Class 10 RD Sharma Solutions- Chapter 1 Real Numbers – Exercise 1.3

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

i) 420

ii) 468

iii) 945

iv) 7325

Solution:

Let us express each of the numbers as a product of prime factors. 

i) 420

Performing prime factorisation of the number, we get,

420 = 2 × 2 × 3 × 5 × 7

ii) 468

Performing prime factorisation of the number, we get,

468 = 2 × 2 × 3 × 3 × 13

iii) 945

Performing prime factorisation of the number, we get,

945 = 3 × 3 × 3 × 5 × 7

iv) 7325

Performing prime factorisation of the number, we get,

7325 = 5 × 5 × 293

Question 2. Determine the prime factorisation of each of the following positive integer:

i) 20570

ii) 58500

iii) 45470971

Solution:

Let us express each of the numbers as a product of prime factors. 

i) 20570

Performing prime factorisation of the number, we get,

20570 = 2 × 5 × 11 × 11 × 17

ii) 58500

Performing prime factorisation of the number, we get,

58500 = 2 × 2 × 3 × 3 × 5 × 5 × 5 × 13

iii) 45470971

Performing prime factorisation of the number, we get,

45470971 = 7 × 7 × 13 × 13 × 17 × 17 × 19

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

Solution:

Both of these numbers have a common factor of 7. Also, every number is divisible by 1.

7 × 11 × 13 + 13 = (77 + 1) × 13 = 78 × 13

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 4 × 3 × 2 + 1) × 5 = 1008 × 5

Composite numbers are those numbers which have at least one more factor other than 1.

Now,

Both of these numbers are even. Therefore, the given two numbers are composite numbers

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

Solution:

Since, 6n = (2 × 3)n

6n = 2n × 3n

Any number can end with 0 if it divisible by 10 or 5 and 2 together. The, prime factorisation of 6n does not contain 5 and 2 as a pair of factors. 

Therefore, 6n can never end with the digit 0 for any natural number n.

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