NCERT Solutions
Class 6 Maths
Playing with Numbers

Ex. 3.6 Q.1
Find the H.C.F. 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
(a) Factors of 18 = 2 * 3 * 3
Factors of 48 = 2 * 2 * 2 * 2 * 3
H.C.F. (18, 48) = 2 * 3 = 6
(b) Factors of 30 = 2 * 3 * 5
Factors of 42 = 2 * 3 * 7
H.C.F. (30, 42) = 2 * 3 = 6
(c) Factors of 18 = 2 * 3 * 3
Factors of 60 = 2 * 2 * 3 * 5
H.C.F. (18, 60) = 2 * 3 = 6
(d) Factors of 27 = 3 * 3 * 3
Factors of 63 = 3 * 3 * 7
H.C.F. (27, 63) = 3 * 3 = 9
(e) Factors of 36 = 2 * 2 * 3 * 3
Factors of 84 = 2 * 2 * 3 * 7
H.C.F. (36, 84) = 2 * 2 * 3 = 12
(f) Factors of 34 = 2 * 17
Factors of 102 = 2 * 3 * 17
H.C.F. (34, 102) = 2 * 17 = 34
(g) Factors of 70 = 2 * 5 * 7
Factors of 105 = 3 * 5 * 7
Factors of 175 = 5 * 5 * 7
H.C.F (70, 105, 175) = 5 * 7 = 35
(h) Factors of 91 = 7 * 13
Factors of 112 = 2 * 2 * 2 * 2 * 7
Factors of 49 = 7 * 7
H.C.F.(91, 112, 49) = 1 * 7 = 7
(i) Factors of 18 = 2 * 3 * 3
Factors of 54 = 2 * 3 * 3 * 3
Factors of 81 = 3 * 3 * 3 * 3
H.C.F.(18, 54, 81) = 3 * 3 = 9
(j) Factors of 12 = 2 * 2 * 3
Factors of 45 = 3 * 3 * 5
Factors of 75 = 3 * 5 * 5
H.C.F.(12, 45, 75) = 1 * 3 = 3