#### 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 = 24×3×5×7
(ii) 468 = 22×32×13
(iii) 945 = 33×5×7
(iv) 7325 = 52×293

2. Determine the prime factorisation of each of the following positive integer:
(i) 20570
(ii) 58500
(iii) 45470971

(i) 20570 = 2×5×124×17
(ii) 58500 = 22×32×53×13
(iii) 45470971 = 72×132×172×19

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

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

We know that 6n = (2×6n = 2n×6n)
Therefore, prime factorization of 6. does not contain both 5 and 2 as a factor.
Hence, 6n can never end with the digit 0 for any natural number n.

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