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# RD Sharma Solutions: Exercise 2.6- Playing With Numbers Notes | EduRev

## Class 6 : RD Sharma Solutions: Exercise 2.6- Playing With Numbers Notes | EduRev

The document RD Sharma Solutions: Exercise 2.6- Playing With Numbers Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
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Q.1. Find the H.C.F of the following numbers using prime factorization method:

(i) 144,198

(ii) 81,117

(iii) 84,98

(iv) 225,450

(v) 170, 238

(vi) 504, 980

(vii) 150, 140, 210

(viii) 84, 120, 138

(ix) 106, 159, 265

Ans:

(i) 144 and 198

Prime factorisation of 144 = 2 × 2 × 2 × 2 ×3 × 3

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

∴ HCF = 2 × 3 × 3 = 18

(ii) 81 and 117

Prime factorisation of 81 = 3 ×3 × 3 × 3

Prime factorisation of 117 = 3 × 3 × 13

∴ HCF = 3 × 3 = 9

(iii) 84 and 98

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

Prime factorisation of 98 = 2 × 7 × 7

∴ HCF = 2 × 7 = 14

(iv) 225 and 450

Prime factorisation of 225 = 3 × 3 × 5 × 5

Prime factorisation of 450 = 2 × 3 × 3 × 5 × 5

∴ HCF = 3 × 3 × 5 × 5 = 225

(v) 170 and 238

Prime factorisation of 170 = 2 × 5 × 17

Prime factorisation of 238 = 2 × 7 × 17

∴ HCF = 2 × 17= 34

(vi) 504 and 980

We have

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

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

∴ HCF = 2 × 2 × 7 = 28

(vii) 150, 140 and 210

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

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

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

∴ HCF = 2 × 5 = 10

(viii) 84, 120 and 138

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

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

Prime factorisation of 138 = 2 × 3 × 23

∴ HCF = 2 × 3 = 6

(ix) 106, 159, and 265

Prime factorisation of 106 = 2 × 53

Prime factorisation of 159 = 3 × 53

Prime factorisation of 265 = 5 × 53

∴ HCF = 53

Q.2.What is the H.C.F of two consecutive

(i) Numbers

(ii) even numbers

(iii) odd numbers

Ans:

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

∴ HCF of two consecutive numbers = 1

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

∴ HCF of two consecutive even numbers = 2

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

∴ HCF of two consecutive odd numbers = 1

Q.3. H.C.F of co-prime numbers 4 and 15 was found as follow:

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?

Ans: 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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## Mathematics (Maths) Class 6

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