NCERT Solutions For Class 6 Maths Chapter 3

Playing With Numbers

NCERT Solutions For Class 6 Maths chapter 3 Exercise 3.6

Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.5 Exercise 3.6 Exercise 3.7

Question 1

Find the HCF 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

Answer 1

(a) 8,48
Factor of 18 = 2 x 3 x 3
Factor of 48 = 2 x 2 x 2 x 2 x 3
Required HCF = 2 x 3 = 6
(b) 30, 42
factor of 30 = 2 x 3 x 5
Factor of 42 = 2 x 3 x 7
Required HCF = 2 x 3 = 6
(c) 18,60
Factor of 18 = 2 x 3 x 3
Factor of 60 = 2 x 2 x 3 x 5
Required HCF =2x 3 =6
(d) 27,63
Factors of 27 = 1 x 3 x 3 x 3.
Factors of 63 =1 x 3 x 3 x 7.
Highest of these common factors is 3 x 3
. H.C.F. of 27 , 63 is 9. (e) 36,84
36 = 2 × 2 × 3 × 3
84 = 2 × 2 × 3 × 7
Required HCF = 2 × 2 × 3 = 12
f) 34,102
34 = 2 × 17
102 = 2 × 3 × 17
Required HCF = 2 × 17 = 34
So,the HCF of 34, 102 is 34

(g) 70,105,175
Factors of 70 = 2 × 5 × 7
Factors of 105 = 3 × 5 × 7
Factors of 175 = 5 × 5 × 7
Required HCF = 5 × 7 = 35
So,the HCF of 70, 105 and 175 is 35 (h) 91,112,49
Factors of 91 = 7 × 13
Factors of 112 = 2 × 2 × 2 × 2 × 7
Factors of 49 = 7 × 7
Required HCF = 7
So,the HCF of 91, 112 and 49 is 7
(i) 18,54,81
Factors of 18 = 2 × 3 × 3
Factors of 54 = 2 × 3 × 3 × 3
Factors of 81 = 3 × 3 × 3 × 3<
Required HCF = 3 × 3 = 9
Therefore the HCF of 18, 54 and 81 is 9
(j) 12, 45, 75
Factors of 12 = 2 × 2 × 3
Factors of 45 = 3 × 3 × 5
Factors of 75 = 3 × 5 × 5
Required HCF = 3
So the HCF of 12, 45 and 75 is 3

Question 2

What is the H.C.F. of two consecutive :
(a) numbers?
(b) even numbers?
(c) odd numbers?

Answer 2

(a) The H.C.F. of two consecutive numbers is 1.
Ex: The HCF of 11 and 12 is 1
(b) The H.C.F. of two consecutive even numbers is 2.
Ex The HCF of 6 and 8 is 2
(c) The H.C.F. of two consecutive odd numbers is 1.
ExThe HCF of 7 and 9 is 1

Question 3

H. C.F. of co-prime numbers 4 and 15 was found as follows factorization: 4 = 2 × 2 and 15 = 3 × 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?