RD Sharma 2018 Solutions for Class 6 Math Chapter 2 Playing With Numbers are provided here with simple step-by-step explanations. These solutions for Playing With Numbers are extremely popular among class 6 students for Math Playing With Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2018 Book of class 6 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2018 Solutions. All RD Sharma 2018 Solutions for class 6 Math are prepared by experts and are 100% accurate.

#### Page No 2.10:

#### Question 1:

What are prime numbers? List all primes between 1 and 30.

#### Answer:

Those numbers with only two factors, i.e., 1 and the number itself, are known as prime numbers.

Examples: 2, 3, 5, 7, 11 and 13

The prime numbers between 1 and 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

#### Page No 2.10:

#### Question 2:

Write all prime numbers between:

(i) 10 and 50

(ii) 70 and 90

(iii) 40 and 85

(iv) 60 and 100

#### Answer:

(i) The prime numbers between 10 and 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

(ii) The prime numbers between 70 and 90 are 71, 73, 79, 83 and 89.

(iii) The prime numbers between 40 and 85 are 41, 43, 47, 53, 59, 61, 67, 71, 73, 79 and 83.

(iv) The prime numbers between 60 and 100 are 61, 67, 71, 73, 79, 83, 89 and 97.

#### Page No 2.10:

#### Question 3:

What is the smallest prime number? Is it an even number?

#### Answer:

The number 2 is the smallest prime number.

It is an even prime number. Except 2, all the other even numbers are composite numbers.

#### Page No 2.10:

#### Question 4:

What is the smallest odd prime? Is every odd number a prime number? If not, give an example of an odd number which is not prime.

#### Answer:

The smallest odd prime number is 3.

No, every odd number is not a prime number. For example, 9 is an odd number but it is not a prime number because its three factors are 1, 3 and 9.

#### Page No 2.10:

#### Question 5:

What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number.

#### Answer:

A number which has more than two factors is called a composite number.

For example, the numbers 4, 6, 8, 9, 10 and 15 are composite numbers.

Yes, a composite number can be an odd number. The smallest odd composite number is 9.

#### Page No 2.11:

#### Question 6:

What are the twin-primes? Write all pairs of twin-primes between 50 and 100.

#### Answer:

**Twin primes:** Two prime numbers are said to be twin primes if there is only one composite number between them.

For example, (3, 5) and (5, 7) are twin primes.

Twin primes between 50 and 100 are (59, 61) and (71, 73).

#### Page No 2.11:

#### Question 7:

What are co-primes? Give examples of five of co-primes. Are co-primes always prime? If no, illustrate your answer by an examples.

#### Answer:

Two numbers are said to be co-primes if they do not have any common factor other than 1.

For example, (2, 3), (3, 4), (4, 5), (5, 7), and (13, 17) are co-primes.

Two co-prime numbers need not be both prime numbers.

e.g., (3, 4), (6, 7) and (4, 13).

#### Page No 2.11:

#### Question 8:

Which of the following pairs are always co-primes?

(i) two prime numbers

(ii) one prime and one composite number

(iii) two composite numbers

#### Answer:

(i) Two prime numbers are always co-primes to each other.

Example: 7 and 11 are co-primes to each other.

(ii) One prime and one composite number are not always co-prime.

Example: 3 and 21 are not co-primes to each other.

(iii) Two composite numbers are not always co-primes to each other.

Example: 4 and 6 are are not co-primes to each other.

#### Page No 2.11:

#### Question 9:

Express each of the following as a sum of two or more primes?

(i) 13

(ii) 130

(iii) 180

#### Answer:

We can write the given numbers as the sums of two or more primes as follows:

(i) 13 = 11 + 2

(ii) 130 = 59 + 71

(iii) 180 = 139 + 17 + 11 + 13 or 79 + 101

#### Page No 2.11:

#### Question 10:

Express each of the following numbers as the sum of two odd primes:

(i) 36

(ii) 42

(iii) 84

#### Answer:

We can express the given numbers as the sums of two odd primes as follows:

(i) 36 = 7 + 29 or 17 + 19

(ii) 42 = 5 + 37 or 13 + 29

(iii) 84 = 17 + 67 or 23 + 61

#### Page No 2.11:

#### Question 11:

Express each of the following numbers as the sum of three odd prime numbers:

(i) 31

(ii) 35

(iii) 49

#### Answer:

We can express the given numbers as the sums of three odd prime numbers as follows:

(i) 31 = 5 + 7 + 19 or 31 = 11 + 13 + 7

(ii) 35 = 5 + 7 + 23 or 35 = 17 + 13 + 5

(iii) 49 = 3 + 5 + 41 or 49 = 7 + 11 + 31

#### Page No 2.11:

#### Question 12:

Express each of the following numbers as the sum of twin primes:

(i) 36

(ii) 84

(iii) 120

#### Answer:

We can express the given numbers as the sums of twin primes which are as follows:

(i) 36 = 17 + 19

(ii) 84 = 41 + 43

(iii) 120 = 59 + 61

#### Page No 2.11:

#### Question 13:

Find the possible missing twins for the following numbers so that they become twin primes:

(i) 29

(ii) 89

(iii) 101

#### Answer:

(i) The possible missing twins for 29 are 27 and 31. Since 31 is a prime and 27 is not, 31 is the missing twin.

(ii) The possible missing twins for for 89 are 87 and 91. Since 87 and 91 are not primes, 89 has no twin.

(iii) The possible missing twins for 101 are 99 and 103. Since 103 is a prime and 99 is not, 103 is the missing twin.

#### Page No 2.11:

#### Question 14:

A list consists of the following pairs of numbers:

(i) 51, 53; 55, 57; 59, 61; 63, 65; 67, 69; 71, 73

Categorize them as pairs of:

(i) co-primes

(ii) primes

(iii) composites

#### Answer:

(i) **Co-primes:** Two natural numbers are said to be co-prime numbers if they have 1 as their only common factor.

Hence, all the given pairs of numbers are co-primes.

(ii) **Primes: **Natural numbers which have exactly two distinct factors, i.e., 1 and the number itself are called prime numbers.

Hence, (59, 61) and (71, 73) are pairs of prime numbers.

(iii) **Composite numbers:** Natural numbers which have more than two factors are called composite numbers.

Hence, (55, 57) and (63, 65) are pairs of composite numbers.

#### Page No 2.11:

#### Question 15:

For a number, greater than 10, to be prime what may be the possible digit in the unit's place?

#### Answer:

For a number (greater than 10) to be a prime number, the possible digit in the unit's place may be 1, 3, 7, or 9.

Example: 11, 13, 17, and 19 are prime numbers greater than 10.

#### Page No 2.11:

#### Question 16:

Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

#### Answer:

The required seven consecutive composite numbers are 90, 91, 92, 93, 94, 95 and 96.

#### Page No 2.11:

#### Question 17:

State true (T) of false (F):

(i) The sum of primes cannot be a prime.

(ii) The product of primes cannot be a prime.

(iii) An even number is composite.

(iv) Two consecutive numbers cannot be both primes.

(v) Odd numbers cannot be composite.

(vi) Odd numbers cannot be written as sum of primes.

(vii) A number and its successor are always co-primes.

#### Answer:

(i) False.

2 + 3 = 5 which is a prime number.

(ii) True.

The product of prime numbers is always a composite number.

(iii) False.

The even number 2 is not a composite number.

(iv) False.

2 and 3 are consecutive numbers and are also prime numbers.

(v) False.

9 is an odd number but it is a composite number as its factors are 1, 3 and 9.

(vi) False.

9 is an odd number: 9 = 7 + 2 where 7 and 2 are prime numbers.

(vii) True.

A number and its successor have only one common factor (i.e., 1).

#### Page No 2.11:

#### Question 18:

Fill in the blanks in the following:

(i) A number having only two factors is called a .......................... .

(ii) A number having more than two factors is called a ..................... .

(iii) 1 is neither .................. nor ........................... .

(iv) The smallest prime number is .................................. .

(v) The smallest composite number is .................................. .

#### Answer:

(i) A number having only two factors is called a __prime number__.

(ii) A number having more than two factors is called a __composite number__.

(iii) 1 is neither __composite__ nor __prime__.

(iv) The smallest prime number is __2__.

(v) The smallest composite number is __4__.

#### Page No 2.15:

#### Question 1:

In which of the following expressions, prime factorization has been done?

(i) 24 = 2 × 3 × 4

(ii) 56 = 1 × 7 × 2 × 2 × 2

(iii) 70 = 2 × 5 × 7

(iv) 54 = 2 × 3 × 9

#### Answer:

(i) 24 = 2 × 3 × 4 is not a prime factorisation as 4 is not a prime number.

(ii) 56 = 1 × 7 × 2 × 2 × 2 is not a prime factorisation as 1 is not a prime number.

(iii) 70 = 2 × 5 × 7 is a prime factorisation as 2, 5, and 7 are prime numbers.

(iv) 54 = 2 × 3 × 9 is not a prime factorisation as 9 is not a prime number.

#### Page No 2.15:

#### Question 2:

Determine prime factorization of each of the following numbers:

(i) 216

(ii) 420

(iii) 468

(iv) 945

(v) 7325

(vi) 13915

#### Answer:

(i) 216

We have:

∴ Prime factorisation of 216 = 2 × 2 × 2 × 3 × 3 × 3

(ii) 420

We have:

∴ Prime factorisation of 420 = 2 × 2 × 3 × 5 × 7

(iii) 468

We have:

∴ Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13

(iv) 945

We have:

∴ Prime factorisation of 945 = 3 × 3 × 3 × 5 × 7

(v) 7325

We have:

∴ Prime factorisation of 7325 = 5 × 5 × 293

(vi) 13915

We have:

∴ Prime factorisation of 13915 = 5 × 11 × 11 × 23

#### Page No 2.15:

#### Question 3:

Write the smallest 4-digit number and express it as a product of primes.

#### Answer:

The smallest 4-digit number is 1000.

1000 = 2 × 500

= 2 × 2 × 250

= 2 × 2 × 2 × 125

= 2 × 2 × 2 × 5 × 25

= 2 × 2 × 2 × 5 × 5 × 5

∴ 1000 = 2 $\times $2 $\times $ 2 $\times $ 5 $\times $ 5 $\times $ 5

#### Page No 2.15:

#### Question 4:

Write the largest 4-digit number and give its prime factorization.

#### Answer:

The largest 4-digit number is 9999.

We have:

Hence, the largest 4-digit number 9999 can be expressed in the form of its prime factors as 3 × 3 × 11 × 101.

#### Page No 2.15:

#### Question 5:

Find the prime factors of 1729. Arrange the factors in ascending order, and find the relation between two consecutive prime factors.

#### Answer:

The given number is 1729.

We have:

Thus, the number 1729 can be expressed in the form of its prime factors as 7 ×13 ×19.**Relation between its two consecutive prime factors:**

The consecutive prime factors of the given number are 7, 13, and 19.

Clearly, 13 − 7 = 6 and 19 − 13 = 6

Here, in two consecutive prime factors, the latter is 6 more than the previous one.

#### Page No 2.15:

#### Question 6:

Which factors are not included in the prime factorization of a composite number?

#### Answer:

1 and the number itself are not included in the prime factorisation of a composite number.

Example: 4 is a composite number.

Prime factorisation of 4 = 2 × 2

#### Page No 2.15:

#### Question 7:

Here are two different factor trees for 60. Write the missing numbers:

#### Answer:

(i) Since 6 = 2 × __ 3__ and 10 = 5 ×

**, we have:**

__2__(ii) Since 60 = 30 ×

__, 30 = 10 ×__

**2****and 10 =**

__3__**×**

__5__**, we have:**

__2__#### Page No 2.20:

#### Question 1:

Test the divisibility of the following numbers by 2:

(i) 6520

(ii) 984325

(iii) 367314

#### Answer:

**Rule****:** A natural number is divisible by 2 if its unit digit is 0, 2, 4, 6, or 8.

(i) Here, the unit's digit = 0

Thus, the given number is divisible by 2.

(ii) Here, the unit's digit = 5

Thus, the given number is not divisible by 2.

(iii) Here, the unit's digit = 4

Thus, the given number is divisible by 2.

#### Page No 2.20:

#### Question 2:

Test the divisibility of the following numbers by 3:

(i) 70335

(ii) 607439

(iii) 9082746

#### Answer:

**Rule****:** A number is divisible by 3 if the sum of its digits is divisible by 3.

(i) Here, the sum of the digits in the given number = 7 + 0 + 3 + 3 + 5 = 18 which is divisible by 3.

Thus, 70,335 is divisible by 3.

(ii) Here, the sum of the digits in the given number = 6 + 0 + 7 + 4 + 3 + 9 = 29 which is not divisible by 3.

Thus, 6,07,439 is not divisible by 3.

(iii) Here, the sum of the digits in the given number = 9 + 0 + 8 + 2 + 7 + 4 + 6 = 36 which is divisible by 3.

Thus, 90,82,746 is divisible by 3.

#### Page No 2.20:

#### Question 3:

Test the divisibility of the following numbers by 6:

(i) 7020

(ii) 56423

(iii) 732510

#### Answer:

**Rule****:** A number is divisible by 6 if it is divisible by 2 as well as 3.

(i) Here, the unit's digit = 0

Thus, the given number is divisible by 2.

Also, the sum of the digits = 7 + 0 + 2 + 0 = 9 which is divisible by 3. So, the given number is divisible by 3.

Hence, 7,020 is divisible by 6.

(ii) Here, the unit's digit = 3

Thus, the given number is not divisible by 2.

Also, the sum of the digits = 5 + 6 + 4 + 2 + 3 = 20 which is not divisible by 3. So, the given number is not divisible by 3.

Since 3,56,423 is neither divisible by 2 nor by 3, it is not divisible by 6.

(iii) Here, the unit's digit = 0

Thus, the given number is divisible by 2.

Also, the sum of the digits = 7 + 3 + 2 + 5 + 1 + 0 = 18 which is divisible by 3. So, the given number is divisible by 3.

Hence, 7,32,510 is divisible by 6.

#### Page No 2.20:

#### Question 4:

Test the divisibility of the following numbers by 4:

(i) 786532

(ii) 1020531

(iii) 9801523

#### Answer:

**Rule****:** A natural number is divisible by 4 if the number formed by its last two digits is divisible by 4.

(i) Here, the number formed by the last two digits is 32 which is divisible by 4.

Thus, 7,86,532 is divisible by 4.

(ii) Here, the number formed by the last two digits is 31 which is not divisible by 4.

Thus, 10,20,531 is not divisible by 4.

(iii) Here, the number formed by the last two digits is 23 which is not divisible by 4.

Thus, 98,01,523 is not divisible by 4.

#### Page No 2.21:

#### Question 5:

Test the divisibility of the following numbers 8:

(i) 8364

(ii) 7314

(iii) 36712

#### Answer:

** Rule:** A number is divisible by 8 if the number formed by its last three digits is divisible by 8.

(i) The given number = 8364

The number formed by its last three digit is 364 which is not divisible by 8.

Therefore, 8,364 is not divisible by 8.

(ii) The given number = 7314

The number formed by its last three digit is 314 which is not divisible by 8.

Therefore, 7,314 is not divisible by 8.

(iii) The given number = 36712

Since the number formed by its last three digit = 712 which is divisible by 8.

Therefore, 36,712 is divisible by 8.

#### Page No 2.21:

#### Question 6:

Test the divisibility of the following numbers by 9:

(i) 187245

(ii) 3478

(iii) 547218

#### Answer:

** Rule:** A number is divisible by 9 if the sum of its digits is divisible by 9.

(i) The given number = 187245

The sum of the digits in the given number = 1 + 8 + 7 + 2 + 4 + 5 = 27 which is divisible by 9.

Therefore, 1,87,245 is divisible by 9.

(ii) The given number = 3478

The sum of the digits in the given number = 3 + 4 + 7 + 8 = 22 which is not divisible by 9.

Therefore, 3,478 is not divisible by 9.

(iii) The given number = 547218

The sum of the digits in the given number = 5 + 4 + 7 + 2 + 1 + 8 = 27 which is divisible by 9.

Therefore, 5,47,218 is divisible by 9.

#### Page No 2.21:

#### Question 7:

Test the divisibility of the following numbers by 11:

(i) 5335

(ii) 70169803

(iii) 10000001

#### Answer:

(i) The given number is 5,335.

The sum of the digit at the odd places = 5 + 3 = 8

The sum of the digits at the even places = 3 + 5 = 8

Their difference = 8 − 8 = 0

∴ 5,335 is divisible by 11.

(ii) The given number is 7,01,69,803.

The sum of the digit at the odd places = 7 + 1 + 9 + 0 = 17

The sum of the digits at the even places = 0 + 6 + 8 + 3 = 17

Their difference = 17 − 17 = 0

∴ 7,01,69,803 is divisible by 11.

(iii) The given number is 1,00,00,001.

The sum of the digit at the odd places = 1 + 0 + 0 + 0 = 1

The sum of the digits at the even places = 0 + 0 + 0 + 1 = 1

Their difference = 1 − 1 = 0

∴ 1,00,00,001 is divisible by 11.

#### Page No 2.21:

#### Question 8:

In each of the following numbers, replace * by the smallest number to make it divisible by 3:

(i) 75 * 5

(ii) 35 * 64

(iii) 18 * 71

#### Answer:

We can replace the * by the smallest number to make the given numbers divisible by 3 as follows:

(i) 75*5 = 75__1__5

As 7 + 5 + 1 + 5 = 18, it is divisible by 3.

(ii) 35*64 = 35__0__64

As 3 + 5 +6 + 4 = 18, it is divisible by 3.

(iii) 18*71 = 18__1__71

As 1 + 8 + 1 + 7 + 1 = 18, it is divisible by 3.

#### Page No 2.21:

#### Question 9:

In each of the following numbers, replace * by the smallest number to make it divisible by 9:

(i) 67 * 19

(ii) 66784 *

(iii) 538 * 8

#### Answer:

(i) Sum of the given digits = 6 + 7 + 1 + 9 = 23

The multiple of 9 which is greater than 23 is 27.

Therefore, the smallest required number = 27 − 23 = 4

(ii) Sum of the given digits = 6 + 6 + 7 + 8 + 4 = 31

The multiple of 9 which is greater than 31 is 36.

Therefore, the smallest required number = 36 − 31 = 5

(iii) Sum of the given digits = 5 + 3 + 8 + 8 = 24

The multiple of 9 which is greater than 24 is 27.

Therefore, the smallest required number = 27 − 24 = 3

#### Page No 2.21:

#### Question 10:

In each of the following numbers. replace * by the smallest number to make it divisible by 11:

(i) 86 * 72

(ii) 467 * 91

(iii) 9 * 8071

#### Answer:

** Rule:** A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.

(i)

**86 $\times $ 72**

Sum of the digits at the odd places = 8 + missing number + 2 = missing number + 10

Sum of the digits at the even places = 6 + 7 = 13

Difference = [missing number + 10 ] − 13 = Missing number − 3

According to the rule, missing number − 3 = 0

**[âˆµ the missing number is a single digit]**

Thus, missing number = 3

Hence, the smallest required number is 3.

(ii)

**467 $\times $ 91**

Sum of the digits at the odd places = 4 + 7 + 9 = 20

Sum of the digits at the even places = 6 + missing number + 1 = missing number + 7

Difference = 20 − [missing number + 7] = 13 − missing number

According to rule, 13 − missing number = 11 [âˆµ the missing number is a single digit]

Thus, missing number = 2

Hence, the smallest required number is 2.

(iii)

**9 $\times $ 8071**

Sum of the digits at the odd places = 9 + 8 + 7 = 24

Sum of the digits at the even places = missing number + 0 + 1 = missing number + 1

Difference = 24 − [missing number + 1] = 23 − missing number

According to rule, 23 − missing number = 22 [âˆµ 22 is a multiple of 11 and the missing number is a single digit]

Thus, missing number = 1

Hence, the smallest required number is 1.

#### Page No 2.21:

#### Question 11:

Given an example of a number which is divisible by

(i) 2 but not by 4.

(ii) 3 but not by 6.

(iii) 4 but not by 8.

(iv) both 4 and 8 but not by 32.

#### Answer:

(i) A number which is divisible by 2 but not by 4 is 6.

(ii) A number which is divisible by 3 but not by 6 is 9.

(iii) A number which is divisible by 4 but not by 8 is 28.

(iv) A number which is divisible by 4 and 8 but not by 32 is 48.

#### Page No 2.21:

#### Question 12:

Which of the following statements are true?

(i) If a number is divisible by 3, it must be divisible by 9.

(ii) If a number is divisible by 9, it must be divisible by 3.

(iii) If a number is divisible by 4, it must be divisible by 8.

(iv) If a number is divisible by 8, it must be divisible by 4.

(v) A number is divisible by 18, if it is divisible by both 3 and 6.

(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.

(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.

(viii) If a number divides three numbers exactly, it must divide their sum exactly.

(ix) If two numbers are co-prime, at least one of them must be a prime number

(x) The sum of two consecutive odd numbers is always divisible by 4.

#### Answer:

(i) False. 12 is divisible by 3 but not by 9.

(ii) True.

(iii) False. 20 is divisible by 4 but not by 8.

(iv) True.

(v) False. 12 is divisible by both 3 and 6 but it is not divisible by 18.

(vi) True.

(vii) False. 10 divides the sum of 18 and 2 (i.e., 20) but 10 divides neither 18 nor 2.

(viii) True.

(ix) False. 4 and 9 are co-primes and both are composite numbers.

(x) True.

#### Page No 2.24:

#### Question 1:

Find the H.C.F of the following numbers using prime factorization method:

(i) 144,198

(ii) 81,117

(iii) 84,98

(iv) 225,450

(v) 170, 238

(vi) 504, 980

(vii) 150, 140, 210

(viii) 84, 120, 138

(ix) 106, 159, 265

#### Answer:

(i) 144 and 198

Prime factorisation of 144 = 2 $\times $ 2 $\times $ 2 $\times $ 2 $\times $3 $\times $ 3

Prime factorisation of 198 = 2 $\times $ 3 $\times $ 3 $\times $ 11

∴ HCF = 2 $\times $ 3 $\times $ 3 = 18

(ii) 81 and 117

Prime factorisation of 81 = 3 $\times $3 $\times $ 3 $\times $ 3

Prime factorisation of 117 = 3 $\times $ 3 $\times $ 13

∴ HCF = 3 $\times $ 3 = 9

(iii) 84 and 98

Prime factorisation of 84 = 2 $\times $ 2 $\times $ 3 $\times $ 7

Prime factorisation of 98 = 2 $\times $ 7 $\times $ 7

∴ HCF = 2 $\times $ 7 = 14

(iv) 225 and 450

Prime factorisation of 225 = 3 $\times $ 3 $\times $ 5 $\times $ 5

Prime factorisation of 450 = 2 $\times $ 3 $\times $ 3 $\times $ 5 $\times $ 5

∴ HCF = 3 $\times $ 3 $\times $ 5 $\times $ 5 = 225

(v) 170 and 238

Prime factorisation of 170 = 2 $\times $ 5 $\times $ 17

Prime factorisation of 238 = 2 $\times $ 7 $\times $ 17

∴ HCF = 2 $\times $ 17= 34

(vi) 504 and 980

We have

Prime factorisation of 504 = 2 $\times $ 2 $\times $ 2 $\times $ 3 $\times $ 3 $\times $ 7

Prime factorisation of 980 = 2 $\times $ 2 $\times $ 5 $\times $ 7 $\times $ 7

∴ HCF = 2 $\times $ 2 $\times $ 7 = 28

(vii) 150, 140 and 210

Prime factorisation of 150 = 2 $\times $ 3 $\times $ 5 $\times $5

Prime factorisation of 140 = 2 $\times $ 2 $\times $ 5 $\times $ 7

Prime factorisation of 210 = 2 $\times $ 3 $\times $ 5 $\times $ 7

∴ HCF = 2 $\times $ 5 = 10

(viii) 84, 120 and 138

Prime factorisation of 84 = 2 $\times $ 2 $\times $ 3 $\times $ 7

Prime factorisation of 120 = 2 $\times $ 2 $\times $ 2 $\times $ 3 $\times $ 5

Prime factorisation of 138 = 2 $\times $ 3 $\times $ 23

∴ HCF = 2 $\times $ 3 = 6

(ix) 106, 159, and 265

Prime factorisation of 106 = 2 $\times $ 53

Prime factorisation of 159 = 3 $\times $ 53

Prime factorisation of 265 = 5 $\times $ 53

∴ HCF = 53

#### Page No 2.24:

#### Question 2:

What is the H.C.F of two consecutive

(i) Numbers

(ii) even numbers

(iii) odd numbers

#### Answer:

(i) The common factor of two consecutive numbers is always 1.

∴ HCF of two consecutive numbers = 1

(ii) The common factors of two consecutive even numbers are 1 and 2.

∴ HCF of two consecutive even numbers = 2

(iii) The common factor of two consecutive odd numbers is 1.

∴ HCF of two consecutive odd numbers = 1

#### Page No 2.24:

#### Question 3:

H.C.F of co-prime numbers 4 and 15 was found as follow:

4=2$\times $2 and 15=3$\times $5

Since there is no common prime factor. So, H.C.F of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F?

#### Answer:

No, it is not correct.

We know that HCF of two co-prime number is 1.

4 and 15 are co-prime numbers because the only factor common to them is 1.

Thus, HCF of 4 and 15 is 1.

#### Page No 2.27:

#### Question 1:

Determine the H.C.F of the following numbers by using Euclid's algorithm (i-x):

(i) 300,450

(ii) 399,437

(iii) 1045,1520

#### Answer:

(i) 300 and 450

Dividend = 450 and divisor = 300

Clearly, the last divisor is 150.

Hence, HCF of the given numbers is 150.

(ii) 399 and 437

We have dividend = 399 and divisor = 437

Clearly, the last divisor is 19.

Hence, HCF of the given numbers is 19.

(iii) 1045 and 1520

We have dividend = 1045 and divisor = 1520

Clearly, the last divisor is 95.

Hence, HCF of the given numbers is 95.

#### Page No 2.27:

#### Question 2:

Show that the following pairs are co-prime:

(i) 59,97

(ii) 875,1859

(iii) 288,1375

#### Answer:

We know that two numbers are co-primes if their HCF is 1.

(i) 59 and 97

Here, dividend = 97 and divisor = 59

Clearly, the last divisor is 1.

Hence, the given numbers are co-primes.

(ii) 875 and 1859

Here, dividend = 1,859 and divisor = 875

Clearly, the last divisor is 1.

Hence, the given numbers are co-primes.

(iii) 288 and 1375

Here, dividend = 288 and divisor = 1,375

Clearly, the last divisor is 1.

Hence, the given numbers are co-primes.

#### Page No 2.27:

#### Question 3:

What is the H.C.F of two consecutive numbers?

#### Answer:

The HCF of two consecutive numbers is 1.**Example:**

D = 4 and d = 5 are two consecutive numbers.

Here, we have dividend = 5 and divisor = 4

Clearly, the last divisor is 1.

Hence, HCF of 4 and 5 is 1.

#### Page No 2.27:

#### Question 4:

Write true (T) of false (F) for each of the following statements:

(i) The H.C.F of two distinct prime numbers is 1.

(ii) The H.C.F of two co-prime number is 1.

(iii) The H.C.F of an even and an odd numbers is 1.

(iv) The H.C.F of two consecutive even numbers is 2.

(v) The H.C.F of two consecutive odd numbers is 2.

#### Answer:

(i) True.

(ii) True.

(iii) False.

HCF of 6 and 9 is 3, not 1.

(iv) True.

(v) False.

HCF of two consecutive odd number is 1.

Example: HCF of 25 and 27 is 1.

#### Page No 2.31:

#### Question 1:

Find the largest number which 615 and 963 leaving remainder 6 in each case.

#### Answer:

We have to find the largest number which divides (615 − 6) and (963 − 6) exactly.

Therefore, the required number = HCF of 609 and 957

Resolving 609 and 957 into prime factors, we have:

609 = 3 × 7 × 29

957 = 3 × 11 × 29

Therefore, HCF of 609 and 957 = 29 × 3 = 87

Hence, the required largest number is 87.

#### Page No 2.31:

#### Question 2:

Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.

#### Answer:

We have to find the greatest number which divides (285 − 9) and (1,249 − 7) exactly.

The required number will be given by the HCF of 276 and 1242.

Resolving 276 and 1242 into prime factors, we have:

276 = 2 × 2 × 3 × 23

1242 = 2 × 3 × 3 × 3 × 23

∴ HCF of 276 and 1242 is 2 × 3 × 23 = 138.

#### Page No 2.31:

#### Question 3:

What is the largest number that divides 626,3127 and 15628 and leaves remainders of 1,2 and 3 respectively?

#### Answer:

We have to find the largest number which divides (626 − 1), (3,127 − 2), and (15,628 − 3) exactly.

The required number will be given by the HCF of 625, 3,125 and 15,625.

Resolving 625, 3125, and 15625 into prime factors, we have:

625 = 5 × 5 × 5 × 5

3,125 = 5 × 5 × 5 × 5 × 5

15,625 = 5 × 5 × 5 × 5 × 5 × 5

Therefore, HCF of 625, 3125 and 15625 = 5 × 5 × 5 × 5 = 625

Hence, the required largest number is 625.

#### Page No 2.31:

#### Question 4:

The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions if the room exactly.

#### Answer:

Given:

Length of the room = 8 m 25 cm = 825 cm

Breadth of the room = 6 m 75 cm = 675 cm

Height of the room = 4 m 50 cm = 450 cm

The longest rod will be given by the HCF of 825, 675 and 450.

Prime factorisation of 825 = 3 × 5 × 5 × 11

Prime factorisation of 675 = 3 × 3 × 3 × 5 × 5

Prime factorisation of 450 = 2 × 3 × 3 × 5 × 5

Therefore, HCF of 825, 675 and 450 = 3 × 5 × 5 = 75

Thus, the required length of the longest rod is 75 cm.

#### Page No 2.31:

#### Question 5:

A rectangular courtyard is 20 m 16 cm long and 15 m 60 cm broad. It is to be paved with square stones of the same size. Find the least possible number of such stones.

#### Answer:

Length of the rectangular courtyard = 20 m 16 cm = 2,016 cm

Breadth of the rectangular courtyard = 15 m 60 cm = 1,560 cm

Least possible side of the square stones used to pave the rectangular courtyard = HCF of (2,016 and 1,560)

Prime factorisation of 2,016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7

Prime factorisation of 1,560 = 2 × 2 × 2 × 3 × 5 × 13

HCF of (2,016, 1,560) = 2 × 2 × 2 × 3 = 24

Least possible side of square stones used to pave the rectangular courtyard is 24 cm.

Number of square stones used to pave the rectangular courtyard

$=\frac{\mathrm{Area}\mathrm{of}\mathrm{rectangular}\mathrm{courtyard}}{\mathrm{Area}\mathrm{of}\mathrm{square}\mathrm{stone}}=\frac{2016\mathrm{cm}\times 1560\mathrm{cm}}{(24\mathrm{cm}{)}^{2}}=5460$

Thus, the least number of square stones used to pave the rectangular courtyard is 5,460.

#### Page No 2.31:

#### Question 6:

Determine the longest tape which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm.

#### Answer:

Given:

Length of the first tape = 7 m = 700 cm

Length of the second tape = 3 m 85 cm = 385 cm

Length of the third tape = 12 m 95 cm = 1,295 cm

The length of the longest tape will be the HCF of 700, 385, and 1,295.

Prime factorisation of 700 = 2 × 2 × 5 × 5 × 7

Prime factorisation of 385 = 5 × 7 × 11

Prime factorisation of 1,295 = 5 × 7 × 37

∴ HCF of 700, 385, and 1,295 = 5 × 7 = 35

∴ Required length of the longest tape = 35 cm

#### Page No 2.31:

#### Question 7:

105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip.

#### Answer:

We have to find the largest possible number of animals. Thus, we will have to find the HCF of 105, 140, and 175.

Prime factorisation of 105 = 3 × 5 × 7

Prime factorisation of 140 = 2 × 2 × 5 × 7

Prime factorisation of 175 = 5 × 5 × 7

∴ Required HCF = 5 × 7 = 35

Hence, 35 animals went in each trip.

#### Page No 2.31:

#### Question 8:

Two brands of chocolates are available in packs of 24 and 15 respectively. If i need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind i would need to buy?

#### Answer:

Let the brand ‘A’ contain 24 chocolates in one packet and brand ‘B’ contain 15 chocolates in one packet.

Equal number of chocolates of each kind can be find out by taking the LCM of the number of chocolates in each packet.

∴ LCM of 15 and 24 is:

Required LCM = 2 × 2 × 2 × 3 × 5 = 120

Therefore, minimum 120 chocolates of each kind should be purchased.

Number of boxes of brand ‘A’ which needs to be purchased = 120 ÷ 24 = 5

Number of boxes of brand ‘B’ which needs to be purchased = 120 ÷ 15 = 8

#### Page No 2.31:

#### Question 9:

During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many each would you need to buy?

#### Answer:

To find the required number of pencils and crayons, we need to find the LCM of 24 and 32.

Prime factorisation of 24 = 2 × 2 × 2 × 3

Prime factorisation of 32 = 2 × 2 × 2 × 2 × 2

∴ Required LCM of 24 and 32 = 2 × 2 × 2 × 2 × 2 × 3 = 96

Thus, number of pencils and crayons needed to be bought is 96 each, i.e. $96\xf724=4$ packs of colour pencils and $96\xf732=3$ packs of crayons.

#### Page No 2.31:

#### Question 10:

Reduce each of the following fractions to the lowest terms:

(i) $\frac{161}{207}$

(ii) $\frac{296}{481}$

#### Answer:

(i) For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF.

Now, we have to find the HCF of 161 and 207.

Prime factorisation of 161 = 7 × 23

Prime factorisation of 207 = 3 × 3 × 23

∴ HCF of 161 and 207 = 23

Now, $\frac{161\xf723}{207\xf723}=\frac{7}{9}$

Hence, $\frac{7}{9}$ is the required fraction.

(ii) For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF.

Now, we have to find the HCF of 296 and 481.

Prime factorisation of 296 = 2 × 2 × 2 × 37

Prime factorisation of 481 = 13 × 37

∴ HCF of 296 and 481 = 37

Now, $\frac{296\xf737}{481\xf737}=\frac{8}{13}$

Hence, $\frac{8}{13}$ is the required fraction.

#### Page No 2.31:

#### Question 11:

A merchant has 120 litres of oil of one kind, 180 liters of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

#### Answer:

The maximum capacity of the required tin is the HCF of the three quantities of oil.

Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5

Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

Prime factorisation of 240 = 2 × 2 × 2 × 2 × 3 × 5

∴ HCF of 120, 180, and 240 = 2 × 2 × 3 × 5 = 60

Hence, the required greatest capacity of the tin must be 60 litres.

#### Page No 2.34:

#### Question 1:

Determine the L.C.M of the numbers given below:

(i) 48,60

(ii) 42,63

(iii) 18,17

(iv) 15,30,90

(v) 56,65,85

(vi) 180,384,144

(vii) 108,135,162

(viii) 28,36,45,60

#### Answer:

(i) Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3

Prime factorisation of 60 = 2 × 2 × 3 × 5

∴ Required LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240

(ii) Prime factorisation of 42 = 2 × 3 × 7

Prime factorisation of 63 = 3 × 3 × 7

∴ Required LCM = 2 × 3 × 3 × 7 = 126

(iii) Prime factorisation of 18 = 2 × 3 × 3

Prime factorisation of 17 = 17

∴ Required LCM = 2 × 3 × 3 × 17 = 306

(iv) Prime factorisation of 15 = 3 × 5

Prime factorisation of 30 = 2 × 3 × 5

Prime factorisation of 90 = 2 × 3 × 3 × 5

∴ Required LCM = 2 × 3 × 3 × 5 = 90

(v) Prime factorisation of 56 = 2 × 2 × 2 × 7

Prime factorisation of 65 = 5 × 13

Prime factorisation of 85 = 5 × 17

∴ Required LCM = 2 × 2 × 2 × 5 × 7 × 13 × 17 = 61,880

(vi) Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

Prime factorisation of 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Prime factorisation of 144 = 2 × 2 × 2 × 2 × 3 × 3

∴ Required LCM = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 5,760

(vii) Prime factorisation of 108 = 2 × 2 × 3 × 3 × 3

Prime factorisation of 135 = 3 × 3 × 3 × 5

Prime factorisation of 162 = 2 × 3 × 3 × 3 × 3

∴ Required LCM = 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1,620

(viii) Prime factorisation of 28 = 2 × 2 × 7

Prime factorisation of 36 = 2 × 2 × 3 × 3

Prime factorisation of 45 = 3 × 3 × 5

Prime factorisation of 60 = 2 × 2 × 3 × 5

∴ Required LCM = 2 × 2 × 3 × 3 × 5 × 7 = 1,260

#### Page No 2.37:

#### Question 1:

What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?

#### Answer:

We have to find prime factorisation of 24, 36, and 54.

Prime factorisation of 24 = 2 × 2 × 2 × 3

Prime factorisation of 36 = 2 × 2 × 3 × 3

Prime factorisation of 54 = 2 × 3 × 3 × 3

∴ Required LCM = 2 × 2 × 2 × 3 × 3 × 3 = 216

Thus, 216 is the smallest number exactly divisible by 24, 36, and 54.

To get the remainder as 5:

Smallest number = 216 + 5 = 221

Thus, the required number is 221.

#### Page No 2.37:

#### Question 2:

What is the smallest number that both 33 and 39 divide leaving remainders of 5?

#### Answer:

We have to find prime factorisation of 33 and 39.

Prime factorisation of 33 = 3 × 11

Prime factorisation of 39 = 3 × 13

∴ Required LCM = 3 × 11 × 13 = 429

Thus, 429 is the smallest number exactly divisible by 33 and 39.

To get the remainder as 5:

Smallest number = 429 + 5 = 434

Thus, the required number is 434.

#### Page No 2.37:

#### Question 3:

Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

#### Answer:

To find the required least number, we have to find the LCM of the numbers from 1 to 10.

We know that 2, 3, 5, and 7 are prime number.

Prime factorisation of 4 = 2 × 2

Prime factorisation of 6 = 2 × 3

Prime factorisation of 8 = 2 × 2 × 2

Prime factorisation of 9 = 3 × 3

Prime factorisation of 10 = 2 × 5

∴ Required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520

#### Page No 2.37:

#### Question 4:

What is the smallest number that, when divide by 35, 56 and 91 leaves remainders of 7 in each case?

#### Answer:

We have to find the prime factorisation of 35, 56, and 91.

Prime factorisation of 35 = 5 × 7

Prime factorisation of 56 = 2 × 2 × 2 × 7

Prime factorisation of 91 = 7 × 13

∴ Required LCM = 2 × 2 × 2 × 5 × 7 × 13 = 3,640

Thus, 3,640 is the smallest number exactly divisible by 35, 56, and 91.

To get the remainder as 7:

Smallest number = 3,640 + 7 = 3,647

Thus, the required number is 3,647.

#### Page No 2.37:

#### Question 5:

In a school there are two sections - section A and section B of Class VI. There are 32 students in section A and 36 in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.

#### Answer:

We have to find the LCM of 32 and 36.

Prime factorisation of 32 = 2 × 2 × 2 × 2 × 2

Prime factorisation of 36 = 2 × 2 × 3 × 3

Required LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

∴ Minimum number of books required = LCM of 32 and 36 = 288 books

#### Page No 2.37:

#### Question 6:

In a morning walk three persons step off together. Their steps measure 80 cm, 85 cm, and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?

#### Answer:

We have to find the LCM of 80 cm, 85 cm, and 90 cm.

Prime factorisation of 80 = 2 × 2 × 2 × 2 × 5

Prime factorisation of 85 = 5 × 17

Prime factorisation of 90 = 2 × 3 × 3 × 5

∴ Required LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17 = 12,240

∴ Required minimum distance = LCM of 80 cm, 85 cm, and 90 cm

= 12,240 cm

= 122 m 40 cm (since 1 m =100 cm)

#### Page No 2.37:

#### Question 7:

Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

#### Answer:

First, we have to find the LCM of 8, 15, and 21.

Prime factorisation of 8 = 2 × 2 × 2

Prime factorisation of 15 = 3 × 5

Prime factorisation of 21 = 3 × 7

Therefore, required LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840.

The number nearest to 1,00,000 and exactly divisible by each of 8, 15, and 21 should also be exactly divisible by their LCM (i.e. 840).

We have to divide 1,00,000 by 840.

Remainder = 40

∴ Number just greater than 1,00,000 and exactly divisible by 840 = 1,00,000 + (840 − 40)

= 1,00,000 + 800 = 1,00,800

∴ Required number = 1,00,800

#### Page No 2.37:

#### Question 8:

A school bus picking up children in a colony of flats stops at every sixth block of flats. Another school bus starting from the same place stops at every eighth blocks of flats. Which is the first bus stop at which both of them will stop?

#### Answer:

First bus stop at which both the buses will stop together = LCM of 6th block and 8th block

Prime factorisation of 6 = 2 × 3

Prime factorisation of 8 = 2 × 2 × 2

∴ Required LCM = 2 × 2 × 2 × 3 = 24

Hence, the first bus stop at which both the buses will stop together will be at the 24th block.

#### Page No 2.37:

#### Question 9:

Telegraph poles occur at equal distances of 220 m along a road and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole?

#### Answer:

We have to find the LCM of 220 m and 300 m.

Prime factorisation of 220 = 2 × 2 × 5 × 11

Prime factorisation of 300 = 2 × 2 × 3 × 5 × 5

∴ Required LCM = 2 × 2 × 3 × 5 × 5 × 11 = 3,300

Hence, 3,300 m far is the next heap that lies at the foot of a pole.

#### Page No 2.37:

#### Question 10:

Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

#### Answer:

First, we have to find the LCM of 28 and 32.

Prime factorisation of 28 = 2 × 2 × 7

Prime factorisation of 32 = 2 × 2 × 2 × 2 × 2

∴ Required LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224

It is given that when we divide the number by 28, the remainder is 8 and when we divide the number by 32, the remainder is 12.

We observe:

28 − 8 = 20

32 − 12 = 20

∴ Required number = 224 − 20 = 204

#### Page No 2.40:

#### Question 1:

For each of the following pairs of numbers, verify the property:

Product of the number = Product of their H.C.F and L.C.M

(i) 25,65

(ii) 117,221

(iii) 35,40

(iv) 87,145

(v) 490,1155

#### Answer:

(i) Given numbers are 25 and 65.

Prime factorisation of 25 = 5 × 5

Prime factorisation of 65 = 5 × 13

HCF of 25 and 65 = 5

LCM of 25 and 65 = 5 × 5 × 13 = 325

Product of the given numbers = 25 × 65 = 1,625

Product of their HCF and LCM = 5 × 325 = 1,625

∴ Product of the number = Product of their HCF and LCM **(Verified)**

(ii) Given numbers are 117 and 221.

Prime factorisation of 117 = 3 × 3 × 13

Prime factorisation of 221 = 13 × 17

HCF of 117 and 221 = 13

LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1,989

Product of the given number = 117 × 221 = 12,857

Product of their HCF and LCM = 13 × 1,989 = 12,857

∴ Product of the number = Product of their HCF and LCM **(Verified)**

(iii) Given numbers are 35 and 40.

Prime factorisation of 35 = 5 × 7

Prime factorisation of 40 = 2 × 2 × 2 × 5

HCF of 35 and 40 = 5

LCM of 35 and 40 = 2 × 2 × 2 × 5 × 7 = 280

Product of the given number = 35 × 40 = 1,400

Product of their HCF and LCM = 5 × 280 = 1,400

∴ Product of the number = Product of their HCF and LCM **(Verified)**

(iv) Given numbers are 87 and 145.

Prime factorisation of 87 = 3 × 29

Prime factorisation of 145 = 5 × 29

HCF of 87 and 145 = 29

LCM of 87 and 145 = 3 × 5 × 29 = 435

Product of the given number = 87 × 145 = 12,615

Product of their HCF and LCM = 29 × 435 = 12,615

∴ Product of the number = Product of their HCF and LCM **(Verified)**

(v) Given numbers are 490 and 1,155.

Prime factorisation of 490 = 2 × 5 × 7 × 7

Prime factorisation of 1,155 = 3 × 5 × 7 × 11

HCF of 490 and 1,155 = 5 × 7 = 35

LCM of 490 and 1,155 = 2 × 3 × 3 × 5 × 7 × 7 × 11 = 16,170

Product of the given number = 490 × 1,155 = 5,65,950

Product of their HCF and LCM = 35 × 16,170 = 5,65,950

∴ Product of the number = Product of their HCF and LCM **(Verified)**

#### Page No 2.40:

#### Question 2:

Find the H.C.F and L.C.F of the following pairs of numbers:

(i) 117,221

(ii) 234,572

(iii) 145,232

(iv) 861,1353

#### Answer:

(i) 117 and 221

Prime factorisation of 117 = 3 × 3 × 13

Prime factorisation of 221 = 13 × 17

∴ Required HCF of 117 and 221 = 13

∴ Required LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1,989

(ii) 234 and 572

Prime factorisation of 234 = 2 × 3 × 3 × 13

Prime factorisation of 572 = 2 × 2 × 11 × 13

Required HCF of 234 and 572 = 2 × 13 = 26

Required LCM of 234 and 572 = 2 × 2 × 3 × 3 × 11 × 13 = 5,148

(iii) 145 and 232

Prime factorisation of 145 = 5 × 29

Prime factorisation of 232 = 2 × 2 × 2 × 29

Required HCF of 145 and 232 = 29

Required LCM of 145 and 232 = 2 × 2 × 2 × 5 × 29 = 1,160

(iv) 861 and 1,353

Prime factorisation of 861 = 3 × 7 × 41

Prime factorisation of 1,353 = 3 × 11 × 41

Required HCF of 861 and 1,353 = 3 × 41 = 123

Required LCM of 861 and 1,353 = 3 × 7 × 11 × 41 = 9,471

#### Page No 2.40:

#### Question 3:

The L.C.M and H.C.F of two numbers are 180 and 6 respectively. If ones of the numbers is 30, find the other number.

#### Answer:

Given:

HCF of two numbers = 6

LCM of two numbers = 180

One of the given number = 30

Product of the two numbers = Product of their HCF and LCM

∴ 30 × other number = 6 × 180

Other number $=\frac{6\times 180}{30}=36$

Thus, the required number is 36.

#### Page No 2.40:

#### Question 4:

The H.C.F of two numbers is 16 and their product is 3072. Find their L.C.M.

#### Answer:

Given:

HCF of two numbers = 16

Product of these two numbers = 3,072

Product of the two numbers = Product of their HCF and LCM

∴ 3,072 = 16 × LCM

LCM = $=\frac{3072}{16}=192$

Thus, the required LCM is 192.

#### Page No 2.40:

#### Question 5:

The H.C.F of two numbers is 145, their L.C.M is 2175. If one number is 725, find the other.

#### Answer:

HCF of two numbers = 145

LCM of two numbers = 2,175

One of the given numbers = 725

Product of the given two numbers = Product of their LCM and HCF

∴ 725 × other number = 145 × 2,175

$\mathrm{Other}\mathrm{number}=\frac{145\mathrm{x}2175}{725}=435$

Thus, the required number is 435.

#### Page No 2.40:

#### Question 6:

Can two numbers have 16 as their H.C.F and 380 as their L.C.M.? Give reason.

#### Answer:

No.

We know that HCF of the given two numbers must exactly divide their LCM.

But 16 does not divide 380 exactly.

Hence, there can be no two numbers with 16 as their HCF and 380 as their LCM.

#### Page No 2.40:

#### Question 1:

Which of the following numbers is a perfect number?

(a) 4

(b) 12

(c) 8

(d) 6

#### Answer:

(d) 6

A number for which the sum of all its factors is equal to twice the number is called a perfect number.

Factors of 6 are 1, 2 ,3, and 6.

Sum of the factors of 6 = 1 + 2 + 3 + 6 = 12 = 2 × 6

Hence, 6 is a perfect number.

#### Page No 2.40:

#### Question 2:

Which of the following are not twin-primes?

(a) 3,5

(b) 5,7

(c) 11,13

(d) 17,23

#### Answer:

(d) 17, 23

Pairs of prime numbers that differ by 2 are called twin primes.

The difference between 17 and 23 is 6.

Hence, 17 and 23 are not twin primes.

#### Page No 2.40:

#### Question 3:

Which of the following are co-primes?

(a) 8,10

(b) 9,10

(c) 6,8

(d) 15,18

#### Answer:

(b) 9, 10

9 = 3 ×3 × 1

10 = 2 × 5 × 1

Though both 9 and 10 are composite numbers, the only factor common to them is 1.

Therefore, 9 and 10 are co-primes.

#### Page No 2.40:

#### Question 4:

Which of the following is a prime number?

(a) 263

(b) 361

(c) 323

(d) 324

#### Answer:

(a) 263

263 = 1 × 263

The number 263 has only two factors, 1 and 263.

Hence, it is a prime number.

#### Page No 2.40:

#### Question 5:

The number of primes between 90 and 100 is

(a) 0

(b) 1

(c) 2

(d) 3

#### Answer:

(b) 1

There is only one prime number between 90 and 100, i.e. 97.

#### Page No 2.40:

#### Question 6:

Which of the following numbers is a perfect number?

(a) 16

(b) 8

(c) 24

(d) 28

#### Answer:

(d) 28

A number for which the sum of all its factors is equal to twice the number is called a perfect number.

Factors of 28 are 1, 2 , 4, 7, 14, and 28.

Sum of factors of 28 = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2 × 28

Hence, 28 is a perfect number.

#### Page No 2.40:

#### Question 7:

Which of the following is a prime number?

(a) 203

(b) 139

(c) 115

(d) 161

#### Answer:

(b) 139

139 = 1 × 139

The number 139 has only two factors, 1 and 139.

Hence, it is a prime number.

#### Page No 2.40:

#### Question 8:

The total number of even prime numbers is

(a) 0

(b) 1

(c) 2

(d) unlimited

#### Answer:

(b) 1

There is only one even number that is prime, i.e. 2.

#### Page No 2.40:

#### Question 9:

Which one of the following is a prime number?

(a) 161

(b) 221

(c) 373

(d) 437

#### Answer:

(c) 373

373 = 1 × 373

The number 373 has only two factors, 1 and 373.

Hence, it is a prime number.

#### Page No 2.40:

#### Question 10:

The least prime is

(a) 1

(b) 2

(c) 3

(d) 5

#### Answer:

(b) 2

2 is the least prime number. It is the only even prime number.

#### Page No 2.40:

#### Question 11:

Which one of the following numbers is divisible by 3?

(a) 27326

(b) 42356

(c) 73545

(d) 45326

#### Answer:

(c) 73,545

Sum of the digits in 73,545 = 7 + 3 + 5 + 4 + 5 = 24

Since 24 is divisible by 3, 73545 is divisible by 3.

#### Page No 2.40:

#### Question 12:

Which of the following numbers is divisible by 4?

(a) 8675231

(b) 9843212

(c) 1234567

(d) 543123

#### Answer:

(b) 98,43,212

Here, the number formed by the last two digits is 12, which is divisible by 4.

Therefore, 98,43,212 is divisible by 4.

#### Page No 2.40:

#### Question 13:

Which of the following numbers is divisible by 6?

(a) 7908432

(b) 68719402

(c) 45982014

(d)

#### Answer:

(a) 79,08,432 and (c) 4,59,82,014

A number divisible by 6 must also be divisible by 3 and 2 as 6 is a multiple of 3 and 2.

In 79,08,432, the sum of the digits = 7 + 9 + 0 + 8 + 4 + 3 + 2 = 33

Since 33 is a multiple of 3, this number is divisible by 3.

Also, since the last digit is 2, it is also divisible by 2.

Therefore, 79,08,432 is divisible by 6.

In number 4,59,82,014, the sum of the digits = 4 + 5 + 9 + 8 + 2 + 0 + 1 + 4 = 33.

Since 33 is a multiple of 3, this number is divisible by 3.

Also, since the last digit is 4, it is also divisible by 2.

So, 4,59,82,014 is also divisible by 6.

#### Page No 2.40:

#### Question 14:

Which of the following numbers is divisible by 8?

(a) 87653234

(b) 78956042

(c) 64298602

(d) 98741032

#### Answer:

(d) 9,87,41,032

In 9,87,41,032, the number formed by the last three digits is 032, which is also divisible by 8.

Hence, 9,87,41,032 is divisible by 8.

#### Page No 2.40:

#### Question 15:

Which of the following numbers is divisible by 9?

(a) 9076185

(b) 92106345

(c) 10349576

(d) 95103476

#### Answer:

(a) 90,76,185

In 90,76,185:

Sum of the digits = 9 + 0 + 7 + 6 + 1 + 8 + 5 = 36

Since 36 is divisible by 9, 9076185 is divisible by 9.

#### Page No 2.41:

#### Question 16:

Which of the following numbers is divisible by 11?

(a) 1111111

(b) 22222222

(c) 3333333

(d) 4444444

#### Answer:

(b) 2,22,22,222

In 2,22,22,222, the difference of the sum of alternate digits 2 + 2 + 2 + 2 = 8 and 2 + 2 + 2 +2 = 8 is zero.

Hence, the number is divisible by 11.

#### Page No 2.41:

#### Question 17:

If 1*548 is divisible by 3, then * can take the value

(a) 0

(b) 2

(c) 7

(d) 8

#### Answer:

(a) 0

Sum of the given digits = 1 + 5 + 4 + 8 = 18

Since 18 is a multiple of 3, the required digit is 0.

#### Page No 2.41:

#### Question 18:

5*2 is a three digit number with * as a missing digit. If the number is divisible by 6, the missing digit is.

(a) 2

(b) 3

(c) 6

(d) 7

#### Answer:

(a) 2

A number divisible by 6 must also be divisible by 3 as 6 is a multiple of 3.

Sum of the given digits = 5 + 2 = 7

We know that multiple of 3 greater than 7 is 9.

∴ 9 − 7 = 2

Therefore, the required digit is 2.

#### Page No 2.41:

#### Question 19:

What least value should be given to * so that the number 6342*1 is divisible by 3?

(a) 0

(b) 1

(c) 2

(d) 3

#### Answer:

(c) 2

Sum of the given digits = 6 + 3 + 4 + 2 + 1 = 16

We know that multiple of 3 greater than 16 is 18.

∴ 18 − 16 = 2

Therefore, the smallest required digit is 2.

#### Page No 2.41:

#### Question 20:

What least value should be given to * so that the number 915*26 is divisible by 9?

(a) 1

(b) 4

(c) 2

(d) 6

#### Answer:

(b) 4

A number is divisible by 9 if the sum of its digits is a multiple of 9.

Sum of the given digits = 9 + 1 + 5 + 2 + 6 = 23

We know that multiple of 9 greater than 23 is 27.

∴ 27 − 23 = 4

Hence, the smallest required digit is 4.

#### Page No 2.41:

#### Question 21:

What least number be assigned to * so that number 653*47 is divisible by 11?

(a) 1

(b) 2

(c) 6

(d) 9

#### Answer:

(a) 1

Sum of the digits at odd places = 6 + 3 + 4 = 13

Sum of the digits at even places = 5 + * + 7 = 12 + *

Difference = 13 − [12 + *] = 1 − *

If 6,53,*47 is divisible by 11, then 1 − * must be zero or multiple of 11.

1 − * = 0 or 11

* = 1 or −10

But * is a digit, so * must be 1.

#### Page No 2.41:

#### Question 22:

What least number be assigned to * so that the number 63576*2 is divisible by 8?

(a) 1

(b) 2

(c) 3

(d) 4

#### Answer:

(c) 3

The given number is divisible by 8 if the number formed by its last three digits is divisible by 8.

Hence, 63,57,6*2 is divisible by 8 if 6*2 is divisible by 8.

Thus, the least value of * will be 3.

#### Page No 2.41:

#### Question 23:

Which one of the following numbers is exactly divisible by 11?

(a) 235641

(b) 245642

(c) 315624

(d) 415624

#### Answer:

(d) 4,15,624

Sum of digits at odd places = 4 + 5 + 2 = 11

Sum of digits at even places = 1 + 6 + 4 = 11

Difference of these two sums = 11 − 11 = 0

Therefore, 4,15,624 is divisible by 11.

#### Page No 2.41:

#### Question 24:

If 1*548 is divisible by 3, which of the following digits can replace *?

(a) 0

(b) 2

(c) 7

(d) 9

#### Answer:

(a) 0

Sum of the given digits = 1 + 5 + 4 + 8 = 18

Since 18 is a multiple of 3, the required digit is 0.

#### Page No 2.41:

#### Question 25:

The sum of the prime numbers between 60 and 75 is

(a) 199

(b) 201

(c) 211

(d) 272

#### Answer:

(d) 272

Prime numbers between 60 and 75 are 61, 67, 71, and 73.

Their sum is given by:

61 + 67 + 71 + 73 = 272

#### Page No 2.41:

#### Question 26:

The HCF of two consecutive natural numbers is

(a) 0

(b) 1

(c) 2

(d) non-existant

#### Answer:

(b) 1

The HCF of any two consecutive natural numbers is 1 because two consecutive natural numbers are always co-prime.

#### Page No 2.41:

#### Question 27:

The HCF of two consecutive even numbers is

(a) 1

(b) 2

(c) 0

(d) non-existant

#### Answer:

(b) 2

HCF of two consecutive even numbers is always 2.**For example:**

HCF of 4 and 6 is 2.

HCF of 10 and 12 is 2 and so on.

#### Page No 2.41:

#### Question 28:

The HCF of two consecutive odd numbers is

(a) 1

(b) 2

(c) 0

(d) non-existant

#### Answer:

(a) 1

We know that the common factor of two consecutive odd numbers is 1.

Thus, HCF of two consecutive odd numbers is 1.

#### Page No 2.41:

#### Question 29:

The HCF of an even number and an odd number is

(a) 1

(b) 2

(c) 0

(d) non-existant

#### Answer:

(d) non-existent**Example:** HCF of 8 and 21 is 1.

HCF of 6 and 9 is 3.

HCF of 9 and 36 is 9.

So there is no fixed number that can be the HCF of an even number and an odd number.

#### Page No 2.41:

#### Question 30:

The LCM of 24,36 and 40 is

(a) 4

(b) 90

(c) 360

(d) 720

#### Answer:

(c) 360

We have:

24 = 2 × 2 × 2 × 3 = 2^{3} × 3

36 = 2 × 2 × 3 × 3 = 2^{2} × 3^{2}

40 = 2 × 2 × 2 × 5 = 2^{3} × 5

Here, 2, 3, and 5 are the prime factors. Highest powers of 2, 3, and 5 are 3, 2, and 1, respectively.

∴ LCM of 24, 36, and 40 = 2^{3} × 3^{2} × 5 = 8 × 9 × 5 = 360

#### Page No 2.41:

#### Question 31:

If *x* and *y* are two co-primes, then their LCM is

(a) *xy*

(b)* x+y*

(c) $\frac{x}{y}$

(d) 1

#### Answer:

(a) xy

The LCM of two co-prime numbers is equal to their product.

Thus, LCM of 'x' and 'y' will be xy.

#### Page No 2.41:

#### Question 32:

If the HCF of two number is 16 and their product is 3072, then their LCM is

(a) 182

(b) 192

(c) 12

(d) None of these

#### Answer:

(b) 192

We know:

HCF × LCM = Product of two numbers

âˆµ 16 × LCM = 3,072

∴ $\mathrm{LC}\mathrm{M}=\frac{3,072}{16}=192$

#### Page No 2.41:

#### Question 33:

The least number divisible by 15,20,24,32 and 36 is

(a) 1440

(b) 1660

(c) 2880

(d) None of these

#### Answer:

(a) 1,440

The least number divisible by 15, 20, 24, 32, and 36 can be found by taking their LCM as:

∴ LCM of 15, 20, 24, 32 and 36 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1,440

Hence, 1,440 is the least number that is divisible by 15, 20, 24, 32 and 36.

#### Page No 2.41:

#### Question 34:

The smallest number which when diminished by 3 is divisible by 11,28,36 and 45 is

(a) 1257

(b) 1260

(c) 1263

(d) None of these

#### Answer:

(d) None of these

Required smallest number = LCM of (11, 28, 36, 45) + 3 = 13,860 + 3 = 13,863

#### Page No 2.41:

#### Question 35:

Three numbers are in the ratio 1:2:3 and their HCF is 6, the numbers are

(a) 4,8,12

(b) 5,10,15

(c) 6,12,18

(d) 10,20,30

#### Answer:

(c) 6, 12, 18

Three numbers are 1× HCF, 2 × HCF, and 3 × HCF, i.e. 1 × 6 = 6, 2 × 6 = 12, and 3 × 6 = 18.

Thus, the numbers are 6, 12, 18.

#### Page No 2.41:

#### Question 36:

The ratio of two numbers is 3:4 and their HCF is 4. Their LCM is

(a) 12

(b) 16

(c) 24

(d) 48

#### Answer:

(d) 48

Two numbers are 3 × HCF and 4 × HCF

i.e. 3 × 4 = 12 and 4 × 4 = 16

LCM of 12 and 16 = 48

#### Page No 2.42:

#### Question 1:

Which of the following numbers is prime?

(a) 23

(b) 51

(c) 38

(d) 26

#### Answer:

(a) 23 = 1 × 23,

23 has only two factors 1 and 23, Therfore, it is a prime number.

(b) 51 = 1 × 3 × 17,

51 has three factors 1, 3 and 17, Therfore, it is a composite number.

(c) 38 = 1 × 2 × 19,

38 has three factors 1, 2 and 19, Therfore, it is a composite number.

(d) 26 = 1 × 2 × 13,

26 has three factors 1, 2 and 13, Therfore, it is a composite number.

Hence, the correct answer is option (a).

#### Page No 2.42:

#### Question 2:

Which of the following numbers are twin primes?

(a) 3,5

(b) 5,11

(c) 3,11

(d) 13,17

#### Answer:

Twin primes are pairs of primes which differ by two.

In (3, 5), the difference between the two primes is 2.

Therefore, (3, 5) are twin primes.

Hence, the correct answer is option (a).

#### Page No 2.42:

#### Question 3:

What smallest digit be written in the blank space of the mumber ......6724 so that the number formed is divisible by 3?

(a) 3

(b) 4

(c) 2

(d) 1

#### Answer:

A number is divisible by 3 if the sum of its digits is divisible by 3.

The sum of digits in 6724 is 6 + 7+ 2 + 4 = 19

For divisble by 3 we have to add 2 in 19 i.e., 2 + 19 = 21, which is divisible by 3.

Hence, the correct answer is option (c).

#### Page No 2.42:

#### Question 4:

Which of the following numbers is divisible by 6?

(a) 1258

(b) 61233

(c) 901352

(d) 1790184

#### Answer:

A number divisible by 2 and 3 is also divisble by 6.

Since, 1790184 is an even number

Therefore, it is divisible by 2.

The sum of digits in 1790184 is 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30, which is divisible by 3.

Therefore, 1790184 is divisible by 6.

Hence, the correct answer is option (d).

#### Page No 2.42:

#### Question 5:

Which of the following numbers is divisible by 11?

(a) 7138965

(b) 10000001

(c) 10834

(d) 901154

#### Answer:

A number is divisible by 11 if the difference of the sums of alternating digits is divisible by 11.

Sum of the digits at odd places = 1 + 0 + 0 + 0 = 1

Sum of the digits at even places = 0 + 0 + 0 + 1 = 1

Required difference, 1 − 1 = 0

Since, 0 is divisible by 11.

Therefore, 10000001 is divisible by 11.

Hence, the correct answer is option (b).

#### Page No 2.42:

#### Question 6:

Which of the following numbers is a perfect number?

(a) 12

(b) 28

(c) 8

(d) 16

#### Answer:

A perfect number is a positive number that equals the sum of its divisors, excluding itself.

Divisors of 12 = 1, 2, 3 ,4 ,6 ,12

Divisors of 28 = 1, 2, 4, 7, 14, 28

Divisors of 8 = 1, 2 ,4 ,8

Divisors of 16 = 1, 2 ,4 ,8 ,16

In 28, the sum of divisors except itself, 1 + 2 + 4 + 7 + 14 is 28.

Hence, the correct answer is option (b).

#### Page No 2.42:

#### Question 7:

Which of the folowing numbers is not divisible by 4?

(a) 78536

(b) 1264

(c) 6421

(d) 7935

#### Answer:

A number is divisible by 4 if the number's last two digits are divisible by 4.

In (a) 78536 and (b) 1264, the last two digits 36 and 64 respectively are divisible by 4.

Therefore, (a) 78536 and (b) 1264 are divisible by 4.

In (c) 6421 and (d) 7935 the last two digits 21 and 35 respectively are not divisible by 4.

Therefore, (c) 6421 and (d) 7935 are not divisible by 4.

#### Page No 2.42:

#### Question 8:

The smallest prime just greater than the HCF of 84 and 144 is

(a) 11

(b) 17

(c) 19

(d) 13

#### Answer:

84 = 1 × 2 × 2 × 3 × 7 = 2^{2} × 3^{1} × 7^{1}

144 = 1 ×2 × 2 × 2 × 2 × 3 × 3 = 2^{4} × 3^{2}

HCF of 84 and 144 = 2^{2} × 3^{1 }= 12

Prime number just greater than 12 is 13.

Hence, the correct answer is option (d).

#### Page No 2.42:

#### Question 9:

(a) factors

(b) multiples

(c) prime factors

(d) None of these

#### Answer:

Thus, every counting number has an infinite number of multiples

Hence, the correct answer is option (b).

#### Page No 2.42:

#### Question 10:

(a) 8

(b) 7

(c) 6

(d) 5

#### Answer:

The sum of digits in 37610*2 is 3 + 7+ 6 + 1 + 0 + 2 = 19

For divisble by 9 we have to add 8 in 19 i.e., 8 + 19 = 27, which is divisible by 9.

Hence, the correct answer is option (a).

#### Page No 2.42:

#### Question 11:

Define a perfect number. Write two perfect numbers

#### Answer:

A perfect number is a positive number that equals the sum of its divisors, excluding itself.

Divisors of 6 = 1, 2, 3

Divisors of 28 = 1, 2, 4, 7, 14, 28

In 6, the sum of divisors except itself, 1 + 2 + 3 is 6.

In 28, the sum of divisors except itself, 1 + 2 + 4 + 7 + 14 is 28.

Two perfect numbers are __6__ and __28.__

#### Page No 2.42:

#### Question 12:

Make a list of seven consecutive numbers, none of which is prime.

#### Answer:

The seven consecutive numbers, none of which is prime are:

90, 91, 92, 93, 94, 95, 96

#### Page No 2.42:

#### Question 13:

The HCF of two numbers is 23 and their product is 55545. Find their LCM

#### Answer:

Product of two numbers = HCF of two numbers × LCM of two numbers

⇒ 55545 = 23 × LCM of two numbers

⇒ LCM of two numbers = $\frac{55545}{23}=2415$

#### Page No 2.42:

#### Question 14:

Find the smallest 5-digit number which is exactly divisible by 20, 25, 30.

#### Answer:

20 = 1 × 2 × 2 × 5 = 2^{2} × 5^{1}

25 = 1 × 5 × 5 × 31 = 5^{2}

30 = 1 × 2 × 3 × 5 = 2^{1 }× 3^{1} × 5^{1}

LCM of 20, 25 and 30 = 2^{2} × 3^{1} × 5^{2} = 300

Smallest five digit number is 10000

Now, if we divide 10000 by 60, we will get 33.33 as quotient.

The integer just greater than 33.33 is 34

∴ Required number = 300 × 34 = 10200

Hence, the smallest 5-digit number which is exactly divisible by 20, 25, 30 is 10200.

#### Page No 2.42:

#### Question 15:

Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case.

#### Answer:

First we subtract the required remainder from 615 and 963.

Thus, we will get 609 and 957.

609 = 3 × 7 × 29 = 3^{1} × 7^{1} × 29^{1}

165 = 3 × 11 × 29 = 3^{1} × 11^{1} × 29^{1}

HCF = 3^{1} × 29^{1} = 87

Hence, the greatest number which divides 615 and 963, leaving the remainder 6 in each case is 87.

#### Page No 2.42:

#### Question 16:

The length, breadth and height of a room are 1050 cm, 750 cm and 425 cm respectively. Find the length of the longest tape which can measure the three dimensions of the room exactly

#### Answer:

1050 = 1 × 2 × 3 × 5 × 5 × 7 = 2^{1} × 3^{1 }× 5^{2} × 7^{1}

750 = 1 × 2 × 3 × 5 × 5 × 5 = 2^{1} × 3^{1 }× 5^{3}

425 = 1 × 5 × 5 × 17 = 5^{2 }× 17^{1}

30 = 1 × 2 × 3 × 5 = 2^{1} × 3^{1 }× 5^{1}

HCF of 1050, 750, and 425 = 5^{2} = 25

Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 25 cm

#### Page No 2.42:

#### Question 17:

Find the greatest number of four digits which is exactly divisible by each of 8, 12, 18 and 30.

#### Answer:

8 = 1 × 2 × 2 × 2 = 2^{3}

12 = 1 × 2 × 2 × 3 = 2^{2 }× 3^{1}

18 = 1 × 2 × 3 × 3 = 2^{1 }× 3^{2}

30 = 1 × 2 × 3 × 5 = 2^{1} × 3^{1 }× 5^{1}

LCM of 8, 12, 18, and 30 = 2^{3} × 3^{2} × 5^{1} = 360

Largest 4-digit number is 9999

Now, if we divide 9999 by 360, we will get 27.78 as quotient.

The integer just less than 27.78 is 27

∴ Required number = 360 × 27 = 9720

Hence, the greatest number of four digits which is exactly divisible by each of 8, 12, 18 and 30 is 9720.

#### Page No 2.42:

#### Question 18:

Find the least number of five digits which is exactly divisible by each of 8, 12, 18, 40 and 45

#### Answer:

8 = 1 × 2 × 2 × 2 = 2^{3}

12 = 1 × 2 × 2 × 3 = 2^{2 }× 3^{1}

18 = 1 × 2 × 3 × 3 = 2^{1 }× 3^{2}

40 = 1 × 2 × 2 × 2 × 5 = 2^{3} × 5^{1}

45 = 1 × 3 × 3 × 5 = 3^{2} × 5^{1}

LCM of 8, 12, 18, 40 and 45 = 2^{3} × 3^{2} × 5^{1} = 360

Smallest five digit number is 10000

Now, if we divide 10000 by 360, we will get 27.78 as quotient.

The integer just greater than 27.78 is 28

∴ Required number = 360 × 28 = 10080

Hence, the least number of five digits which is exactly divisible by each of 8, 12, 18, 40 and 45 is 10080.

#### Page No 2.42:

#### Question 19:

Reduce $\frac{289}{391}$ to the lowest terms

#### Answer:

$\frac{289}{391}=\frac{17\times 17}{17\times 23}\phantom{\rule{0ex}{0ex}}=\frac{17}{23}$

#### Page No 2.42:

#### Question 20:

Three tankers contain 403 lltres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.

#### Answer:

The maximum capacity of three containers is equal to the HCF of 403, 434 and 465.

403 = 1 × 13 × 31 = 13^{1} × 31^{1}

434 = 1 × 2 × 7 × 31 = 2^{1} × 7^{1} × 31^{1}

465 = 1 × 3 × 5 × 31 = 3^{1} × 5^{1} × 31^{1}

HCF of 403, 434 and 465 = 31^{1} = 31

Hence, the maximum capacity of a container that can measure the diesel of three containers exact number of times is 31 litres

#### Page No 2.42:

#### Question 21:

A number which has only two factors is called a ..............

#### Answer:

A number which has only two factors is called a __prime number__

#### Page No 2.43:

#### Question 22:

The smallest composite number is ......

#### Answer:

A composite number is a positive integer which is not prime (i.e., which has factors other than 1 and itself).

Since, the factors of 4 are 1, 2 ,4.

Hence, the smallest composite number is__ 4.__

#### Page No 2.43:

#### Question 23:

Two perfect numbers are ......... and ...........

#### Answer:

A perfect number is a positive number that equals the sum of its divisors, excluding itself.

Divisors of 6 = 1, 2, 3

Divisors of 28 = 1, 2, 4, 7, 14, 28

In 6, the sum of divisors except itself, 1 + 2 + 3 is 6.

In 28, the sum of divisors except itself, 1 + 2 + 4 + 7 + 14 is 28.

Therefore, two perfect numbers are __6__ and __28.__

#### Page No 2.43:

#### Question 24:

The HCF of two consecutive odd numbers is ..........

#### Answer:

Since, the common factor in two consecutive odd numbers is only 1.

Hence, the HCF of two consecutive odd numbers is 1.

#### Page No 2.43:

#### Question 25:

The prime triplet is ................

#### Answer:

A set of three prime numbers which form an arithmetic sequence with common difference two is called a prime triplet.

Since, 3, 5 and 7 are satisfying the above condition.

Hence, the prime triplet is (3, 5, 7).

#### Page No 2.43:

#### Question 1:

The greatest five digit number exactly divisible by 9 and 13 is

(a) 99945

(b) 99918

(c) 99964

(d) 99972

#### Answer:

LCM of 9 and 13 = 9 × 13 = 117

Largest 5-digit number is 99999

Now, if we divide 99999 by 117, we will get 854.69 as quotient.

The integer just less than 854.69 is 854

∴ Required number = 117 × 854 = 99918

Hence, the correct answer is option (b).

#### Page No 2.43:

#### Question 2:

From the numbers 2, 3, 4, 5, 6, 7, 8, 9 how many pairs of co-primes can be formed?

(a) 19

(b) 18

(c) 20

(d) 21

#### Answer:

We can form 19 pairs of co primes from the 2, 3, 4, 5, 6, 7, 8, 9 which are given below,

(2, 3), (2, 5), (2, 7), (2, 9), (3, 4),(3, 5), (3, 7), (3, 8), (4, 5), (4, 7), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (7, 8), (7, 9) and (8, 9)

Hence, the correct answer is option (a).

#### Page No 2.43:

#### Question 3:

If the number 2345 *a* 60*b* is exactly divisible by 3 and 5, then the maximum value of* a* + *b* is

(a) 12

(b) 13

(c) 14

(d) 15

#### Answer:

A number is divisible by 5 if its last digit is either 0 or 5 out of which 5 is maximum.

∴ *b* = 5

A number is divisible by 3 if the sum of its digits is divisible by 3

2 + 3 + 4 + 5 + 6 + 0 + 5 = 25

So, we can add maximum 8 to 25 which will give us 33 which is divisible by 3.

∴ *a* = 8

Now, *a* + *b* = 8 + 5 = 13

Hence, the correct answer is option (b).

#### Page No 2.43:

#### Question 4:

The HCF of 100 and 101 is .

(a) 1

(b) 7

(c) 37

(d) None of these

#### Answer:

100 = 1 × 2 × 2 × 5 × 5

101 = 1 × 101

Since, 100 is a composite number and 101 is a prime number.

Thus, their HCF is 1.

Hence, the correct answer is option (a).

#### Page No 2.43:

#### Question 5:

The LCM of 100 and 101 is

(a) 10100

(b) 1001

(c) 10101

(d) None of these

#### Answer:

100 = 1 × 2 × 2 × 5 × 5

101 = 1 × 101

Since, 100 is a composite number and 101 is a prime number.

Thus, their LCM = 100 × 101 = 10100

Hence, the correct answer is option (a).

#### Page No 2.43:

#### Question 6:

The greatest four digit number which when divided by 18 and 12 leaves a remainder of 4 in each case is

(a) 9976

(b) 9940

(c) 9904

(d) 9868

#### Answer:

18 = 1 × 2 × 3 × 3 = 2^{1 }× 3^{2}

12 = 1 × 2 × 2 × 3 = 2^{2 }× 3^{1}

LCM of 18 and 12 = 2^{2} × 3^{2} = 36

Largest 4-digit number is 9999

Now, if we divide 9999 by 36, we will get 277.75 as quotient.

The integer just less than 277.75 is 277

∴ Required number = (36 × 277) + 4 = 9972 + 4 = 9976

Hence, the correct answer is option (a).

#### Page No 2.43:

#### Question 7:

The GCD of two numbers is 17 and their LCM is 765. How many pairs of values can the numbers assume?

(a) 1 (b) 2 (c) 3 (d) 4

#### Answer:

GCD of two numbers is 17

So, the numbers can be 17*a* and 17*b*.

Now, 17*a* × 17*b* = 17 × 765

⇒ *ab* = 45

So, we can get two pairs*a* = 5 and *b* = 9 or *a* = 9 and *b* = 5

Thus, the numbers are 17 × 5 = 85 and 17 × 9 = 153.

Also, we can get the other pair 17 × 1 = 17 and 765.

Hence, the correct answer is option (b).

#### Page No 2.43:

#### Question 8:

The number of factors of 1080 is

(a) 32

(b) 28

(c) 24

(d) 36

#### Answer:

1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2^{3} × 3^{3} × 5^{1}

Thus, the total number of factors ig given by

(3 + 1)(3 + 1)(1 + 1) = 32

Hence, the correct answer is option (a).

#### Page No 2.43:

#### Question 9:

The HCF of first 100 natural numbers is

(a) 2

(b) 100

(c) 1

(d) None of these

#### Answer:

The HCF of first 100 natural numbers is 1 because there are some prime numbers like 2, 3, 5 and so on which can't have common factor other than 1.

Hence, the correct answer is option (c).

#### Page No 2.43:

#### Question 10:

The least number exactly divisible by 36 and 24 is

(a) 144

(b) 72

(c) 64

(d) 324

#### Answer:

36 = 2 × 2 × 3 × 3 = 2^{2} × 3^{2}

24 = 2 × 2 × 2 × 3 = 2^{3} × 3^{1}

LCM of 36 and 24 = 2^{3} × 3^{2} = 72

Hence, the correct answer is option (b).

#### Page No 2.43:

#### Question 11:

Find the HCF of all natural numbers from 200 to 478.

#### Answer:

The HCF of all natural numbers from 200 to 478 is 1 because there are some prime numbers like 211, 233 and so on which can't have common factor other than 1.

#### Page No 2.43:

#### Question 12:

If *x* is prime, *y* is a composite number such that* x *+ *y* = 240 and their LCM is 4199. Find *x* and *y*.

#### Answer:

We know that the LCM of a prime number and a composite number is equal to their product.

So, *xy* = 4199

Now, *x* + *y* = 240

⇒ *y* = 240 − *x*

Substituting the value of *y* in *xy* = 4199, we will get*x*( 240 − *x*) = 4199

⇒ 240*x *− *x*^{2} = 4199

⇒ * **x*^{2} − 240*x *+ 4199 = 0

⇒ *x*^{2} − 19*x *− 221*x* + 4199 = 0

⇒*x*(*x* − 19) − 221(*x *− 19) = 0

⇒ (*x* − 19)(*x *− 221) = 0

⇒ (*x* − 19) = 0 or (*x *− 221) = 0

⇒* x *= 19, 221

∴ *x* = 19 and *y* = 221 or *x* = 221 and *y* = 19

#### Page No 2.43:

#### Question 13:

The LCM of two numbers is 1024 and one of them is a prime number. Find their HCF.

#### Answer:

LCM of two numbers is 1024 = 2^{10}

Since, the other is prime number.

Hence, the other must be 2.

HCF of 2 and 1024 is 2.

#### Page No 2.43:

#### Question 14:

The HCF of two numbers is 4 and their LCM is 400. How many pairs of values can the numbers assume?

#### Answer:

HCF of two numbers is 4

So, the numbers can be 4*a* and 4*b*.

Now, 4*a* × 4*b* = 4 × 400

⇒ *ab* = 100

So, we can get the pairs*a* = 25 and *b* = 4*a* = 4 and *b* = 25

Thus, the numbers are 4 × 25 = 100 and 4 × 4 = 16.

Also, we can get the other pair 4 × 1 = 4 and 400.

Hence, there are two pairs.

#### Page No 2.44:

#### Question 15:

Find the greatest number that can divide 101 and 115 leaving remainders 5 and 7 respectively.

#### Answer:

First we will subtract 5 and 7 from 101 and 115 recpectively.

Now, we have 101 − 5 = 96 and 115 − 7 = 108

96 = 2 × 2 × 2 × 2 × 2 × 3 = 2^{5} × 3^{1}

108 =2 × 2 × 3 × 3 × 3 = 2^{2} × 3^{3}

HCF of 96 and 108= 2^{2} × 3^{1} = 12

Hence, the greatest number that can divide 101 and 115 leaving remainders 5 and 7 respectively is 12.

#### Page No 2.44:

#### Question 16:

Find the least three digit number which when divided by 20, 30, 40 and 50 leaves remainder 10 in each case.

#### Answer:

20 = 1 × 2 × 2 × 5 = 2^{2} × 5^{1}

30 = 1 × 2 × 3 × 5 = 2^{1 }× 3^{1} × 5^{1}

40 = 1 × 2 × 2 × 2 × 5 = 2^{3 }× 5^{1}

50 = 1 × 2 × 5 × 5 = 2^{1} × 5^{2}

LCM of 20, 30, 40 and 50 = 2^{3} × 3^{1} ×^{}5^{2} = 600

∴ Required number = 600 + 10 = 610.

Hence, the least three digit number which when divided by 20, 30, 40 and 50 leaves remainder 10 in each case is 610.

#### Page No 2.44:

#### Question 17:

Find the largest number that divides 59 and 54 leaving remainders 3 and 5 respectively

#### Answer:

First we will subtract 3 and 5 from 59 and 54 recpectively.

Now, we have 59 − 3 = 56 and 54 − 5 = 49

56 = 2 × 2 × 2 × 7 = 2^{3} × 7^{1}

49 = 7 × 7 = 7^{2}

HCF of 56 and 49 = 7^{1} = 7

Hence, the largest number that divides 59 and 54 leaving remainders 3 and 5 respectively is 7.

#### Page No 2.44:

#### Question 18:

Can two numbers have 12 as their HCF and 512 as their LCM? Justify your answer

#### Answer:

HCF of two numbers is a factor of the LCM of those numbers

Thu, we cannot have two numbers whose HCF is 12 and LCM is 512.

Because, when we divide 512 by 12, we get a remainder of 42.68

Thus, 12 is not a factor of 512.

Hence, we cannot have two numbers of whose HCF is 12 and LCM is 512.

#### Page No 2.44:

#### Question 19:

Write all prime numbers between 50 and 100.

#### Answer:

The prime numbers between 50 and 100 are given below:

53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

#### Page No 2.44:

#### Question 20:

Find the least 5-digit number which is exactly divisible by 20, 25 and 30.

#### Answer:

20 = 1 × 2 × 2 × 5 = 2^{2} × 5^{1}

25 = 1 × 5 × 5 × 31 = 5^{2}

30 = 1 × 2 × 3 × 5 = 2^{1 }× 3^{1} × 5^{1}

LCM of 20, 25 and 30 = 2^{2} × 3^{1} × 5^{2} = 300

Least five digit number is 10000

Now, if we divide 10000 by 60, we will get 33.33 as quotient.

The integer just greater than 33.33 is 34

∴ Required number = 300 × 34 = 10200

Hence, the least 5-digit number which is exactly divisible by 20, 25, 30 is 10200.

#### Page No 2.44:

#### Question 21:

The least number which when divided by 6, 9, 12 and 18 leaves no remainder is

#### Answer:

6 = 2 × 3 = 2^{1} × 3^{1}

9 = 3 × 3 = 3^{2}

12 = 2 × 2 × 3 = 2^{2} × 3^{1}

18 = 2 × 3 × 3 = 2^{1} × 3^{2}

LCM of 6, 9, 12 and 18 = 2^{2} × 3^{2} = 36

Hence, the least number which when divided by 6, 9, 12 and 18 leaves no remainder is 36

#### Page No 2.44:

#### Question 22:

If the product of two numbers is 360 and their HCF is 6, then their LCM is

#### Answer:

Product of two numbers = HCF of two numbers × LCM of two numbers

⇒ 360 = 6 × LCM of two numbers

⇒ LCM of two numbers = $\frac{360}{6}=60$

#### Page No 2.44:

#### Question 23:

The least number which when divided by 5, 7 and 8 leaves 3 as remainder in each case is

#### Answer:

LCM of 5, 7 and 8 is 5 × 7 × 8 = 280.

The least number which when divided by 5, 7 and 8 leaves 3 as remainder in each case is given by

280 + 3

= 283

#### Page No 2.44:

#### Question 24:

The LCM of two numbers is 26. The possible values of HCF are .......

#### Answer:

26 = 1 × 2 × 13

So, the possible values of HCF are__ 1, 2 and 13.__

#### Page No 2.44:

#### Question 25:

#### Answer:

Divisors of 6 = 1, 2, 3

Divisors of 28 = 1, 2, 4, 7, 14, 28

In 6, the sum of divisors except itself, 1 + 2 + 3 is 6.

In 28, the sum of divisors except itself, 1 + 2 + 4 + 7 + 14 is 28.

Therefore, two perfect numbers are

__6__and

__28.__

#### Page No 2.5:

#### Question 1:

Define

(i) factor

(ii) multiple

Give four examples of each.

#### Answer:

(i)** Factor:** A factor of a number is an exact divisor of that number.

For example, 4 exactly divides 32. Therefore, 4 is a factor of 32.**Examples of factors are:**

- 2 and 3 are factors of 6 because 2 $\times $ 3 = 6
- 2 and 4 are factors of 8 because 2 $\times $ 4 = 8
- 3 and 4 are factors of 12 because 3 $\times $ 4 = 12
- 3 and 5 are factors of 15 because 3 $\times $ 5 = 15

**Multiple:**When a number 'a' is multiplied by another number 'b', the product is the multiple of both the numbers 'a' and 'b'.

**Examples of multiples:**

- 6 is a multiple of 2 because 2 $\times $ 3 = 6
- 8 is a multiple of 4 because 4 $\times $ 2 = 8
- 12 is a multiple of 6 because 6 $\times $ 2 = 12
- 21 is a multiple of 7 because 7 $\times $ 3 = 21

#### Page No 2.5:

#### Question 2:

Write all factors of each of the following numbers:

(i) 60

(ii) 76

(iii) 125

(iv) 729

#### Answer:

(i) 60 = 1 $\times $ 60

60 = 2 $\times $ 30

60 = 3 $\times $ 20

60 = 4 $\times $ 15

60 = 5 $\times $ 12

60 = 6 $\times $ 10

∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

(ii) 76 = 1 $\times $ 76

76 = 2 $\times $ 38

76 = 4 $\times $ 19

∴ The factors of 76 are 1, 2, 4, 19, 38 and 76.

(iii) 125 = 1 $\times $ 125

125 = 5 $\times $ 25

∴ The factors of 125 are 1, 5, 25 and 125.

(iv) 729 = 1 $\times $ 729

729 = 3 $\times $ 243

729 = 9 $\times $ 81

729 = 27 $\times $ 27

#### Page No 2.5:

#### Question 3:

Write first five multiples of each of the following numbers:

(i) 25

(ii) 35

(iii) 45

(iv) 40

#### Answer:

(i) The first five multiples of 25 are as follows:

25 $\times $ 1 = 25

25 $\times $ 2 = 50

25 $\times $ 3 = 75

25 $\times $ 4 = 100

25 $\times $ 5 = 125

(ii) The first five multiples of 35 are as follows:

35 $\times $ 1 = 35

35 $\times $ 2 = 70

35 $\times $ 3 = 105

35 $\times $ 4 = 140

35 $\times $ 5 = 175

(iii) The first five multiples of 45 are as follows:

45 $\times $ 1 = 45

45 $\times $ 2 = 90

45 $\times $ 3 = 135

45 $\times $ 4 = 180

45 $\times $ 5 = 225

(iv) The first five multiples of 40 are as follows:

40 $\times $ 1 = 40

40 $\times $ 2 = 80

40 $\times $ 3 = 120

40 $\times $ 4 = 160

40 $\times $ 5 = 200

#### Page No 2.5:

#### Question 4:

Which of the following numbers have 15 as their factor?

(i) 15625

(ii) 123015

#### Answer:

(i)15 is not a factor of 15,625 because it is not a divisor of 15,625.

(ii) 15 is a factor of 1,23,015 because it is a divisor of 1,23,015.

i.e., 8,201 $\times $ 15 = 1,23,015

#### Page No 2.5:

#### Question 5:

Which of the following numbers are divisible by 21?

(i) 21063

(ii) 20163

#### Answer:

We know that a given number is divisible by 21 if it is divisible by each of its factors.

The factors of 21 are 1, 3, 7 and 21.

(i) Sum of the digits of the given number = 2 + 1 + 0 + 6 + 3 = 12 which is divisible by 3.

Hence, 21,063 is divisible by 3.

Again, a number is divisible by 7 if the difference between twice the one's digit and the number formed by the other digits is either 0 or a multiple of 7.

2,106 − (2 $\times $ 3) = 2,100 which is a multiple of 7.

Thus, 21,063 is divisible by 21.

(ii) Sum of the digits of the given number = 2 + 0 + 1 + 6 + 3 = 12 which is divisible by 3.

Hence, 20,163 is divisible by 3.

Again, a number is divisible by 7 if the difference between twice the one's digit and the number formed by the other digits is either 0 or multiple of 7.

2016 − (2 $\times $ 3) = 2010 which is not a multiple of 7.

Thus, 20,163 is not divisible by 21.

#### Page No 2.5:

#### Question 6:

Without actual division show that 11 is a factor of each of the following numbers:

(i) 1111

(ii) 11011

(iii) 110011

(iv) 1100011

#### Answer:

(i) 1,111

The sum of the digits at the odd places = 1 + 1 = 2

The sum of the digits at the even places = 1 + 1 = 2

The difference of the two sums = 2 − 2 = 0

∴ 1,111 is divisible by 11 because the difference of the sums is zero.

(ii) 11,011

The sum of the digits at the odd places = 1 + 0 + 1 = 2

The sum of the digits at the even places = 1 + 1 = 2

The difference of the two sums = 2 − 2 = 0

∴ 11,011 is divisible by 11 because the difference of the sums is zero.

(iii) 1,10,011

The sum of the digits at the odd places = 1 + 0 + 1 = 2

The sum of the digits at the even places = 1 + 0 + 1 = 2

The difference of the two sums = 2 − 2 = 0

∴ 1,10,011 is divisible by 11 because the difference of the sums is zero.

(iv) 11,00,011

The sum of the digits at the odd places = 1 + 0 + 0 + 1 = 2

The sum of the digits at the even places = 1 + 0 + 1 = 2

The difference of the two sums = 2 − 2 = 0

∴ 11,00,011 is divisible by 11 because the difference of the sums is zero.

#### Page No 2.5:

#### Question 7:

Without actual division show that each of the following numbers is divisible by 5:

(i) 55

(ii) 555

(iii) 5555

(iv) 50005

#### Answer:

A number will be divisible by 5 if the unit's digit of that number is either 0 or 5.

(i) In 55, the unit's digit is 5. Hence, it is divisible by 5.

(ii) In 555, the unit's digit is 5. Hence, it is divisible by 5.

(iii) In 5,555, the unit's digit is 5. Hence, it is divisible by 5.

(iv) In 50,005, the unit's digit is 5. Hence, it is divisible by 5.

#### Page No 2.5:

#### Question 8:

Is there any natural number having no factor at all?

#### Answer:

No, because each natural number is a factor of itself.

#### Page No 2.5:

#### Question 9:

Find numbers between 1 and 100 having exactly three factors.

#### Answer:

The numbers between 1 and 100 having exactly three factors are 4, 9, 25, and 49.

The factors of 4 are 1, 2 and 4.

The factors of 9 are 1, 3 and 9.

The factors of 25 are 1, 5 and 25.

The factors of 49 are 1, 7 and 49.

#### Page No 2.5:

#### Question 10:

Sort out even and odd numbers:

(i) 42

(ii) 89

(iii) 144

(iv) 321

#### Answer:

A number which is exactly divisible by 2 is called an even number.

Therefore, 42 and 144 are even numbers.

A number which is not exactly divisible by 2 is called an odd number.

Therefore, 89 and 321 are odd numbers.

#### Page No 2.7:

#### Question 1:

Find the common factors of:

(i) 15 and 25

(ii) 35 and 50

(iii) 20 and 28

#### Answer:

(i) 15 and 25

15 = 1 $\times $ 15

15 = 3 $\times $ 5

i.e., the factors of 15 are 1, 3, 5 and 15.

Again, 25 = 1 $\times $ 25

25 = 5 $\times $ 5

i.e., the factors of 25 are 1, 5 and 25.

Therefore, the common factors of the two numbers are 1 and 5.

(ii) 35 and 50

35 = 1 $\times $ 35

35 = 5 $\times $ 7

i.e., the factors of 35 are 1, 5, 7 and 35.

Again, 50 = 1 $\times $ 50

50 = 2 $\times $ 25

50 = 5 $\times $ 10

i.e., the factors of 50 are 1, 2, 5, 10, 25 and 50.

Therefore, the common factors of the two numbers are 1 and 5.

(iii) 20 and 28

20 = 1 $\times $ 20

20 = 2 $\times $ 10

20 = 4 $\times $ 5

i.e., the factors of 20 are 1, 2 , 4, 5, 10 and 20.

Again, 28 = 1 $\times $ 28

28 = 2 $\times $ 14

28 = 7 $\times $ 4

i.e., the factors of 28 are 1, 2, 4, 7, 14 and 28.

Therefore, the common factors of the two numbers are 1, 2 and 4.

#### Page No 2.7:

#### Question 2:

Find the common factors of:

(i) 5, 15 and 25

(ii) 2, 6 and 8

#### Answer:

(i) 5, 15 and 25

Factors of 5 are 1 and 5

Factors of 15 are 1, 3, 5 and 15

Factors of 25 are 1, 5 and 25

Therefore, the common factors of 5, 15, and 25 are 1 and 5.

(ii) 2, 6 and 8

Factors of 2 are 1 and 2

Factors of 6 are 1, 2, 3 and 6

Factors of 8 are 1, 2, 4 and 8

Therefore, the common factors of 2, 6 and 8 are 1 and 2.

#### Page No 2.7:

#### Question 3:

Find first three common multiples of 6 and 8.

#### Answer:

Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, …

Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, …

Therefore, the first three common multiples of 6 and 8 are 24, 48 and 72.

#### Page No 2.7:

#### Question 4:

Find first two common multiples of 12 and 18.

#### Answer:

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, …

Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198, …

Therefore, the first two common multiples of 12 and 18 are 36 and 72.

#### Page No 2.7:

#### Question 5:

A number is divisible by both 7 and 16. By which other number will that number be always divisible?

#### Answer:

Since the number is divisible by 7 and 16, they are the factors of that number.

So, the number will be divisible by the common factor of 7 and 16.

The factors of 7 are 1 and 7.

The factors of 16 are 1, 2, 4, 8, and 16.

Therefore, the common factor of 7 and 16 is 1 and the number is divisible by 1.

#### Page No 2.7:

#### Question 6:

A number is divisible by 24. By what other numbers will that number be divisible?

#### Answer:

Since the number is divisible by 24, it will be divisible by all the factors of 24.

The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

Hence, the number is also divisible by 1, 2, 3, 4, 6, 8 and 12.

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