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## Karnataka State Syllabus Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6

Question 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

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

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

Common factors = 1, 2, 3, 6

H.C.F of 18 & 48 = \(\boxed { 6 } \)

b) 30, 42

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

42 = 1, 2, 3, 4, 6, 7, 14, 21, 42

Common factors = 1, 2, 3, 6

∴ H.C.F. = \(\boxed { 6 } \)

∴ H.C.F of 30 & 42 is \(\boxed { 6 } \)

c) 18 & 60

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

60 = 1, 2, 3, 4, 6, 10, 12, 15, 30, 60

Common factors = 1, 2, 3, 6

∴ H.C.F. of 18 & 60 is \(\boxed { 6 } \)

d) 27 & 63

27 = 1, 3, 9, 27

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

∴ Common factors = 1, 3, 9

∴ H.C.F. of 27 & 63 is = 9.

e) 36 & 84

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

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

∴ Common factors = 1, 2, 3, 4, 6, 12

∴ H.C.F. of 36 & 84 is 12.

f) 34 & 102

34 = 1, 2, 17, 34

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

∴ Common factors = 1, 2, 17, 34

∴ H.C.F. of 34 & 102 is 34.

g) 70, 105, 175

70 = __1__, 2, __5__, 7, 10, 14, __35__, 70

105 = __1__, 3, __5__, __7__, 10, 14, __35__, 105

175 = __1__, __5__, __7__, 25, __35__, 175

∴ Common factors 1, 5, 7, 35

∴ H.C.F. of 70, 105, 175 is 35

h) 91, 112, 49

91 = __1__, __7__, 13, 91

112 = __1__, 2, 4, __7__, 8, 14, 16, 28, 56, 112

49 = __1__, __7__, __49__

∴ Common factors = 1, 7

∴ H.C.F. of 91, 112, 49 is 7

i) 12, 45, 75

12 = __1__, 2, 3, 4, 6, 12

45 = __1__, 3, 5, 9, 15, 45

75 = __1__, 3, 5, 15, 26, 75

∴ Common factors 1, 3

∴ H.C.F of 12, 45, 75 is 3.

Question 2.

What is the HCF of two consecutive

(a) numbers?

(b) even numbers?

(c) odd numbers?

Find the HCF of the following:

(i) 24 and 36

(ii) 15, 25 and 30

(iii) 8 and 12

(iv) 12, 16 and 28

Solution:

(i) 1 e.g., HCF of 2 and 3 is 1.

(ii) 2 e.g., HCF of 2 and 4 is 2.

(iii) 1 e.g., HCF of 3 and 5 is 1.

To do the prime factorization of 84, successively divide 84 by prime numbers.

Question 3.

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

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:

No. The answer is not correct. 1 is the correct HCF.