# GSEB Solutions Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6

Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 6 Maths Chapter 3 Playing With Numbers Ex 3.6

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
Solution:
(a) 18 and 48
Factors of 18 are: 1, 2, 3, 6, 9 and 18
Factors of 48 are: 1,2, 3,4, 6, 8, 12, 16, 24 and 48
Their common factors are: 1, 2, 3 and 6
Since, the highest common factor is 6,
HCF of 18 and 48 = 6 (b) 30 and 42
Factors of 30 are: 1, 2, 3, 5, 6, 10, 15 and 30
Factors of 42 are: 1, 2, 3, 6, 7, 14, 21 and 42
Their common factors are: 1, 2, 3, 6
Since, the highest common factor is 6,
HCF of 30 and 42 = 6

(c) 18 and 60
Factors of 18 are: 1, 2, 3, 6, 9 and 18
Factors of 60 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60
Their common factors are: 1, 2, 3, 6
Since, the highest common factor is 6,
HCF of 18 and 60 = 6

(d) 27 and 63
Factors of 27 are: 1, 3, 9 and 27
Factors of 63 are: 1, 3, 7, 9, 21 and 63
Their common factors are: 1, 3 and 9
Since, the highest common factor is 9,
HCF of 27 and 63 = 9 (e) 36 and 84
Factors of 36 are: 1,2,3,4,6,9, 12, 18 and 36
Factors of 84 are: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
Their common factors are: 1, 2, 3, 4, 6 and 12
Since, the highest common factor is 12
HCF of 36 and 84 = 12

(f) 34 and 102
Factors of 34 are: 1, 2, 17 and 34
Factors of 102 are: 1, 2, 3, 6, 17, 34, 51 and 102
Common factors are: 1, 2, 17 and 34
Since, the highest common factor is 34,
HCF of 34 and 102 = 34

(g) 70, 105 and 175
Factors of 70 are: 1, 2, 5, 7, 10, 14, 35 and 70.
Factors of 105 are: 1, 3, 5, 7, 15, 21, 35 and 105
Factors of 175 are: 1, 5, 7, 25, 35 and 175
Their common factors are: 1, 5, 7 and 35.
Since, the highest common factor is 35,
HCF of 70, 105 and 175 = 35

(h) 91, 112 and 49
Factors of 91 are: 1, 7, 13 and 91
Factors of 112 are: 1, 2, 4, 7, 8, 14, 16, 28, 56 and 112
Factors of 49 are: 1, 7 and 49
Their common factors are: 1 and 7
Since, the highest common factor = 7,
HCF of 91, 112 and 49 = 7 (i) 18, 54 and 81
Factors of 18 are: 1, 2, 3, 6, 9 and 18
Factors of 54 are: 1,2,3,6,9, 18,27 and 54
Factors of 81 are: 1, 3, 9, 27 and 81
Their common factors are: 1, 3 and 9
Since, their highest common factor is 9
HCF of 18, 54 and 81 – 9

(j) 12, 45 and 75
Factors of 12 are: 1, 2, 3, 4, 6 and 12
Factors of 45 are: 1, 3, 5, 9, 15 and 45
Factors of 75 are: 1, 3, 5, 15, 25 and 75
Their common factors are: 1 and 3
Since, the highest common factor is 3
HCF of 12, 45 and 75 = 3 Question 2.
What is the HCF of two consecutive
(a) numbers ?
(b) even numbers ?
(c) odd numbers ?
Solution:
(a) The common factor of two consecutive numbers is always 1.
HCF of two consecutive numbers = 1

(b) The common factors of two consecutive even numbers are 1 and 2.
The HCF of two consecutive even numbers = 2

(c) The common factor of two consecutive odd numbers is 1.
HCF of two consecutive odd numbers = 1 Question 3.
HCF 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 HCF of 4 and 15 is 0. Is the answer correct ? If not, what is the correct HCF?
Solution:
cannot be a factor of any number.
1 is a factor of every number.
No common factor means, 1 is the common factor.
The given answer is incorrect. The correct
answer is: HCF of 4 and 15 = 1