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Playing With Numbers

    Exercise 3.6

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6th class > Playing With Numbers > Exercise 3.6

Exercise 3.6

1. Find the HCF of the following numbers?

a) 18,48= × = (Enter the common factors of the given number)

(b) 30,42 = × = (Enter the common factors of the given number)

(c) 18,60 = × = (Enter the common factors of the given number)

(d) 27,63 = × = (Enter the common factors of the given number)

(e) 36,84= × = (Enter the common factors of the given number)

(f) 34,102= × = (Enter the common factors of the given number)

(g) 70,105,175= × = (Enter the common factors of the given number)

(h) 91,112,49= (Enter the common factors of the given number)

(i) 18,54,81= (Enter the common factors of the given number) , {.reveal(when="blank-22")}(j) 12,45,75= (Enter the common factors of the given number)

2. What is the HCF of two consecutive (a) numbers?(b) even numbers?(c) odd numbers?

Instructions

(a) Numbers: Since consecutive numbers differ by 1, they have common factor(s) other than 1.
Example: Factors of 4 : 1, ,
Factors of 5: 1,
Common factor is .
(b) Even Numbers: Two consecutive even numbers can be represented as 2n and 2n+2 where n is a integer.
Example- Factors of 4: 1, 2, while factors of 6 are: 1, 2, and 6.
The only common factor between any two consecutive even numbers is .
(c) Odd Numbers: Two consecutive odd numbers can be represented as 2n+1 and 2n+3 where n is an integer.
Example- Factors of 3 are 1, while factors of 5 are: 1, .
Consecutive odd numbers are always odd and differ by . Since they are consecutive, they have common factor(s) other than 1.

3. 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.

Since 4 and 15 have no common prime factors, they are .

The correct HCF of two co-prime numbers is always , not .

Conclusion: The correct HCF of 4 and 15 is 1. Thus, the above given statement is .