RD Sharma Solutions Class 6 Playing With Numbers Exercise 2.6

RD Sharma Solutions Class 6 Chapter 2 Exercise 2.6

RD Sharma Class 6 Solutions Chapter 2 Ex 2.6 PDF Free Download

Exercise: 2.6

1.) Find the H.C. F of the following numbers using prime factors using prime factorization method:

Answer:

(i) 144 and 198

Prime factorization of 144 = 2 x 2 x 2 x 3 x 3

Prime factorization of 198 = 2 x 3 x 3 x11

Therefore, HCF = 2 x 2 x 3 = 18

(ii) 81 and 117

Prime factorization of 81 = 3 x 3 x 3 x 3

Prime factorization of 117 = 3 x 3 x13

Therefore, HCF = 3 x 3 = 9

(iii) 84 and 98

Prime factorization of 84 = 2 x 2 x 3 x 7

Prime factorization of 98 = 2 x 7 x 7

Therefore, HCF = 2 x 7 = 14

(iv) 225 and 450

Prime factorization of 225 = 3 x 3 x 5 x 5

Prime factorization of 198 = 2 x 3 x 3 x 5 x 5

Therefore, HCF = 3 x 3 x 5 x 5 = 225

(v) 170 and 238

Prime factorization of 170 = 2 x 5 x 17

Prime factorization of 238 = 2 x 7 x 17

Therefore, HCF = 2 x 17 = 34

(vi) 504 and 980

Prime factorization of 504 = 2 x 2 x 2 x 3 x 3 x 7

Prime factorization of 980 = 2 x 2 x 5 x 7 x 7

Therefore, HCF = 2 x 2 x 7 = 28

(vii) 150, 140 and 210

Prime factorization of 150 = 2 x 3 x 5 x 5

Prime factorization of 140 = 2 x 2 x 5 x 7

Prime factorization of 210 = 2 x 3 x 5 x 7

Therefore, HCF = 2 x 5 = 10

(viii) 84, 120 and 138

Prime factorization of 84= 2 x 2 x 3 x 7

Prime factorization of 120 = 2 x 2 x 2 x 3 x 5

Prime factorization of 138 = 2 x 3 x 23

Therefore, HCF = 2 x 3 = 6

(ix) 106, 159 and 265

Prime factorization of 106 = 2 x 53

Prime factorization of 159= 2 x 53

Prime factorization of 265 = 5 x 53

Therefore, HCF = 53

2.) What is the H.C.F of two consecutive?

Answer:

(i) The common factor of two consecutive numbers is always 1.

Therefore, HCF of two consecutive numbers = 1

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

Therefore, HCF of two consecutive even numbers = 2

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

Therefore, HCF of two consecutive odd numbers = 1

3.) H.C.F of co-primes numbers 4 and 15 was found as follows:

4 = 2 x 2 and 15 = 3 x 5

Since there is no common prime factor. So, H.C.F of 4 and 15 is 0. Is the answer correct? If not what is the correct H.C.F?

Answer:

No, it is not correct.

We know that HCF of two co-prime number is 1.

4 and 15 are co-prime numbers because the only factor common to them is 1.

Thus, HCF of 4 and 15 is 1.