Real Numbers ⇒⇒ Exercise 1.1

Question 1 (i)

Express each number as a product of its prime factors: 140

Solution :

The given number is 140

On prime factorizing, we get :

$$ { 2 × 70} $$

$$⇒ { 2 × 2 × 35 } $$

$$⇒ { 2 × 2 × 5 × 7} $$

$$⇒ { 2^2 × 5 × 7} $$

The prime factor of 140 = $ { 2^2 × 5^1 × 7^1}$

Question 1 (ii)

Express each number as a product of its prime factors: 156

Solution :

The given number is 156

On prime factorizing, we get :

$$ { 2 × 78} $$

$$⇒ { 2 × 2 × 39 } $$

$$⇒ { 2 × 2 × 3 × 13} $$

$$⇒ { 2^2 × 3 × 13} $$

The prime factor of 156 = $ { 2^2 × 3^1 × 13^1}$

Question 1 (iii)

Express each number as a product of its prime factors: 3825

Solution :

The given number is 3825

On prime factorizing, we get :

$$ { 3 × 1275} $$

$$⇒ { 3 × 3 × 425} $$

$$⇒ { 3 × 3 × 5 × 85} $$

$$⇒ { 3 × 3 × 5 × 5 × 17} $$

$$⇒ { 3^2 × 5^2 × 17} $$

The prime factor of 3825 = $ { 3^2 × 5^2 × 17^1}$

Question 1 (iv)

Express each number as a product of its prime factors: 5005

Solution :

The given number is 5005

The number can be as a product of its prime factors as follows :

$$ { 5 × 1001} $$

$$⇒ { 5 × 7 × 143} $$

$$⇒ { 5 × 7 × 11 × 13 } $$

The prime factor of 5005 = $ { 5^1 × 7^1 × 11^1 × 13^1}$

Question 1 (v)

Express each number as a product of its prime factors: 7429

Solution :

The given number is 7429

The number can be as a product of its prime factors as follows :

$$ { 17 × 437} $$

$$⇒ { 17 × 19 × 23} $$

The prime factor of 7429 = $ { 17^1 × 19^1 × 23^1}$

Question 2

Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. × H.C.F. = product of two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Solution :

(i) The given numbers : 26 and 91
The number can be expressed as the product of prime factors as

$$ 26 = 2 × 13 $$

$$ 91 = 7 × 13 $$

HCF (26 , 91) = 13

LCM (26 , 91) = 2 × 7 × 13 = 182

Product of two numbers = $$ 26 × 91 = 2366 $$

HCF × LCM = $$ 13 × 182 = 2366 $$

Hence, product of two numbers = HCF × LCM

(ii) The given numbers : 510 and 92
The number can be expressed as the product of prime factors as

$$ 510 = 2 × 3 × 5 × 17 $$

$$ 92 = 2 × 2 × 23 $$

$$ 92 = 2^2 × 23 $$

HCF HCF (510 , 92) = 2.

LCM (510 , 92)= $ 2^2 × 3 × 5 × 17 × 23 $ = 23460

Product of two numbers = $$ 510 × 92 = 46920 $$

HCF × LCM = $$ 2 × 23460 = 46920 $$

Hence, product of two numbers = HCF × LCM

(iii) The given numbers : 336 and 54
The number can be expressed as the product of prime factors as

$$ 336 = 2 × 2 × 2 × 2 × 3 × 7 $$

$$ 336 = 2^4 × 3 × 7 $$

$$ 54 = 2 × 3 × 3 × 3 $$

$$ 54 = 2 × 3^3 $$

HCF (336, 54) = 2 × 3 = 6.

LCM (336, 54) = $ 2^4 × 3^3 × 7 $ = 3024

Product of two numbers = $$ 336 × 54 = 18144 $$

HCF × LCM = $$ 6 × 3024 = 18144 $$

Hence, product of two numbers = HCF × LCM

Question 3 (i)

Find the L.C.M. and H.C.F. of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21

Solution :

(i) The given numbers : 12, 15 and 21
let us write prime factors of the given numbers

$$ 12 = 2 × 2 × 3 $$

$$ 12 = 2^2 × 3 $$

$$ 15 = 3 × 5 $$

$$ 21 = 3 × 7 $$

To find the H.C.F., we take the product of the common prime factors with the lowest power. The only common prime factor is 3, and its lowest power is $ 3^1 $

Therefore, HCF (12, 15, 21) = 3

To find the L.C.M., we take the product of all prime factors with their highest power. The prime factors are 2, 3, 5, and 7, and their highest powers are $ 2^2 × 3^1 × 5^1 × 7^1 $.

LCM (12, 15, 21) = $ 2^2 $× 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420.

Question 3 (ii)

Find the L.C.M. and H.C.F. of the following integers by applying the prime factorisation method.
(ii) 17, 23, and 29

Solution :

(i) The given numbers : 17, 23, and 29
let us write prime factors of the given numbers

$$ 17 = 1 × 17 = 17^1 $$

$$ 23 = 1 × 23 = 23^1$$

$$ 29 = 1 × 29 = 29^1$$

To find the H.C.F., we take the product of the common prime factors with the lowest power. Here, 1 is the only common factor.

Therefore, HCF (17, 23, 29) = 1

To find the L.C.M., we take the product of all prime factors with their highest power. The L.C.M. of prime numbers is their product.

LCM (17, 23, 29) = 17 × 23 × 29 = 11339.

Question 3 (iii)

Find the L.C.M. and H.C.F. of the following integers by applying the prime factorisation method.
(iii) 8, 9 and 25

Solution :

(i) The given numbers : 8, 9 and 25
let us write prime factors of the given numbers

$$ 8 = 2 × 2 × 2 = 2^3 $$

$$ 9 = 3 × 3 = 3^2$$

$$ 25 = 5 × 5 = 5^2$$

To find the H.C.F., we take the product of the common prime factors with the lowest power. The numbers have no common prime factors.

Therefore, HCF ( 8, 9, 25) = 1

To find the L.C.M., we take the product of all prime factors with their highest power. The prime factors are 2, 3, and 5, with their highest powers being.

LCM ( 8, 9, 25) = $ 2^3 × 3^2 × 5^2 $ = 8 × 9 × 25 = 1800

Question 4

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

Solution :

Given that HCF (306, 657) = 9

We know that, L.C.M. × H.C.F. = Product of the two numbers

$$ L.C.M × 9 = 306 × 657 $$

$$ L.C.M = {{306 × 657} \over 9} $$

$$ L.C.M = {{201042} \over 9} $$

$$ L.C.M = {22338} $$

Therefore, LCM (306, 657) = 22338

Question 5

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

Solution :

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5 .

The prime factorization of 6 is 2 × 3. Therefore, the prime factorization of $ 6^n = 2^n × 3^n $.

Since the prime factorization of $ 6^n$ does not contain the prime factor 5, it is not divisible by 10.

Consequently, for any natural number n, $ 6^n$ cannot end with the digit 0

Question 6

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

Solution :

A composite number is a positive integer that has at least one divisor other than 1 and itself.

It can be observed that

7×11×13+13
=13(7×11+1)
=13(77+1)
=13×78
=13×13×6×1
=13×13×2×3×1

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

Since, it has factors other than 1 and itself. Therefore, it is a composite number.

(II) It can be observed that

7×6×5×4×3×2×1+5
=5×(7×6×4×3×2×1+1)
=5×(1008+1)
=5×1009×1

Because this number can be expressed as a product of three integers, 5 and 1009 and 1 .

Since, it has factors other than 1 and itself. Therefore, it is a composite number.

Question 7

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 :

It can be observed that time taken by Sonia is more than Ravi to complete one round.

As they are going in the same direction, we have to find the least common multiple (LCM) of the time.

This is because they will only meet at the starting point again when a multiple of Sonia's time (18 minutes) and a multiple of Ravi's time (12 minutes) are equal. The first time this happens is their LCM.

LCM of 12 and 18

18= 2 × $ 3^2 $

And, 12 = $ 2^2 $ × 3

To calculate the LCM, we take the highest power of each prime factor present in either factorization.

LCM of (12 , 18 ) = 2 × 2 × 3 × 3 = 36

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