Chapter - 3 - EXERCISE 3.6 - Playing with Numbers (NCERT Class 6th Maths Solution)

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


Solution/Explanation:
(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) = 2 x 2 x 3 = 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 x 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


2. What is the HCF of two consecutive (a) numbers? (b) even numbers? (c) odd numbers?


Solution/Explanation:

(a) H.C.F. of two consecutive numbers be 1. 
(b) H.C.F. of two consecutive even numbers be 2. 
(c) H.C.F. of two consecutive odd numbers be 1.

3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation : 
4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF? 

Solution/Explanation:

No, the answer is not correct. 1 is the correct HCF.

As 

4 = 2  2  1

15 = 3  5  1

1 is common in both so HCF is 1.