In this chapter, we provide RD Sharma Class 6 ex 2.6 Solutions Chapter 2 Playing With Numbers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 2.6 Solutions Chapter 2 Playing With Numbers Maths pdf, Now you will get step by step solution to each question.

RD Sharma Solutions for Class 6 Chapter 2 Playing With Numbers Ex 2.6 Download PDF

Chapter 2: Playing With Numbers – Exercise 2.6

Question: 1

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

Solution:

(i) 144 and 198

Prime factorization of 144 = 2 × 2 × 2 × 3 × 3

Prime factorization of 198 = 2 × 3 × 3 ×11

Therefore, HCF = 2 × 2 × 3 = 18

(ii) 81 and 117

Prime factorization of 81 = 3 × 3 × 3 × 3

Prime factorization of 117 = 3 × 3 ×13

Therefore, HCF = 3 × 3 = 9

(iii) 84 and 98

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

Prime factorization of 98 = 2 × 7 × 7

Therefore, HCF = 2 × 7 = 14

(iv) 225 and 450

Prime factorization of 225 = 3 × 3 × 5 × 5

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

Therefore, HCF = 3 × 3 × 5 × 5 = 225

(v) 170 and 238

Prime factorization of 170 = 2 × 5 × 17

Prime factorization of 238 = 2 × 7 × 17

Therefore, HCF = 2 × 17 = 34

(vi) 504 and 980

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

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

Therefore, HCF = 2 × 2 × 7 = 28

(vii) 150, 140 and 210

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

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

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

Therefore, HCF = 2 × 5 = 10

(viii) 84, 120 and 138

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

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

Prime factorization of 138 = 2 × 3 × 23

Therefore, HCF = 2 × 3 = 6

(i×) 106, 159 and 265

Prime factorization of 106 = 2 × 53

Prime factorization of 159 = 2 × 53

Prime factorization of 265 = 5 × 53

Therefore, HCF = 53

Question: 2

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

Solution:

(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

Question: 3

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

4 = 2 × 2 and 15 = 3 × 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?

Solution:

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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