Real Number Exercise 1.2

Table of Contents

1. Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Find the LCM and HCF of the following integers by applying the prime factorization method.


(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

(i) The prime factorisation of 26 and 91 is,
26 = 2 × 13 and 91 = 7 × 13
 LCM (26, 91) = 2 × 7 × 13 = 182
HCF (26, 91) = 13   Now, LCM × HCF = 182 × 13 = 2366 and 26 × 91 = 2366
i.e., LCM × HCF = Product of two numbers.
(ii) The prime factorisation of 510 and 92 is,
510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
LCM (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
HCF (510, 92) = 2
Now, LCM × HCF = 23460 × 2 = 46920
and 510 × 92 = 46920
i.e., LCM × HCF = Product of two numbers.
(iii) The prime factorisation of 336 and 54 is,
336 = 2 × 2 × 2 × 2 × 3 × 7 and 54 = 2 × 3 × 3 × 3
 LCM (336, 54) = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
and HCF(336, 54) = 2 × 3 = 6
Now, LCM × HCF = 3024 × 6 = 18144
Also, 336 × 54 = 18144
Thus, LCM × HCF = Product of two numbers.

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

Since, LCM × HCF = Product of the numbers
LCM × 9 = 306× 657
⇒ LCM =306 × 657/ 9
= 22338
Thus, LCM of 306 and 657 is 22338

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

 

Here, n is a natural number and let 6n end with digit 0.
6n is divisible by 5.
But the prime factors of 6 are 2 and 3. i.e., 6 = 2 × 3
⇒ 6n = (2 × 3)n
i.e., In the prime factorization of 6n, there is no factor 5.
So, by the fundamental theorem of Arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order in which the prime factorization occurs.
Our assumption that 6n ends with digit 0, is wrong.
Thus, there does not exist any natural number n for which 6n ends with zero.

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

We have
7 × 11 × 13 + 13 = 13((7 × 11) + 1) = 13(78), which cannot
be a prime number because it has factors 13 and 78.
Also, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5[7 × 6 × 4 × 3 × 2 × 1 + 1],
which is also not a prime number because it has a factor 5
Thus, 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

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?

 

Time taken by Sonia to drive one round of the field = 18 minutes
Time taken by Ravi to drive one round of the field = 12 minutes
LCM of 18 and 12 gives the exact number of minutes after which they meet again at the starting point.
Now, 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
Thus, they will meet again at the starting point after 36 minutes

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