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 

Answer

(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 


Answer

(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. 

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

Answer

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. 

Answer

Let a = 3×5×7 + 7 = 73×5 + 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.

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