EXERCISE 3.3

1. If a number has 0 in its ones place then it is divisible by 10.

2. A number which has either 0 or 5 in its ones place is divisible by 5.

3. A number is divisible by 2 if it has any of the digits 0, 2, 4, 6 or 8 in its ones place.

4. The sum of the digits is a multiple of 3, then the number is divisible by 3.

5. If a number is divisible by 2 and 3 both then it is divisible by 6 also.

6. A number with 3 or more digits is divisible by 4 if the number formed by its last two digits is divisible by 4.

7. A number with 4 more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.

8. If the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.

9. Find the difference between the sum of the digits at odd places ( from right ) and the sum of the digits at even places ( from right ) of a number. If the difference is 0 or divisible by 11, then the number is divisible by 11.

### PROBLEMS WITH SOLUTIONS OF PLAYING WITH SOLUTIONS OF 6TH CLASS MATHS

Problem 1

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.

a. 128 990, b. 990, c. 1586, d. 275, e. 6686, f. 639210, g. 429714, h. 2856, i. 3060, j. 406839

Solutions :

a. 128

128 is divisible by 2, 4, 8

128 is not divisible by 3, 5, 6, 9, 10, 11.

b. 990

990 is divisible by 2, 3, 5, 6, 9, 10, 11.

990 is not divisible by 4, 8

c. 1586

1586 is divisible by 2

1586 is not divisible by 3, 4, 5, 6, 8, 9, 10, 11

d. 275

275 is divisible by 5, 11

275 is not divisible by 2, 3, 4, 6, 8, 9, 10

e. 6686

6686 is divisible by 2

6686 is not divisible by 3, 4, 5, 6, 8, 9, 10, 11

f. 639210

639210 is divisible by 2, 3, 5, 6, 10, 11

639210 is not divisible by 4, 8, 9

g. 429714

924714 is divisible by 2, 3, 6, 9

924714 is not divisible by 4, 5, 8, 10, 11

h. 2856

2856 is divisible by 2, 3, 4, 6, 8

2856 is not divisible by 5, 9, 10, 11

i. 3060

3060 is divisible by 2, 3, 4, 5, 6, 9, 10

3060 is not divisible by 8, 11

j. 406839

406839 is divisible by 3

406839 is not divisible by 2, 4, 5, 6, 8, 9, 10, 11

Problem 2

2. Using divisibility tests, determine which of the following numbers divisible by 4 and 8.
a. 572, b. 726352, c. 5500, d. 6000

e. 12159, f. 14560, g. 2108

h. 31795072, i. 1700, j. 215

Solutions :

a. 572

572 is divisible by 4 as its last two digits are divisible by 4. ( 72 ÷ 4 = 18 ).

572 is not divisible by 8 as its last three digits are not divisible by 8.

b. 726352

72635is divisible by 4 as its last two digits are divisible by 4. ( 52 ÷ 4 = 13 ).

726352 is divisible by 8 as its last three digits are divisible by 8. ( 325 ÷ 8 = 44).

c. 5500

5500 is divisible by 4 as its last two digits are zero

5500 is not divisible by 8 as its last three digits are not divisible by 8.

d. 6000

6000is divisible by 4 as its last two digits are zero.

60 is divisible by 8 as its last three digits are zero.

e. 12159

12159 is not divisible by 4 as its last two digits are not divisible by 4.

12159 is not divisible by 8 as its last three digits are not divisible by 8.

f. 14560

14560 is divisible by 4 as its last two digits are divisible by 4. ( 60 ÷ 4 = 15).

14560cis divisible by 8 as its last three digits are divisible by 8. ( 560 ÷ 8 = 70 ).

g. 21084

21084 is divisible by 4 as its last two digits are divisible by 4. ( 84 ÷ 4 = 21).

21084 is not divisible by 8 as its last three digits are not divisible by 8. ( o84÷ 4=21 ).

h. 31795072

31795072 is divisible by 4 as its last two digits are divisible by 4.( 72 ÷ 4 = 18 ).

31795072 is divisible by 8 as its last three digits are divisible by 8. ( o72 ÷ 4 = 18 ).

i. 1700

1700 is divisible by 4 as its last two digits are zero

1700 is not divisible by 8 as its last three digits are not divisible by 8.

j. 2150

2150 is not divisible by 4 as its last two digits are not divisible by 4.

2150 is not divisible by 8 as its last three digits are not divisible by 8.

Problem 3

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

297144 is divisible by 2 as its ones place is an even number.

297144 is divisible by 3 as sum of its digits is divisible by 3. ( 2 + 9 + 7 + 1 + 4 + 4 = 27 )

Since the number is divisible by both 2 and 3, therefore, it is also divisible by 6.

b. 1258

1258 is divisible by 2 as its ones place is an even number.

1258 is not divisible by 3 as sum of its digits is not divisible by 3. ( 1 + 2 + 5 + 8= 16 )

Since the number is divisible by 2 but not divisible by 3, therefore it is not divisible by 6.

c. 4335

4335 is not divisible by 2 as its ones place is not an even number.

4335 is divisible by 3 as sum of its digits is divisible by 3. ( 4 + 3 + 3 + 5= 15 )

Since the number is not divisible by 2 but divisible by 3, therefore, the number is not divisible by 6.

d. 61233

61233 is not divisible by 2 as its ones place is not an even number.

61233 is divisible by 3 as sum of its digits is divisible by 3. ( 6 + 1 + 2 + 3 + 3 = 15 )

Since the number is not divisible by 2 but divisible by 3, therefore, the number is not divisible by 6.

e. 901352

901352 is divisible by 2 as its ones place is an even number.

901352 is not divisible by 3 as sum of its digits is not divisible by 3.

Since the number is divisible by 2 but not divisible by 3, therefore, the number is not divisible by 6.

f. 438750

438750 is divisible by 2 as its ones place is zoro.

438750 is divisible by 3 as sum of its digits is divisible by 3. ( 4 + 3 + 8 + 7 + 5 + 0 =27 )

Since the number is divisible by 2 and 3, therefore, the number is divisible by 6.

g. 1790184

1790184 is divisible by 2 as its ones place is an even number.

1790184 is divisible by 3 as sum of its digits is divisible by 3. ( 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30 )

Since the number is divisible by 2 and 3, therefore, the number is divisible by 6.

h. 12583

12583 is not divisible by 2 as its ones place is not an even number.

12583 is not divisible by 3 as sum of its digits is not divisible by 3. ( 1 + 2 + 5 + 8 + 3 = 19 )

Since the number is divisible by 2 but not divisible by 3, therefore, the number is not divisible by 6.

i. 639210

639210 is divisible by 2 as its ones place is zero.

639210 is divisible by 3 as sum of its digits is divisible by 3. ( 6 + 3 + 9 + 2 + 1 + 0 = 21 )

Since the number is divisible by 2 and 3, therefore, the number is divisible by 6.

j. 17892

17892 is divisible by 2 as its ones place is an even number.

17892 is not divisible by 3 as sum of its digits is not divisible by 3.

Since the number is divisible by 2 but not divisible by 3, therefore, the number is not divisible by 6.

Problem 4

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. 90115

Solutions:

a. 5445

The sum of the digits at odd places = 5 + 4 = 9

The sum of thi digits at even places = 4 + 5 = 9

The difference of both the sums = 9 – 9 = 0

Since the difference is o, therefore, the number is divisible by 11.

b. 10824

The sum of the digits at odd places = 4 + 8 + 1 = 13

The sum of the digits at even places = 2 + 0 = 2

The difference of both the sums = 13 – 2 = 11

Since the difference is 11, therefore, the number is divisible by 11.

c. 7138965

The sum of the digits at odd places = 5 + 9 + 3 + 7 = 24

The sum of the digits at even places = 6 + 8 + 1 = 15

The difference of both the sums = 24 – 15 = 9

Since the difference is o nor 11, therefore, the number is not divisible by 11.

d. 70169308

The sum of the digits at odd places = 8 + 3 + 6 + 0 = 17

The sum of the digits at even places = 0 + 9 + 1 + 7 = 17

The difference of both the sums = 17 – 17 = 0

Since the difference is o, the number is divisible by 11.

e. 10000001

The sum of the digits at odd places = 1 + 0 + 0 + 0 = 1

The sum of the digits at even places = 0 + 0 + 0 + 1 = 1

The difference of both the sums = 1- 1 = 0

Since the difference is o, the number is divisible by 11.

Problem 5

5. Write the smallest digit and the greatest digit in the blank space of each of the following number formed is divisible by 3 :

a. _ 6724, b. 4765 _ 2

Solution :

a. _6724

If the blank is x, for the smallest digit number

The sum of the digits = x + 6 + 7 + 2 + 4 = x + 19

If x= 0, the sum of the digits = 0 + 6 + 7 + 2 + 4 = 19 is not divisible by 3

If x= 1, the sum of the digits = 1 + 6 + 7 + 2 + 4 = 20 is not divisible by 3

If x= 2, the sum of the digits = 2 + 6 + 7 + 2 + 4 = 21 is divisible by 3

Therefore the smallest digit is 2 in the blank space.

The number is 26724 and is divisible by 3

For the greatest digit number

If x= 9, the sum of the digits = 9 + 6 + 7 + 2 + 4 = 28 is not divisible by 3

If x= 8, the sum of the digits = 8 + 6 + 7 + 2 + 4 = 27 is divisible by 3

Therefore the greatest digit is 8 in the blank space.

The number is 86724.It is divisible by 3.

a. 4765_2

If blank is x. For smallest digit number.

The sum of the digits = 4 + 7 + 6 + 5 + x + 2 = x + 24

If x= 0, the sum of the digits = 4 + 7 + 6 + 5 + 0 + 2 = 24. It is divisible by 3

Therefore the smallest digit is 0 in the blank space.

The number is 476502. It is divisible by 3

For the greatest digit number

If x= 9, the sum of the digits = 4 + 7 + 6 + 5 + 9 + 2= 33. It is divisible by 3

Therefore the greatest digit is 9 in the blank space.

The number is 476592. It is divisible by 3

Problem 6

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

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

The sum of the digits at even places = 8 + x + 9 = x + 17

The difference of the two sums = ( x + 17 ) – 14 = x + 17 – 14 = x + 3

The difference is either o or divisible by 11, then the number is divisible by 11

Therefore x + 3 =11

x = 11 – 3 = 8

The digit in the blank is 8.

Since the number is 928389.

b. 8_9484

The sum of the digits at odd places = 4 + 4 + x = x + 8

The sum of the digits at even places = 8 + 9 + 8 = 25

The difference of the two sums = 25 – ( x + 8 ) = – x – 8 + 25 = – × + 17

The difference is either o or divisible by 11, then the number is divisible by 11

Therefore – x + 17 = 11

– x = – 17 + 11= – 6

Therefore, x = 6

The digit in the blank is 6.

Since the number is 869484

### Ncert solutions for class 6 maths,playing with numbers, Factors solutions

EXERCISE 3.4

Two numbers having only 1 as a common factor are called co – prime numbers.

Problem 1

1. Find the common factors of :

a. 20 and 28, b. 15 and 25

c. 35 and 50, d. 56 and 120

Solutions :

a. 20 and 28

The factors of 20 = 1, 2, 4, 5, 10, 20

The factors of 28 = 1, 2, 4, 7, 14, 28

The common factors of 20 and 28 =1, 2, 4

b. 15 and 25

The factors of 15 = 1, 3, 5, 15

The factors of 25 = 1, 5, 25

The common factors of 15and 25 = 1, 5

c. 35 and 50

The factors of 35 = 1, 5, 7, 35

The factors of 50= 1, 2, 5, 10, 25, 50

The common factors of 35 and 50 = 1, 5

d. 56 and 120

The factors of 56 = 1, 2, 4, 7, 5, 8, 28, 56

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

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

Problem 2

2. Find the common factors of :

a. 4, 8 and 12, b. 5, 15 and 25

Solutions :

a. 4, 8 and 12

The factors of 4 = 1, 2, 4

The factors of 8 = 1, 2, 4, 8

The factors of 12 = 1, 2, 4, 6, 12

The common factors of 4, 8 and 12 = 1, 2, 4

b. 5, 15 and 25

The factors of 5 = 1, 5

The factors of 15 = 1, 3, 5, 15

The factors of 25 = 1, 5, 25

The common factors of 5, 15 and 25 = 1, 5

Problem 3

3. Find first three common multiples of :

a. 6 and 8, b. 12 and 18

Solutions :

a. 6 and 8

The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, …….

The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, …

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

b. 12 and 18

The multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, ……

The multiples of 18 = 18, 36, 54, 72, 90, 108, ….

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

Problem 4

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

Solutions :

The multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99.

The multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96.

The common multiples of 3 and 4 = 12, 24, 36, 48, 60, 72, 84, 96.

Problem 5

5. Which of following numbers are co – prime ?

a. 18 and 35, b. 15 and 37

c. 17 and 68, d. 30 and 415

e. 216 and 215, f. 81 and 16

Solutions :

a. 18 and 35

The factors of 18 = 1, 2, 3, 6, 9, 18

The factors of 35 = 1, 5, 7, 35

The common factors of 18 and 35 = 1

Therefore, both have only one common factor 1 as they are co – prime numbers.

b. 15 and 37

The factors of 15 = 1, 3, 5, 15

The factors of 37 = 1, 37

The common factors of 15 and 37 = 1

Therefore, both have only one common factor 1 as they are co – prime numbers.

c. 30 and 415

The factors of 30 = 1, 2, 3, 5, 6, 15, 30

The factors of 415 = 1, 5, 83, 415

The common factors of 30 and 415 = 1, 5

Therefore, both have more than one common factor, as they are not co – prime numbers.

d. 17 and 68

The factors of 17 = 1, 17

The factors of 68 = 1, 2, 4, 17, 34, 68

The common factors of 17 and 68 = 1, 17

Therefore, both have more than one common factor as they are not co-prime numbers.

e. 216 and 215

The factors of 216 = 1, 2, 3, 4, 6, 8, 12, 24, 27, 36, 54, 72, 108, 216

The factors of 215 = 1, 5, 43, 215

The common factors of 216 and 215 = 1

Therefore, both have only one common factor 1 as they are co – prime numbers.

f. 81 and 16

The factors of 81 = 1, 3, 9, 27, 81

The factors of 16 = 1, 2, 4, 8, 16

The common factors of 81 and 16 = 1

Therefore, both have only one common factor 1 as they are co – prime numbers.

Problem 6

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

Solution :

The first common multiple of both 5 and 12 = 60

5 × 12 = 60

The number must be divisible by 60.

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

Solution :

12 is divisible by 12

The factors of 12 = 1, 2, 3, 4, 6, 12

Therefore, the number also be divisible by 1, 2, 3, 4, 6

Ncert solutions for class 6 maths chapter 3, playing with numbers solutions, factors

EXERCISE 3 .5

Important points to remember

1. If a number is divisible by another number then it is divisible by each of the factors of that number.

2. If a number is divisible by two co – prime numbers then it is divisible by their product also.

3. If two given numbers are divisible by a number, then their sum is also divisible by that number.

4. If two given numbers are divisible by a number, then their difference is also divisible by that number.

Problem 1

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.

Problem 2

2. Here are two different factor trees for 60. Write the missing numbers. Solution : Problem 3

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

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

Problem 4

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

Answer ; The greatest four digit number is 9999.

9999 = 3 × 333 = 3 × 1111 = 3 × 11 × 101

The prime factors of 9999 = 3 × 3 × 11 × 101

Problem 5

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

Answer : The smallest 5 digit number is 10000.

10000 = 2 × 5000

10000 = 2 × 2 × 2500

10000 = 2 × 2 × 2 × 1250

10000 = 2 × 2 × 2 × 2 × 625

10000 = 2 × 2 × 2 × 2 × 5 × 125

10000 = 2 × 2 × 2 × 2 × 5 × 5 × 25

10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5× 5

Therefore prime factors of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Problem 6

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

Solution :

1729 = 7 × 247 = 7 × 13 × 19

Therefore the prime factors of 1729 = 7 × 13 × 19

Relation = The difference of two consecutive prime factors is 6.

Problem 7

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

Answer : Among the three consecutive numbers, there must be one even or odd number and one multiple of 3. Therefore ,the product must be multiple of 6.

Examples : 2 × 3 × 4 =24, 5 × 6 × 7 = 210

Problem 8

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

Solution :

The sum of two consecutive odd numbers is divisible by 4.

Examples :

3 + 5 = 8 and 8 is divisible by 4.

5 + 7 = 12 and 12 is divisible by 4.

7 + 9 = 16 and 16 is divisible by 4.

Problem 9

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

a. 24 = 2 × 3 × 4, b. 56 = 7 × 2 × 2 × 2

c. 70 = 2 × 5 × 7, d. 54 = 2 × 3 × 9

In the expressions b and c the prime factorization has been done.

Problem 10

10. Determine if 25110 is divisible by 45. ( Hint : 5 and 9 are co – prime numbers. Test the divisibility of the number by 5 and 9.

25110 is divisible by 45.

The prime factorization of 45 = 5 × 9

25110 is divisible by 5 because 0 is at its ones place.

25110 is divisible by 9 because the sum of its digits is divisible by 9.

Therefore, 25110 must be divisible by 5 × 9 = 45

Problem 11

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

We can not say that 12 is divisible by both 6 and 4 but 12 is not divisible by 24.

Problem 12

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

Answer : 2 × 3 × 5 × 7 = 210

## Cbse solutions for class 6 maths chapter 6, playing with numbers,

EXERCISE 3.6

The highest common factor ( H C F ) of two or more given numbers is the highest or greatest of their common factors. It is also known as Greatest Common Divisor( G C D ).

Problem 1

1. Find the H C F 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

Solutions :

a. 18, 48

The HCF of 18, 48 can be found by prime factorization.

The factors of 18

18 = 2 × 9

18 = 2 × 3 ×3

The factors of 48

48 = 2 × 24

48 = 2 × 2 × 12

48 = 2 × 2 × 2 × 6

48 = 2 × 2 × 2 × 2 × 3

The common factor of 18 and 48 is 2 x 3.

Thus, HCF of 18 and 48 is 6.

b. 30, 42

The HCF of 30, 42 can be found by prime factorization.

The factors of 30

30 = 2 × 15

30 = 2 × 3 ×5

The factors of 42

42 = 2 × 21

42 = 2 × 3 × 7

The common factor of 30 and 42 is 2 x 3.

Thus, HCF of 30 and 42 is 6.

c. 18, 6o

The HCF of 18, 60 can be found by prime factorization.

The factors of 18

18 = 2 × 9

18 = 2 × 3 ×3

The factors of 60

60 = 2 × 30

48 = 2 × 2 × 15

48 = 2 × 2 × 3 × 5

The common factor of 18 and 60 is 2 x 3.

Thus, HCF of 18 and 60 is 6.

d. 27, 63

The HCF of 27, 63 can be found by prime factorization.

The factors of 27

27 = 3 × 9

18 = 3 × 3 ×3

The factors of 63

63 = 3 × 21

63 = 3 × 3 × 7

The common factor of 27 and 63 is 3 x 3.

Thus, HCF of 27 and 63 is 9.

e. 36, 84

The HCF of 36, 84 can be found by prime factorization.

The factors of 36

36 = 2 × 18

36 = 2 × 2 × 9

36 = 2 × 2 × 3 × 3

The factors of 84

84 = 2 × 42

84 = 2 × 2 × 21

84 = 2 × 2 × 3 × 7

The common factor of 36 and 84 is 2 x 3 × 3.

Thus, HCF of 36 and 84 is 12.

f. 34, 102

The HCF of 34, 102 can be found by prime factorization.

The factors of 34

34 = 2 × 17

The factors of 102

102 = 2 × 51

102 = 2 × 3 × 17

The common factor of 34 and 102 is 2 x 17 .

Thus, HCF of 34 and 102 is 34.

g. 70, 105, 175

The HCF of 70, 105 and 175 can be found by prime factorization.

The factors of 70

70 = 2 × 35

70 = 2 × 5 ×7

The factors of 105

105 = 3 × 35

105 = 3 × 5 × 7

The factors of 175

175 = 5 × 35

175 = 5 × 5 × 7

The common factor of 70, 105 and 175 is 5 × 7.

Thus, HCF of 70, 105 and 175 is 35.

h. 91, 112, 49

The HCF of 91, 112 and 49 can be found by prime factorization.

The factors of 91

91 = 7 × 13

The factors of 112

112 = 2 × 56

112 = 2 × 2 × 28

112 = 2 × 2 × 2 × 14

112 = 2 × 2 × 2 × 2 × 7

The factors of 49

49 = 7 × 7

The common factor of 91, 112 and 49 is 7.

Thus, HCF of 91, 112 and 49 is 7.

i. 18, 54, 81

The HCF of 18, 54 and 81 can be found by prime factorization.

The factors of 18

18 = 2 × 9

18 = 2 × 3 ×3

The factors of 54

54 = 2 × 27

54 = 2 × 3 × 9

54 = 2 × 3 × 3 × 3

The common factor of 18, 54 and 81 is 3 x 3.

Thus, HCF of 18, 54 and 81 is 9.

j. 12, 45, 75

The HCF of 12, 45 and 75 can be found by prime factorization.

The factors of 12

12 = 2 × 6

12 = 2 × 2 ×3

The factors of 45

45 = 3 × 15

45 = 3 × 3 × 5

The factor of 75

75 = 3 × 25

75 = 3 × 5 × 5

The common factor of 12, 45 and 75 is 3.

Thus, HCF of 12, 45 and 75 is 3.

Problem 2

2. What is the HCF of two consecutive
a. numbers ? b. even numbers ?

c. odd numbers. ?

Solutions :

a. Numbers

The two consecutive numbers be 2 and 3

The factors of 2 = 1 , 2

The factors of 3 = 1, 3

The common factor = 1

Therefore,the H C F of two consecutive numbers be 1.

b. Even numbers

The two consecutive even numbers be 4 and 6

The factors of 4 =1, 2 ,4

The factors of 6 = 1, 2, 3, 6

The common factor = 1 × 2 =1

Therefore, the H C F of two consecutive even numbers be 2.

c. Odd numbers

The two consecutive odd numbers be 5 and 7

The factors of 5 = 1, 5

The factors of 7 = 1, 7

The common factor = 1

Therefore, the HCF of two consecutive odd numbers be 1.

Problem 3

3. HCF of co-prime numbers 4 and 15 was found as follows by factorization : 4 = 2 × 2 and 15 = 3 × 5 since 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 :

It is not the correct answer.

4 = 1 × 2 × 2

15 = 1 × 3 × 5

The common factor is 1

The HCF of the co -prime numbers be 1.

## Ncert solutions for class vi maths chapter 3,playing with numbers,Factors

EXERCISE 3.7

The lowest common multiple ( L C M ) of two or more given numbers is the lowest or smallest or least of their common multiples.

Problem 1

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

Solution :

The maximum weight of the fertilizer = The HCF of weights of 75 kgs and 69 kgs

Prime factorization of 75 = 3 × 5 ×5

Prime factorization of 69 = 3 × 23

Common factors of 75 and 69 = 3

Therefore HCF of 75 and 69 = 3

So the maximum weight of required fertilizer is 3 kgs.

Problem 2

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 :

The required minimum distance = the L C M of 63 cm, 70 cm and 77cm.

2|63, 79, 77

3|63, 35, 77

3|21, 15, 77

5|7, 5, 77

7|7, 1, 77

11|1, 1, 11

|1, 1, 1

The L C M of 63, 70, 77 = 2 × 3 × 3 × 5 × 7 × 11

= 6930

Therefore the required minimum distance = 6930 cm.

Problem 3

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 measurement of longest tape = the HCF of 825 cm, 675 cm and 450 cm.

The prime factorization of 825 = 3 × 5 × 5 × 11

The prime factorization of 675 = 3 × 3 × 3 × 5 × 5

The prime factorization of 450 = 2 × 3 × 3 × 5 × 5

Therefore the H C F of 825, 675 and 450 = 75

So the required longest tape = 75 cm.

Problem 4

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

Solution :

The L C M of 6, 8 and 12 =

2| 6, 8, 12

2|3, 4, 6

2|3, 2, 3

3|3, 1, 3

1, 1, 1

Therefore the L C M of 6, 8 and 12 = 2 × 2 × 2 × 3 = 24

The smallest three digit number = 100

We divide 100 by 24, then we get the number.

24 × 4 =96

100 ÷ 24 = 96 +4

24 × 5 = 120. It is the three digit number

So the required number = 100 – 4 + 24 = 120

Therefore 120 is smallest three digit number which is exactly divisible by 6, 8, 12.

Problem 5

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

Solution :

The L C M of 8, 10 and 12 =

2| 8, 10, 12

2|4, 5, 6

2|2, 5, 3

3|1, 5, 3

5|1, 5, 1

|1, 1, 1

Therefore the L C M of 8, 10 and 12 = 2 × 2 × 2 × 3 × 5 = 120

The greatest three digit number = 999

We divide 999 by 120, then we get the number.

120 × 8 = 960

999 ÷ 120 = 960 +39

So the required number = 999 – 39 = 960

Therefore 960 is greatest three digit number which is exactly divisible by 8 10, 12.

Problem 6

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 :

The L C M of 48 seconds, 72 seconds and 108 seconds =

2|48, 72, 108

2|24, 36, 54

2|12, 18, 27

2|6, 9, 27

3|3, 9, 27

3|1, 3, 9

3|1, 1, 3

|1, 1, 1

The L C M = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds

After 432 seconds the lights change simultaneously.

432 seconds = 7 minutes and 12 seconds

( 432 ÷ 60, 1 minute = 60 seconds )

Therefore, the time = 7 a.m + 7 minutes and 12 seconds = 7: 07 : 12 a.m

Problem 7

7. Three tankers contain 403 litres,465 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 required tanker = HCF of 403, 434 and 465

403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

HCF = 31

Therefore, A container of capacity 31 litres can measure the diesel of 3 containers exact number of times.

Problem 8

8. Find the least number which divided by 6, 15 and 18.

Solution :

The L C M of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90

Therefore the required number is 5 more than 90

The required least number = 90 + 5 = 95

Problem 9

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

Solution :

The L C M of 18, 24 and 32

Therefore the L C M of 18, 24 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

The smallest four digit number = 1000

We divide 1000 by 288, we get the number

288 × 3 = 864, 1000 – 864 = 136

288 × 4 = 1152, it is the 4 – digit number

Therefore the required number is 1000 = 1000 – 136 + 288 = 1152

Problem 10

10. Find the L C M 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 L C M. Is L C M the product of two numbers in each case ?

Solutions :

a. 9 and 4

The L C M of 9 and 4 =

2|9, 4

2|9, 2

3|9, 1

3|3, 1

|1, 1

Therefore, the L C M of 9 and 4 = 2 × 2 × 3 × 3 = 36

b. 12 and 5

The L C M of 12 and 5 =

2|12, 5

2|6, 5

3|3, 5

5|1, 5

|1, 1

Therefore, the L C M of 12 and 5 = 2 × 2 × 3 × 5 = 60

c. 6 and 5

The L C M of 6 and 5 =

2|6, 5

3|3, 5

5|1, 5

|1, 1

Therefore, the L C M of 6 and 5 = 2 × 3 × 5 = 30

d.15 and 4

The L C M of 15 and 4 =

2 |15, 4

2|15, 2

3|15, 1

5|5, 1

|1, 1

Therefore, the L C M of 15 and 4 = 2 × 2 × 3 × 5 = 60

Here, in each case the L C M is multiples 2, 3 and 6

Yes, in each case the L C M = the product of two numbers.

Problem 11

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 ?

Solutions :

a. 5 and 20

The L C M of 5 and 12 =

2|5, 20

2|5, 10

5|5, 5

|1, 1

Therefore, the L C M of 5 and 12 = 2 × 2 × 5 = 20

b. 6 and 18

The L C M of 6 and 18 =

2|6, 18

3|3, 9

3|1, 3

|1, 1

Therefore, the L C M of 6 and 18 = 2 × 3 × 3 = 18

c. 12 and 48

The L C M of 12 and 48 =

2|12, 48

2|6, 24

2|3, 12

2|3, 6

3|3, 3

|1, 1

Therefore, the L C M of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48

d. 9 and 45

The L C M of 9 and 45 =

Therefore, the L C M of 9 and 45 = 3 × 3 × 5 = 45

The L C M of the given numbers in each case is the larger of the two numbers.

### Cbse solutions for class vi maths chapter 3, playing with numbers, factors.

Note – Observe the solutions of playing with numbers and try them in your own methods.

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