# Gujarat Board Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Gujarat Board Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Question 1.

Find the LCM and HCF of the following integers by applying the prime factorisation method.

- 12, 15 and 21
- 17, 23 and 29
- 8, 9 and 25

Solution:

1. Prime factorisation of 12 = 2 × 2 × 3

Prime factorisation of 15 = 3 × 5

Prime factorisation of 21 = 3 × 7

HCF of 12, 15 and 21 = 3

LCM of 12, 15 and 21

= 2 × 2 × 3 × 5 × 7 = 420

2. Prime factorisation of 17 = 17 × 1

Prime factorisation of 23 = 23 × 1

Prime factorisation of 29 = 29 × 1

HCF of 12, 15, 29 = 1

LCM of 17, 23, 29 = 17 × 23 × 29

= 11339

3. Prime factorisation of 8 = 2 × 2 × 2

Prime factorisation of 9 = 3 × 3

Prime factorisation of 23 = 23 × 1

HCF of 8, 9 and 23 = 1

LCM of 8, 9, 23 = 2 × 2 × 2 × 3 × 3 × 23

= 1656

Question 2.

Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

Given that HCF (306, 657) = 9

LCM × HCF = Product of two numbers

LCM × 9 = 306 × 657

LCM = 306×657/9

LCM = 34 × 657

LCM = 22338

∴ LCM of (306, 657) = 22338.

Question 3.

Check whether 6^{n} can end with the digit 0 for any natural number n.(CBSE)

Solution:

6^{n} will end with the digit zero if 6^{n} is divisible by 2 and 5.

But 6^{n} = (2 × 3)^{n} = 2^{n} × 3^{n}

i.e. in the factorisation of 6^{n}, no factor is of 5. Therefore by the fundamental theorem of arithmetic every composite number can be expressed a product of primes and this factorisation is unique apart from the order in which the prime factors occur.

Therefore our assumption is wrong that 6^{n} ends in zero, thus there does not exist any natural number n for which 6^{n} ends with zero.

Question 4.

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

Solution:

Given that 7 × 11 × 13 + 13

= (7 × 11 × 1 + 1) × 13

= (77 + 1) × 13

= 78 × 13

Composite number because it is product of more than two factors.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= (7 × 6 × 4 × 3 × 2 × 2 + 1) × 5

= (1008 + 1) × 5

= 1009 × 5

Product of more than two factors, which is a composite number.

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