## NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers Exercise 3.6

**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) 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 x 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 x 3 = 6.

(c) 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 x 3 = 6.

(d) Given numbers are 27 and 63.

Prime factorisations of 27 and 63 are:

Here, the common factor is 3 (occurring twice).

Hence, the HCF = 3 x 3 = 9.

(e) 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 x 2 x 3 = 12.

(f) Given numbers are 34 and 102.

Prime factorisations of 34 and 102 are:

Here, the common factors are 2 and 17.

Thus, HCF is 2 x 17 = 34.

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

Prime factorisatios of 70, 105 and 175 are:

Here, common factors are 5 and 7.

Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35.

(h) 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) Given numbers are 18, 54 and 81.

Prime factorisations of 18, 54 and 81 are:

Here, the common factor is 3 (occurring twice).

Thus, the HCF = 3 x 3 = 9.

(j) 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 = 1.

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

Hence, the HCF = 1 x 2 = 2.

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

Hence, the HCF = 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 x 2 and 15 = 3 x 15. 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, answer is not correct.

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

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

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