These NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.6
Question 1.
FindtheH.C.F. ofthe 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
Answer:
(a) Factors of 18 = 2 x 3 x 3
Factors of 48 = 2 x 2 x 2 x 2 x 3
H.C.F. (18, 48) = 2 x 3=6
(b) Factors of 30 = 2 x 3 x 5
Factors of 42 = 2 x 3 x 7
H.C.F. (30, 42) = 2 x 3 = 6
(c) Factors of 18 = 2 x 3 x 3
Factors of 60 = 2 x 2 x 3 x 5
H.C.F. (18, 60) = 2 x 3 = 6
(d) Factors of 27 = 3 x 3 x 3
Factors of 63 = 3 x 3 x 7
H.C.F. (27, 63) = 3 x 3=9
(e) Factors of 36 = 2 x 2 x 3 x 3
Factors of 84 = 2 x 2 x 3 x 7
H.C.F. (36, 84) = 2x2x3 = 12
(f) Factors of 34 = 2 x 17
Factors of 102 = 2 x 3 x 17
H.C.F. (34, 102) = 2 x 17 = 34
(g) Factors of 70 = 2 x 5 x 7
Factors of 105 = 3 x 5 x 7
Factors of 175 = 5 x 5 x 7
H.C.F. = 5 x 7 = 35
(h) Factors of 91 = 7 x 13
Factors of 112 = 2 x 2 x 2 x 2 x 7
Factors of 49 = 7 x 7
H.C.F. = 1 x 7 = 7
(i) Factors of 18 = 2 x 3 x 3
Factors of 54 = 2 x 3 x 3 x 3
Factors of 81 = 3 x 3 x 3 x 3
H.C.F. = 3×3 = 9
(j) Factors of 12 = 2 x 2 x 3
Factors of 45 = 3 x 3 x 5
Factors of 75 = 3 x 5 x 5
H.C.F. = 1 x 3 = 3
The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.
Question 2.
What is the H.C.F. of two consecutive:
(a) numbers?
(b) even numbers?
(c) odd numbers?
Answer:
(a) H.C.F. of two consecutive numbers is 1.
(b) H.C.F. of two consecutive even numbers be is.
(c) H.C.F. of two consecutive odd numbers be is.
Question 3.
H.C.F. of co-prime numbers 4 and 15 was found as follows by factorisation:
4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F.?
Answer:
No, The correct H.C.F. is 1.