**NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7**

**NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6 ****(Rationalised Contents)**

**Ex 3.6 Class 6 Maths 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) The given numbers are 18 and 48.

Prime factorisations of 18 and 48 are:

Here, the common factors are 2 and 3.

Hence, the HCF = 2 × 3 = 6

(b) The given numbers are 30 and 42.

Prime factorisations of 30 and 42 are:

Here, the common factors are 2 and 3.

Hence, the HCF = 2 × 3 = 6

(c) The given numbers are 18 and 60.

Prime factorisations of 18 and 60 are:

Here, the common factors are 2 and 3.

Hence, the HCF of 18 and 60 = 2 × 3 = 6

(d) The given numbers are 27 and 63.

Prime factorisations of 27 and 63 are:

Here, the common factor is 3 × 3.

Hence, the HCF = 3 × 3 = 9

(e) The given numbers are 36 and 84.

Prime factorisations of 36 and 84 are:

Here, the common factors are 2, 2 and 3.

Hence, the HCF = 2 × 2 × 3 = 12

(f) The given numbers are 34 and 102.

Prime factorisations of 34 and 102 are:

Here, the common factors are 2 and 17.

Thus, the HCF = 2 × 17 = 34

(g) The given numbers are 70, 105 and 175.

Prime factorisations 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) The given numbers are 91, 112 and 49.

Prime factorisations of 91, 112 and 49 are:

Here, the common factor is 7.

Hence, the HCF = 7

(i) The given numbers are 18, 54 and 81.

Prime factorisations of 18, 54 and 81 are:

Here, the common factor is 3 × 3.

Thus, the HCF = 3 × 3 = 9

(j) The given numbers are 12, 45 and 75.

Prime factorisations of 12, 45 and 75 are:

Here, the common factor is 3.

Hence, the HCF = 3

**Ex 3.6 Class 6 Maths 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.

Hence, the HCF of two consecutive numbers = 1

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

Hence, the HCF of two consecutive even numbers = 1 × 2 = 2

(c) The common factor of two consecutive odd numbers is 1.

Hence, the HCF of two consecutive odd numbers = 1

**Ex 3.6 Class 6 Maths 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 factors, so HCF of 4 and 15 is 0.

Is the answer correct? If not, what is the correct HCF?

**Solution:
**No, the answer is not correct.

Reason: 0 is not the prime factor of any number.

1 is always the prime factor of co-prime numbers.

Hence, the correct HCF of 4 and 15 is 1.

**Related Links:**

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