Well-designed AP 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise 3.6 offers step-by-step explanations to help students understand problem-solving strategies.
Playing with Numbers Class 6 Exercise 3.6 Solutions – 6th Class Maths 3.6 Exercise Solutions
Question 1.
Find the HCF of the following numbers:
a) 18,48
Solution:
Given numbers are 18 and 48.
Prime factorisation of 18 and 48 are:
Here,
common factors are 2 and 3.
Hence, the HCF = 2 × 3 = 6
b) 30,42
Solution:
The given numbers are 30 and 42.
Prime factorisation of 30 and 42 are:
Here
common factors are 2 and 3.
Hence, the HCF of 30 and 42 = 2 × 3 = 6
c) 18, 60
Solution:
Given numbers are 18 and 60.
Prime factorisation of 18 and 60 are :
common factors are 2 and 3.
Hence, the HCF of 18 and 60 = 2 × 3 = 6
d) 27,63
Solution:
Given numbers are 27 and 63.
Prime factorisation of 27 and 63 are :
Here,
common factor is 3 (occurring twice)
Hence, the HCF = 3 × 3 = 9
e) 36,84
Solution:
Given numbers are 36 and 84.
Prime factorisation of 36 and 84 are :
Here,
common factors are 2,2 and 3.
Hence, the HCF = 2 × 2 × 3 = 12
f) 34,102
Solution:
Given numbers are 34 and 102.
Prime factorisation of 34 and 102 are :
Here,
common factors are 2 and 17.
Hence, the HCF = 2 × 17 = 34
g) 70,105,175
Solution:
Given numbers are 70, 105 and 175
Prime factorisation of 70,105 and 175 are :
Here, common factors are 5 and 7.
Hence, the HCF of 70, 105 and 175 is 5 × 7 = 35
h) 91,112,49
Solution:
Given numbers are 91, 112 and 49.
Prime factorisation of 91,112 and 49 are:
Here, common factor is 7 .
∴ HCF of 91,112 and 49 is 7.
i) 18,54,81
Solution:
Given numbers are 18, 54 and 81 .
Prime factorisation of 18,54 and 81 are:
Here, common factor is 3 (occurring twice)
Hence, the HCF = 3 × 3 = 9
j) 12, 45, 75
Solution:
Given numbers are 12, 45 and 75.
Prime factorisation of 12,45 and 75 are:
Here, common factors is 3.
Hence, the HCF = 3
Question 2.
What is the HCF of two consecutive
a) numbers?
Answer:
The HCF of two consecutive numbers = 1
b) even numbers?
Answer:
The HCF of two consecutive even numbers = 2
c) odd numbers?
Answer:
The 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 × 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:
No, given answer is not correct.
Reason : ‘0’ is not a factor of any nonzero numbers. 1 is a factor of each and every number.
∴ HCF of 4 and 15 is 1.
Hence, the correct HCF of 4 and 15 is 1.