NCERT Solutions Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.6

 

Playing with Numbers

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

Answer:

a). 18, 48

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

Factors of 48= 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Common factors = 1, 2, 3, 6

HCF of 18, 48 = 6

b). 30, 42

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Factors of 42= 1, 2, 3, 6, 7, 14, 21, 42

Common factors = 1, 2, 3, 6

HCF of 30, 42 = 6

c). 18, 60

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

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

Common factors = 1, 2, 3, 6

HCF of 18, 48 = 6

d). 27, 63

Factors of 27 = 1, 3, 9, 27

Factors of 63= 1, 3, 7, 9, 21, 63

Common factors = 1, 3, 9

HCF of 18, 48 = 9

e). 36, 84

Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Common factors = 1, 2, 3, 4, 6, 12

HCF of 36, 84 = 12

f). 34, 102

Factors of 34 = 1, 2, 17, 34

Factors of 102= 1, 2, 3, 6, 17, 34, 51, 102

Common factors = 1, 2, 17, 34

HCF of 34, 102 = 34

g). 70, 105, 175

Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70

Factors of 105 = 1, 3, 5, 7, 15, 21, 35, 105

Factors of 175 = 1, 5, 7, 25, 35, 175

Common factors = 1, 5, 7, 35

HCF of 70, 105, 175 = 35

h) 91, 112, 49

Factors of 91 = 1, 7, 13, 91

Factors of 112 = 1, 2, 4, 7, 8, 14, 16, 28, 56, 112

Factors of 49 = 1, 7, 49

Common factors = 1, 7

HCF of 91, 112, 49 = 7

i). 18, 54, 81

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

Factors of 54= 1, 2, 3, 6, 9, 18, 27, 54

Factors of 81= 1, 3, 9,  27, 81

Common factors = 1, 3, 9

HCF of 18, 54, 81 = 9

j). 12, 45, 75

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

Factors of 45= 1, 3, 5, 9, 15, 45

Factors of 75= 1, 3, 5, 15, 25, 75

Common factors = 1, 3

HCF of 12, 45, 75 = 3

2). What is the HCF of two consecutive

a). numbers

b). Even numbers

c). odd numbers

a). the HCF of two consecutive numbers is 1.

b). the HCF of two consecutive even numbers is 2.

c). the HCF of two consecutive odd numbers is 1.

3). HCF of co-prime numbers 4 and 15 was found as follows by factorization.  4 = 2 X 2 and 15 = 3 X 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?

No, the answer is not correct.

The correct HCF of 4 and 15 is 1.

 

 

 

 

 

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