Playing With Numbers

## EXERCISE 3.6

CBSE Class 6 Maths

**Q1: 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

Answer:

(a) Factors of 18 = 2 × 3 × 3

Factors of 48 = 2 × 2 × 2 × 2 × 3

HCF (18, 48) = 2 × 3 = 6

(b) Factors of 30 = 2 × 3 × 5

Factors of 42 = 2 × 3 × 7

HCF (30, 42) = 2 × 3 = 6

(c) Factors of 18 = 2 × 3 × 3

Factors of 60 = 2 × 2 × 3 × 5

HCF (18, 60) = 2 × 3 = 6

(d) Factors of 27 = 3 × 3 × 3

Factors of 63 = 3 × 3 × 7

HCF (27, 63) = 3 × 3 = 9

(e) Factors of 36 = 2 × 2 × 3 × 3

Factors of 84 = 2 × 2 × 3 × 7

HCF (36, 84) = 2 × 2 × 3 = 12

(f) Factors of 34 = 2 × 17

Factors of 102 = 2 × 3 × 17

HCF (34, 102) = 2 × 17 = 34

**Q2: What is the HCF of two consecutive**

**(a) numbers?**

**(b) even numbers?**

**(c) odd numbers?**

Answer:

(a) HCF of two consecutive numbers be 1.

(b) HCF of two consecutive even numbers be 2.

(c) HCF of two consecutive odd numbers be 1.

**Q3: HCF of co-prime numbers 4 and 15 was found as follows by factorisation :**

**4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?**

Answer: It is incorrect. Since there is no common factor, the correct HCF is 1.

__Other Questions on HCF__

**Q4: State True (T) or FALSE (F)**

**⒜**The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple.

**⒝**HCF of two or more numbers may be one of the numbers.

**⒞**Any two consecutive numbers are coprime.

**⒟**The HCF of two numbers is smaller than the smaller of the numbers.

Answer:

⒜ False

⒝ True

⒞ True

⒟ False

**Q5: Find the largest number that divides 92 and 74 leaving 2 as remainder.**

Answer: Number divides 92 and leaves remainder 2.

∴ The number divides 90 ( = 92 - 2) completely.

The number also divides 74 leaving 2 as remainder

∴ The number divides 72 ( = 74 - 2) completely.

Factors of 90 = 3 × 3 × 5 × 2

Factors of 72 = 3 × 3 × 2 × 2 × 2

HCF of (90, 72) is = 3 × 3 × 2 = 18

∴ The required number is 18.

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