## Topic outline

• ### Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.9)

• Playing With Numbers

Exercise 2.9 (Lowest Common Multiple – L.C.M)

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

Solution:

i. 48, 60

We have,

 2 48, 60 2 24, 30 3 12, 15 2 4,   5 2 2,   5 5 1,   5 1,   5

Therefore, L.C.M of given numbers = 2x2x2x2x3x5 = 240

ii. 42, 63

We have,

 21 42, 63 2 2,   3 3 1,   3 1,   1

Therefore, L.C.M of given numbers = 21x2x3 = 126

iii. 18, 17

We have,

 17 18, 17 2 18,  1 3 9,   1 3 3,   1 1,   1

Therefore, L.C.M of given numbers = 17x2x3x3 = 306

iv. 15, 30, 90

We have,

 3 15, 30, 90 5 5,  15, 30 3 1,   3,  6 2 1,   1,  2 1,   1,   1

Therefore, L.C.M of given numbers = 3x5x3x2 = 90

v. 56, 65, 85

We have,

 5 56, 65, 85 2 56, 13, 17 2 28, 13, 17 2 14, 13, 17 7 7,  13, 17 13 1,  13, 17 17 1,   1,  17 1,   1,  1

Therefore, L.C.M of given numbers = 2x2x2x5x7x13x17 = 61880

vi. 180, 384, 144

We have,

 2 180, 384, 144 2 90,  192,  72 2 45,   96,   36 2 45,   48,   18 3 45,   24,    9 3 15,    8,    3 2 5,     8,    1 2 5,     4,    1 2 5,     2,    1 5 5,     1,    1 1,     1,    1

Therefore, L.C.M of given numbers = 2x2x2x2x2x2x2x3x3x5 = 5760

vii. 108, 135, 162

We have,

 3 108, 135, 162 3 36,   45,  54 3 12,   15,  18 2 4,     5,    6 2 2,     5,    3 3 1,     5,    3 5 1,     5,    1 1,     1,    1

Therefore, L.C.M of given numbers = 2x2x3x3x3x3x5 = 1620

viii. 28, 36, 45, 60

We have,

 2 28, 36, 45, 60 2 14, 18, 45, 30 3 7,   9,  45, 15 3 7,   3,  15,  5 5 7,   1,   5,   5 7 7,   1,   1,   1 1,   1,   1,   1

Therefore, L.C.M of given numbers = 2x2x3x3x5x7= 1260