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Playing with Numbers ( Exercise – 3.6)

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Playing with Numbers


Exercise 3.6


Ex 3.5 Question 1.


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
Solution:

(a) 18 and 48.
Factors of 18 = 2 x 3 x 3

Factors of 48 = 2 x 2 x 2 x 2 x 2 x 3

Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.

(b) 30 and 42.

Factors of 30 = 2 x 3 x 5

Factors of 42 = 2 x 3 x 7

Here, the common factors are 2 and 3.
Hence, the HCF = 2 x 3 = 6.

(c) 18 and 60.

Factors of 18 = 2 x 3 x 3

Factors of 60 = 2 x 2 x 3 x 5

Here, the common factors are 2 and 3.
Hence, the HCF of 18 and 60 = 2 x 3 = 6.

(d) 27 and 63.

Factors of 27 = 3 x 3 x 3

Factors of 63 = 3 x 3 x 7

Here, the common factor is 3 (occurring twice).
Hence, the HCF = 3 x 3 = 9.

(e) 36 and 84.

Factors of 36 = 2 x 2 x 3 x 3

Factors of 84 = 2 x 2 x 3 x 7

Here, the common factors are 2, 2, and 3.
Hence, the HCF = 2 x 2 x 3 = 12.

(f) 34 and 102.

Factors of 34 = 2 x 17

Factors of 102 = 2 x 3 x 17

Here, the common factors are 2 and 17.
Thus, HCF is 2 x 17 = 34.

(g)70, 105, and 175.

Factors of 70 = 2 x 5 x 7

Factors of 105 = 3 x 5 x 7

Factors of 175 = 5 x 5 x 7

Here, common factors are 5 and 7.
Hence, the HCF of 70, 105 and 175 is 5 x 7 = 35.

(h)91, 112, and 49.

Factors of 91 = 7 x 13

Factors of 112 = 2 x 2 x 2 x 2 x 7

Factors of 49 = 7 x 7

Here, the common factor is 7.
Hence, the HCF = 7.

(i) 18, 54, and 81.

Factors of 18 = 2 x 3 x 3

Factors of 54 = 2 x 3 x 3 x 3

Factors of 81 = 3 x 3 x 3 x 3

Here, the common factor is 3 (occurring twice).
Thus, the HCF = 3 x 3 = 9.

(j) 12, 45, and 75.

Factors of 12 = 2 x 2 x 3

Factors of 45 = 3 x 3 x 5

Factors of 75 = 3 x 5 x 5

Here, the common factor is 3.
Hence, the HCF = 3.


Ex 3.6 Question 2.


What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?
Solution:
(a) Common factor of two consecutive numbers is always 1.
Hence, the HCF = 1.
(b) Common factors of two consecutive even numbers are 1 and 2.
Hence, the HCF = 1 x 2 = 2.
(c) Common factor of two consecutive odd numbers is 1.
Hence, the HCF = 1.


Ex 3.6  Question 3.


HCF of co-prime numbers 4 and 15 was found as follows by factorization:
4 = 2 x 2 and 15 = 3 x 15. Since there are no common prime factors, so HCF of 4 and 15 is 0.
Is the answer correct? If not, what is the correct HCF?
Solution:
No, the answer is not correct.
The correct HCF of 4 and 15 is 1.


 

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