RD Sharma Solutions -Ex-2.6, Playing With Numbers, Class 6, Maths Class 6 Notes | EduRev

RD Sharma Solutions for Class 6 Mathematics

Created by: Abhishek Kapoor

Class 6 : RD Sharma Solutions -Ex-2.6, Playing With Numbers, Class 6, Maths Class 6 Notes | EduRev

The document RD Sharma Solutions -Ex-2.6, Playing With Numbers, Class 6, Maths Class 6 Notes | EduRev is a part of the Class 6 Course RD Sharma Solutions for Class 6 Mathematics.
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Q. 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

 

Q. 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

 

Q. 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.

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