**Playing With Numbers Class 6 Ex. 3.6**

**Ncert Class 6 Math Free Solution.**

**Exercise 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**

**Solution1 :-**

(a) Given numbers are 18 and 48.

The factors of 18 = 2 x 3 x 3

The factors of 48 = 2 X 2 X 2 X 2 X 3

The common factors are 2 and 3.

Hence, the H.C.F (18, 48) = 2 x 3 = 6.

(b) Given numbers are 30 and 42.

The factors of 30 = 2 X 3 X 5

The factors of 42 = 2 x 3 x 7

The common factors are 2 and 3.

Hence, the H.C.F (30, 42) = 2 x 3 = 6.

(c) Given numbers are 18 and 60.

The factors of 18 = 2 x 3 x 3

The factors of 60 = 2 x 2 x 3 x 5

The common factors are 2 and 3.

Hence, the H.C.F (18, 60) = 2 x 3 = 6.

(d) Given numbers are 27 and 63.

The factors of 27 = 3 x 3 x 3

The factors of 63 = 3 X 3 X 7

The common factors are 2 and 3.

Hence, the H.C.F (27, 63) = 3 x 3 = 9.

(e) Given numbers are 36 and 84.

The factors of 36 = 2 X 2 X 3 X 3

The factors of 84 = 2 x 2 x 3 x 7

The common factors are 2 and 3.

Hence, the H.C.F (36, 84) = 2 x 2 x 3 = 12.

(f) Given numbers are 34 and 102.

The factors of 34 = 2 x 17

The factors of 102 = 2 x 3 x 17

The common factors are 2 and 17.

Hence, the H.C.F (34, 102) = 2 x 17 = 34.

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

The factors of 70 = 2 x 5 x 7

The factors of 105 = 3 x 5 x 7

The factors of 175 = 5 x 5 x 7

The common factors are 5 and 7.

Hence, the H.C.F (70, 105, 175) = 5 x 7 = 35.

(h) The given numbers are 91, 112 and 49.

The factors of 91 = 7 X 13

The factors of 112 = 2 x 2 x 2 x 2 x 7

The factors of 49 = 7 x 7

The common factors areĀ 7.

Hence, the H.C.F (91, 112, 49) = 7

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

The factors of 18 = 2 x 3 x 3

The factors of 54 = 2 X 3 X 3 X 3

The factors of 81 = 3 X 3 X 3 X 3

The common factors are 3 and 3.

Hence, the H.C.F (18, 54, 81) = 3 x 3 = 9.

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

The factors of 12 = 2 X 2 X 3

The factors of 45 = 3 x 3 x 5

The factors of 75 = 3 x 5 x 5

The common factors are 3.

Hence, the H.C.F (12, 45, 75) = 3.

**Question 2 :- What is the HCF of two consecutive**

**(a) numbers?**

**(b) even numbers?**

**(c) odd numbers?**

**Solution 2:-**

(a) H.C.F of two consecutive numbers is always 1.

(b) H.C.F of two consecutive even numbers are 1 and 2.

So, the H.C.F = 1 x 2 = 2.

(c) H.C.F of two consecutive odd numbers is 1.

So, the H.C.F = 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 factors, so HCF of 4 and 15 is 0. ****Is the answer correct? If not, what is the correct HCF?**

**Solution 3:-**

No, answer is not correct.

0 is not a prime factor of any number, for this reason.

The prime factor of a co-prime number is always 1.

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

**Playing With Numbers Class 6 Exercise. 3.6 for Free**