Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 1 Real Numbers Exercise 1.2

Question 1.

Express each of the following numbers as a product of its prime factors.

(i) 140

Answer:

140

∴ 140 = 2 × 2 × 5 × 7 = 2^{2} × 5 × 7

(ii) 156

Answer:

156

∴ 156 = 2 × 2 × 3 × 13 = 2^{2} × 3 × 13

(iii) 3825

Answer:

3825

∴ 3825 = 3 × 3 × 5 × 5 × 17

= 3^{2} × 5^{2} × 17

(iv) 5005

Answer:

5005

∴ 5005 = 5 × 7 × 11 × 13

(v) 7429

Answer:

7429

∴ 7429 = 17 × 19 × 23

Question 2.

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

(i) 12, 15 and 21

Answer:

12, 15, 21

12 = 2 × 2 × 3, 15 = 3 × 5

21 = 3 × 7

H.C.F of 12, 15, 21 is 3.

L.C.M of 12, 15, 21 = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23, and 29

Answer:

17, 23, 29

17 = 1 × 17, 23 = 1 × 23

29 = 1 × 29

H.C.F of 17, 23 and 29 is 1.

L.C.M of 17, 23, 29 is 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Answer:

8, 9, 25

8 = 1 × 2 × 2 × 2, 9 = 1 × 3 × 3

25 = 1 × 5 × 5

H.C.F. = 1

L.C.M = 2 × 2 × 2 × 3 × 3 × 5 × 5

= 1800

(iv) 72 and 108

Answer:

72, 108

72 = 2 × 36

= 2 × 2 × 18

= 2 × 2 × 2 × 9

= 2 × 2 × 2 × 3 × 3

108 = 2 × 54

= 2 × 2 × 27

= 2 × 2 × 3 × 9

= 2 × 2 × 3 × 3 × 3

H.C.F. = 2 × 2 × 3 × 3 = 36

L.C.M. = 2 × 2 × 2 × 3 × 3 × 3 = 216

(v) 306 and 657

Answer:

306, 657

H.C.F. = 9

L.C.M. = 9 × 34 × 73

= 22338

Question 3.

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

Answer:

If the number 6^{n} for any n, were to end with digit ‘0’ then it would be divisible by 5.

The prime factorisation of 6n would contain the prime 5.

6^{n} = (2 × 3)^{n}

Prime factorisation of 6^{n} does not contain 5 as a factor, so 6^{n} can never end with the digit 0 for any natural number.

Question 4.

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

Answer:

Given number is 7 × 11 × 13 + 13

= 13 (7 × 11 + 1)

= (7 × 11 + 1) × 13 distributive law

= (77 + 1) × 13

= 78 × 13

= (2 × 3 × 13) × 13

= 2 × 3 × 13^{2}

= Product of prime factors

Hence the given number is a composite number.

Given number is

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 (1008 + 1)

= 5 × 1009

= Product of prime numbers

Hence the given number is a composite number.

Question 5.

How will you show that (17 × 11 × 2)+ (17 × 11 × 5) is a composite number? Explain.

Answer:

Given number is

(17 × 11 × 2) + (17 × 11 × 5)

= 17 × 11 × (2 + 5)

= 17 × 11 × 7 = Product of primes

We know that every composite number can be expressed as a product of primes.

Question 6.

Which digit would occupy the units place of 6^{100}.

Answer:

6^{100} = (2 × 3)^{100}

So the last digit of 6^{100} is 6.