Topic outline

    • 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.