# Mathematics (RD Sharma Solution) - Class 6 / Grade 6

- Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.6)
### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.6)

Tags: RD Sharma Class 6 Math Solution Chapter 2 Playing with Numbers Exercise 2.6, Solution of RD Sharma for class 6, RD Sharma solutions Grade VI, Solution of RD Sharma exercises for sixth standard, R.D. Sharma Mathematics solutions and practice pages, Playing With Numbers Exercise 2.6 (Higest Common Factor-H.C.F)Playing With Numbers

Exercise 2.6 (Highest Common Factor – H.C.F)

Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. 144, 198

First, we find the prime factorization of each of the given numbers by division method.

2

144

2

72

2

36

2

18

3

9

3

2

198

3

99

3

33

3

11

Thus 144 = 2x2x2x2x3x3

198 = 2x3x3x3x11

Therefore, HCF of 144 and 198 is 2x3x3 (Common factors)= 18

ii. 81, 117

First, we find the prime factorization of each of the given numbers by division method.

3

81

3

27

3

9

3

3

117

3

39

13

Thus 81 = 3x3x3x3

117 = 3x3x13

Therefore, HCF of 81 and 117 is 3x3 (Common factors)= 9

iii. 84, 98

First, we find the prime factorization of each of the given numbers by division method.

2

84

2

42

3

21

7

2

98

7

49

7

Thus 84= 2x2x3x7

98 = 2x7x7

Therefore, HCF of 84 and 98 is 2x7 (Common factors)= 14

iv. 225, 450

First, we find the prime factorization of each of the given numbers by division method.

3

225

3

75

5

25

5

2

450

3

225

3

75

5

25

5

Thus 225 = 3x3x5x5

450 = 2x3x3x5x5

Therefore, HCF of 225 and 450 is 3x3x5x5 (Common factors)= 225

v. 170, 238

First, we find the prime factorization of each of the given numbers by division method.

2

170

5

85

17

2

238

7

119

17

Thus 170 = 2x5x17

238 = 2x7x17

Therefore, HCF of 170 and 238 is 2x17 (Common factors)= 34

vi. 504, 980

First, we find the prime factorization of each of the given numbers by division method.

2

504

2

252

2

126

3

63

3

21

7

2

980

2

490

5

245

7

49

7

Thus 504 = 2x2x2x3x3x7

980 = 2x2x5x7x7

Therefore, HCF of 504 and 980 is 2x2x7 (Common factors)= 28

vii.150, 140, 210

First, we find the prime factorization of each of the given numbers by division method.

2

150

3

75

5

25

5

2

140

2

70

5

35

7

2

210

3

105

5

35

7

Thus 150 = 2x3x5x5

140 = 2x2x5x7

210 = 2x3x5x7

Therefore, HCF of 150, 140 and 210 is 2x5 (Common factors)= 10

viii. 84, 120, 138

First, we find the prime factorization of each of the given numbers by division method.

2

84

2

42

3

21

7

2

120

2

60

2

30

3

15

5

2

138

3

69

23

Thus 84 = 2x2x3x7

120 = 2x2x2x3x5

138 = 2x3x23

Therefore, HCF of 84, 120 and 138 is 2x3 (Common factors)= 6

ix. 106, 159, 265

First, we find the prime factorization of each of the given numbers by division method.

2

106

53

3

159

53

5

265

53

Thus 106 = 2x53

159 = 3x53

265 = 5x53

Therefore, HCF of 106, 159 and 265 is 53 (Common factors)= 53

Question 2 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

i. H.C.F of two consecutive numbers is 1.

ii. H.C.F of two consecutive even numbers is 2.

iii. H.C.F of two consecutive odd numbers is 1

Question 3 (Refer Book - Mathematics Class VI R.D. Sharma)

Solution:

No, the answer is not correct. The correct H.C.F is 1.