# R.D. Sharma Solutions Class 10th: Ch 1 Real Numbers Exercise 1.3

#### Chapter 1 Real Numbers R.D. Sharma Solutions for Class 10th Math Exercise 1.3

**Exercise 1.3**

**Level 1**

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

(i) 420

(ii) 468

(iii) 945

(iv) 7325

(i) 420 = 2

(i) 420

(ii) 468

(iii) 945

(iv) 7325

**Answer**(i) 420 = 2

^{4}×3×5×7
(ii) 468 = 2

^{2}×3^{2}×13
(iii) 945 = 3

^{3}×5×7
(iv) 7325 = 5

^{2}×2932. Determine the prime factorisation of each of the following positive integer:

(i) 20570

(ii) 58500

(iii) 45470971

**Answer**

(i) 20570 = 2×5×12

^{4}×17

(ii) 58500 = 2

^{2}×3^{2}×5^{3}×13
(iii) 45470971 = 7

^{2}×13^{2}×17^{2}×19**Answer**

Both the numbers have common factor 7 and 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

We know that,

Composite numbers have at least more than one common factor other than 1.

Hence, we see that both numbers are even and therefore both numbers are composite numbers.

4. Check whether 6

4. Check whether 6

^{n}can end with the digit 0 for any natural number n.

**Answer**

We know that 6

^{n}= (2×6^{n}= 2^{n}×6^{n})
Therefore, prime factorization of 6. does not contain both 5 and 2 as a factor.

Hence, 6

^{n}can never end with the digit 0 for any natural number n.5. Explain why 3×5×7 + 7 is a composite number.

**Answer**

^{3}×5 + 1 = 7×16

It can be seen that a has two more factors 7 and 16 other than 1 and the number itself.

Therefore, it is a composite number.