The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of primes. This concept is used to solve the problems of this exercise. The RD Sharma Solutions Class 10 is a wonderful resource for students to rely on for clearing doubts quickly and also to know the correct procedure to solve problems. Students can make use of the RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3 PDF, which is available below, to score high marks in the Class 10 board exam.
RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.3
Access Answers to Maths RD Sharma Solutions for Class 10 Chapter 1 Real Numbers Exercise 1.3
1. Express each of the following integers as a product of its prime.
(i) 420
Solution:
420 = 2 x 2 x 3 x 5 x 7
∴ 420 = 22x 3 x 5 x 7.
(ii) 468
Solution:
468 = 2 x 2 x 3 x 3 x 13
∴ 468 = 22x 32 x 13.
(iii) 945
Solution:
945 = 3 x 3 x 3 x 5 x 7
∴ 945 = 33x 5 x 7.
(iv) 7325
Solution:
7325 = 5 x 5 x 293
∴ 7325 = 52 x 293
2. Determine the prime factorisation of each of the following positive integers:
(i) 20570
Solution:
20570 = 2 x 5 x 11 x 11 x 17
∴ 20570 = 2 x 5 x 112x 17
(ii) 58500
Solution:
58500 = 2 x 2 x 3 x 3 x 5 x 5 x 5 x 13
∴ 58500 = 22 x 32x 53x 13.
(iii) 45470971
Solution:
45470971 = 7x7x13x13x17x17x19
∴ 45470971 = 72x132x172x19.
3. Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Solution:
So, basically, there are two types of numbers, i.e., prime numbers and composite numbers.
Understanding that,
Prime numbers are those numbers having 1 and the number itself as factors.
And composite numbers are those numbers having factors other than 1 and itself.
It’s seen that,
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) [taking 13 out as common]
= 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6
So, the given expression has 6 and 13 as its factors. Therefore, we can conclude that it is a composite number.
Similarly,
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1) [taking 5 out- common]
= 5 x (1008 + 1)
= 5 x 1009
Since 1009 is a prime number, the given expression has 5 and 1009 as its factors other than 1 and the number itself. Hence, it is a composite number.
4. Check whether 6n can end with the digit 0 for any natural number n.
Solution:
In order to check whether 6n can end with the digit 0 for any natural number n, let us find the factors of 6.
It’s seen that the factors of 6 are 2 and 3.
So, 6n = (2 x 3)n
6n =2n x 3n
Since the prime factorisation of 6 does not contain 5 and 2 as its factor together, we can conclude that 6n can never end with the digit 0 for any natural number n.
5. Explain why 3 × 5 × 7 + 7 is a composite number.
Solution:
Basically, there are two types of numbers, i.e., prime numbers and composite numbers.
Understanding that,
Prime numbers are those numbers having 1 and the number itself as factors.
And composite numbers are those numbers having factors other than 1 and itself.
It’s seen that,
3 × 5 × 7+ 7 = 7 × (3 × 5 + 1) = 7 × (15 + 1) = 7 × 16
Since the given expression has 7 and 16 as its factors, we can conclude that it is a composite number.
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