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

• Last Updated : 13 Jan, 2021

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

i) 420

ii) 468

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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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