## Topic outline

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

• 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.