AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2

### 10th Class Maths 1st Lesson Real Numbers Ex 1.2 Textbook Questions and Answers

Question 1.

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

i) 140

ii) 156

iii) 3825

iv) 5005

v) 7429

Answer:

i) 140

∴ 140 = 2 × 2 × 5 × 7 = 2^{2} × 5 × 7

ii) 156

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

iii) 3825

∴ 3825 = 3 × 3 × 5 × 5 × 17 = 3^{2} × 5^{2} × 17

iv) 5005

∴ 5005 = 5 × 7 × 11 × 13

v) 7429

∴ 7429 = 17 × 19 × 23

Question 2.

Find the L.C.M and H.C.F of the following integers by the prime factorization method.

i) 12, 15 and 21

ii) 17, 23 and 29

iii) 8, 9 and 25

iv) 72 and 108

v) 306 and 657

Answer:

i) 12, 15 and 21

12 = 2 × 2 × 3 = 22 × 3

15 = 3 × 5

21 = 3 × 7

L.C.M = 2^{2} × 3 × 5 × 7 = 420

H.C.F = 3

ii) 17, 23 and 29

The given numbers 17, 23 and 29 are all primes.

L.C.M = their product

= 17 × 23 × 29 = 11339

∴ H.C.F = 1

iii) 8, 9 and 25

8 = 2 × 2 × 2 = 2^{3}

9 = 3 × 3 = 3^{2}

25 = 5 × 5 = 5^{2}

L.C.M = 2^{3} × 3^{2} × 5^{2} = 1800

(or)

8, 9 and 25 are relatively prime, therefore L.C.M is equal to their product,

(i.e.,) L.C.M = 8 × 9 × 25 = 1800

H.C.F = 1

iv) 72 and 108

72 = 2^{3} × 3^{2}

108 = 2^{2} × 3^{3}

L.C.M = 2^{3} × 3^{3} = 8 × 27 = 216

H.C.F = 2^{2} × 3^{2} = 4 × 9 = 36

v) 306 and 657

306 = 2 × 3^{2} × 17

657 = 32 × 73

L.C.M = 2 × 3^{2} × 17 × 73 = 22338

H.C.F = 3^{2} = 9

Question 3.

Check whether 6^{n} can end with the digit ‘0’ for any natural number n.

Answer:

Given number = 6^{n} = (2 × 3)^{n}

The prime factors here are 2 and 3 only.

To be end with 0; 6^{n} should have a prime factor 5 and also 2.

So, 6^{n} can’t end with zero.

Question 4.

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

Answer:

Given numbers are 7 × 11 × 13

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

⇒ 13(7 × 11 + 1) and

5(7 × 6 × 4 × 3 × 2 × 1 + 1)

⇒ 13 K and 5 L, where K = 78 and L = 7 × 6 × 4 × 3 × 2 × 1 + 1 = 1009

As the given numbers can be written as product of two numbers, they are composite.

Question 5.

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

Answer:

(17 × 11 × 2) + (17 × 11 × 5)

= (17 × 11) (2 + 5)

= (17 × 11) (7)

= 187 × 7

Now the given expression is written as a product of two integers and hence it is a composite number.

Question 6.

What is the last digit of 6100?

Answer: We know that

6^{1} = 6

6^{2} = 36

6^{3} = 216

6^{4} = 1296

6^{5} = 7776

We see that 6^{n} for any positive integer n ends is 6.

i.e., unit digit is always 6.

∴ Unit digit of 6^{100} is 6.