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Chapter 1: Real Numbers

    Introduction

    The Fundamental Theorem of Arithmetic

    Revisiting Irrational Numbers

    Exercise 1.1

    Exercise 1.2

    Summary

    Enhanced Curriculum Support

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Chapter 1: Real Numbers > Exercise 1.1

Exercise 1.1

Express each number as a product of its prime factors:

(i)

(i) 140

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 140 = × 2 × 5 ×

= × 5 × 7

(ii)

(ii) 156

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 156 = 2 × 2 × ×

= × 3 × 13

(iii)

(iii) 3825

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 3825 = × 3 × 5 × ×

= × × 17

(iv)

(iv) 5005

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 5005 = × 7 × × 13

(v)

(v) 7429

Solution

[Note: give the numbers in ascending order only]

Prime factorization of 7429 = × 19 ×

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i)

(i) 26 and 91

Solution

LCM × HCF = Product of the two numbers.

[Note: give the numbers in ascending order only]

Prime factors of 26 = 2 ×

Prime factors of 91 = × 13

HCF of 26 and 91 =

LCM of 26 and 91 = 2 × 7 × 13

= 14 × 13 =

Product of these two numbers = 26 × 91

=

LCM × HCF = 182 × 13 =

Thus, the product of two numbers = LCM × HCF = 2366.

(ii)

(ii) 510 and 92

Solution

[Note: give the numbers in ascending order only]

Prime factors of 510 = × 3 × × 17

Prime factors of 92 = × 2 ×

HCF of the two numbers =

LCM of the two numbers = 2 × 2 × 3 × 5 × 17 × 23 =

Product of these two numbers = 510 × 92 =

LCM x HCF = 2 × 23460 =

Thus, the product of two numbers = LCM × HCF = 46920

(iii)

(iii) 336 and 54

Solution

[Note: give the numbers in ascending order only]

Prime factors of 336 = × 2 × 2 × 2 × ×

Prime factors of 54 = × 3 × 3 ×

HCF of the two numbers =

LCM of the two numbers = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7

= 24 × 33 × 7 =

Product of these two numbers = 336 × 54 =

LCM x HCF = 3024 × 6 =

Thus, the product of two numbers = LCM × HCF = 18144

Hint

LCM × HCF = Product of the two numbers.

  • Step:1

To find the LCM and HCF of the given pairs of the integers, first, find the prime factors of the given pairs of numbers.

  • Step:2

Find the LCM

  • Step:3

Find the HCF

  • Step:4

Now, we have to verify LCM × HCF = product of the two numbers.

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

Solution

Given:

HCF (306, 657) =

number1=

number2=

We have to find, LCM (306, 657)

We know that LCM × HCF = product of two given integers.

Substitute these values in the above formula and find the value of the unknown i.e. LCM.

LCM × 9 = 306 × 657

LCM = 306×6579

LCM = × 657

LCM =

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

Solution

To determine if 6ncan end with the digit 0, we analyze the properties of the number 6n.

Condition for a number to end with 0:

A number ends with the digit 0 if and only if it is divisible by .(Smallest composite number possible.)

For a number to be divisible by 10, it must have both and as prime factors.(Enter the factors in the ascending order.)

Prime factorization of 6n:

The number 6 is composed of the prime factors and .(Enter the factors in the ascending order.)

Prime factors of 6n = 2×3n = 2n × 3n

Absence of the factor 5:

We can clearly observe, 5 is not present in the prime factors of 6n. That means 6n will not be divisible by .

Therefore, 6n cannot end with the digit 0 for any natural number n.

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

Solution

Now, simplify 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5.

On simplifying them, and prove that both the numbers have more than factors.

So, if the number has more than two factors, it will be composite.

Factorizing the expression, 7 × 11 × 13 + 13

= (7 × 11 + 1)

= 13( + 1)

= 13 × 78

= 13 × ×

= 13 × 13 × × (Hint:Prime factors)

The given number has 2, 3, 13, and 1 as its factors.

Therefore, it is a composite number.

Now,factorize the Expression:7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= × (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × ( + 1)

= 1 × 5 × 1009

1009 be factorized further.

Therefore, the given expression has 1,5,1009 as its factors. Hence, it is a composite number.

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution

Given:

  • Sonia takes minutes to drive one round of the field.

* Ravi takes minutes for the same.

* They both start at the same point and at the same time and go in the same direction.

Time taken by is more than to complete one round.

Now, we have to find after how many minutes will they meet again at the same point.

For this, there will be a number that is divisible by both 18 and 12, and that will be the time when both meet again at the starting point.

To find this we have to take of both numbers.

Let's find LCM of 18 and 12 by prime factorization method.

[Note: give the numbers in ascending order only]

18 = × ×

18= 2 × 32

12 = × ×

12= 22 × 3

LCM of 12 and 18 =22 × 32= 2 × 2 × 3 × 3 =

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.