RD Sharma Solutions Class 10 Real Numbers Exercise 1.3

RD Sharma Class 10 Solutions Chapter 1 Ex 1.3 PDF Free Download

EXERCISE 1.3

1. Express each of the following integers as a product of its prime.

(i) 420

(ii) 468

(iii) 945

(iv) 7325

Solution:

(i) 420

Class 10 Real Numbers Exercise 1.3-1

420 = 2 x 2 x 3 x 5 x 7

420 = 22x 3 x 5 x 7

(ii) 468

Class 10 Real Numbers Exercise 1.3-2

468 = 2 x 2 x 3 x 3 x 13

468 = 22x 33 x 13

(iii) 945

Class 10 Real Numbers Exercise 1.3-3

945 = 3 x 3 x 3 x 5 x 7

945 = 33x 5 x 7

(iv) 7325

Class 10 Real Numbers Exercise 1.3-4

7325 = 5 x 5 x 293

7325 = 55 x 293

2. Determine the prime factorization of each of the following positive integer :

(i) 20570

(ii) 58500

(iii) 45470971

Solution:

(i) 20570

Class 10 Real Numbers Exercise 1.3-5

20570 = 2 x 5 x 11 x 11 x 17

20570 = 2 x 5 x 112x 17

(ii) 58500

Class 10 Real Numbers Exercise 1.3-6

58500 = 2 x 2 x 3 x 3 x 5 x 5 x 5 x 13

58500 = 22 x 32x 53x 13

(iii) 45470971

Class 10 Real Numbers Exercise 1.3-7

45470971 = 7x7x13x13x17x17x19

45470971 = 72x132x172x19

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

Solution:

There are two types of numbers, prime numbers and composite numbers.

Prime numbers: Prime numbers are those numbers having 1 and the number itself as factors.

Composite numbers: Composite numbers are those numbers having factors other than 1 and itself.

We can see that,

7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) [taking 13 out- common]

= 13 x (77 + 1)

= 13 x 78

= 13 x 13 x 6

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

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1) [taking 5 out- common]

= 5 x (1008 + 1)

= 5 x 1009

Since, 1009 cannot be factorised further. The given expression has 5 and 1009 as its factors other than 1 and the number itself. Hence, it is a composite number.

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

Solution:

To check whether 6n can end with the digit 0 for any natural number n, let us find the factors of 6.

The factors of 6 are 2 and 3

So, 6n= (2 x 3)n

6n=2n x 3n

Since the prime factorization of 6 does not contain 5 and 2 as a its factor, together. We can conclude that 6n can never end with the digit 0 for any natural number n.

5. Explain why 3 × 5 × 7 + 7 is a composite number.

Solution:

There are two types of numbers, prime numbers and composite numbers.

Prime numbers: Prime numbers are those numbers having 1 and the number itself as factors.

Composite numbers: Composite numbers are those numbers having factors other than 1 and itself.

We can see from the equation that,

3 × 5 × 7+ 7 = 7 × (3 × 5 + 1) = 7 × (15 + 1) = 7 × 16

Since the given expression has 7 and 16 as its factors. We can conclude that it is a composite number.

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