Exercise 1.2

Question 1: Express each number as product of its prime factors:  (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution 1: (i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7 

(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13

 (iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17

 (iv) 5005 = 5 ×7× 11 × 13

 (v) 7429 = 17 × 19 × 23

Question 2: Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution 2: (i) 26 and 91

 26 = 2 × 13

 91 = 7 × 13

 HCF = 13

 LCM = 2 × 7 × 13 = 182

 Product of the two numbers = 26 × 91 = 2366

 HCF × LCM = 13 × 182 = 2366 

Hence, product of two numbers = HCF × LCM

(ii) 510 and 92510 = 2 × 3 × 5 × 17

 92 = 2 × 2 × 23

 HCF = 2

 LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460

 Product of the two numbers = 510 × 92 = 46920 

HCF × LCM = 2 × 23460 = 46920

 Hence, product of two numbers = HCF × LCM 

(iii) 336 and 54336 = 2 × 2 × 2 × 2 × 3 × 7 

336 = 24× 3× 7

54 = 2 × 3 × 3 × 3 

 HCF = 2 × 3 = 6 

 LCM = 24 × 33 × 7 = 3024  

Product of the two numbers = 336 × 54 = 18144

 HCF × LCM = 6 × 3024 = 18144 

Hence, product of two numbers = HCF × LCM 

Question 3: Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12,15 and 21 (ii) 17,23 and 29 (iii) 8,9 and 25

Solution 3: (i) 12,15 and 21 

12 = 22× 3

 15 = 3 × 5

 21 = 3 × 7

 HCF = 3

 LCM = 22× 3 × 5 × 7 = 420

(ii) 17,23 and 29

 17 = 1 × 17 

23 = 1 × 23

 29 = 1 × 29

 HCF = 1

 LCM = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

 8 = 2 × 2 × 2

 9 = 3 × 3 

25 = 5 × 5

 HCF = 1

 LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution 4: HCF (306, 657) = 9

 We know that, LCM × HCF = Product of two numbers

 ∴LCM × HCF = 306 × 657

 LCM = (306 x 657 )/HCF( 306 657)   

 LCM = 22338 

Question 5: Check whether 6n can end with the digit 0 for any natural number n.

Solution 5: If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

 Prime factorisation of6 ⁿ= (2 ×3)ⁿ

 It can be observed that 5 is not in the prime factorisation of 6n. Hence, for any value of n, 6ⁿ will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.

Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution 6:  Numbers are of two types - prime and composite.

 A prime number can be divided by 1 and only itself, whereas a composite number have factors other than 1 and itself.

 It can be observed that

 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)

 = 13 × (77 + 1) = 13 × 78 = 

13 ×13 × 6 

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

 = 5 × (1008 + 1) = 5 ×1009 1009 cannot be factorised further. ,

 the given expression has 5 and 1009 as its factors. Hence, it is a composite number. 

Question 7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? 

Solution 7:  .  Required time can be obtained  by finding the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively

 i.e., LCM of 18 minutes and 12 minutes. 

18 = 2 ×3 ×3 And, 12 = 2 ×2 ×3

 LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

 Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes. 
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