RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers

Question 10:  Sort out even and odd numbers:

(i) 42

(ii) 89

(iii) 144

(iv) 321 

Solution 10:

Even:- A number which is exactly divisible by 2. Therefore, 42 and 144 are even numbers. 

Odd:- A number which is not exactly divisible by 2. Therefore, 89 and 321 are odd numbers.

Exercise 2.2 

Question 1:  Find the common factors of:

(i) 15 and 25

(ii) 35 and 50

(iii) 20 and 28 

Solution 1:

(i) 15 and 25

Factors of 15 are 1, 3, 5 and 15.

Factors of 25 are 1, 5 and 25.

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

(ii) 35 and 50

The factors of 35 are 1, 5, 7 and 35.

The factors of 50 are 1, 2, 5, 10, 25 and 50.

Hence, the common factors of 35 and 50 are 1 and 5.

(iii) 20 and 28

The factors of 20 are 1, 2 , 4, 5, 10 and 20.

The factors of 28 are 1, 2, 4, 7, 14 and 28.

Hence, the common factors of 20 and 28 are 1, 2 and 4.

Question 2:  Find the common factors of:

(i) 5, 15 and 25

(ii) 2, 6 and 8 

Solution 2:

(i) 5, 15 and 25

The Factors of 5 are 1 and 5.

The Factors of 15 are 1, 3, 5 and 15.

The Factors of 25 are 1, 5 and 25.

Hence, 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

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

Question 3:  Find first three common multiples of 6 and 8. 

Solution 3:

Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84………. so on

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

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

Question 4:  Find first two common multiples of 12 and 18. 

Solution 4:

Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144…… so on

Multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198, 216…… so on

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

Question 5:  A number is divisible by both 7 and 16. By which other number will that number be always divisible? 

Solution  5:

Factors of 7 are 1 and 7.

Factors of 16 are 1, 2, 4, 8, and 16.

Hence, the common factor of 7 and 16 is only 1.

Question 6:  A number is divisible by 24. By what other numbers will that number be divisible? 

Solution  6:  If a number is divisible by 24, then it will be divisible by all the factors of 24, and 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. 

Exercise 2.3

Question 1:  What are prime numbers? List all the prime numbers between 1 and 30. 

Solution  1:  A number with only two factors 1 and number itself are called prime number. i.e. 2, 3, 5, 7, 11 and 13

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

Question 2:  Write all the prime numbers between:

(i) 10 and 50

(ii) 70 and 90

(iii) 40 and 85

(iv) 60 and 100 

Solution 2:

(i) 10 and 50

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

(ii) 70 and 90

71, 73, 79, 83 and 89 are the prime numbers between 70 and 90.

(iii) 40 and 85

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

(iv) 60 and 100

61, 67, 71, 73, 79, 83, 89 and 97are the prime numbers between 60 and 100.

Question 3:  What is the smallest prime number? Is it an even number? 

Solution 3:   2 is the smallest prime number.

Yes, it is an even number. Except 2, all other even numbers are composite numbers.

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.

If yes, write the smallest odd composite number. 

Solution 4:   3 is the smallest odd prime number.

Every odd number is not a prime number. i.e. 9 is an odd number but it is not a prime number because its three factors are 1, 3 and 9.

Question 5:  What are composite numbers? Can a composite number be odd? 

Solution 5:   Composite Number:- A number which has more than two factors. Example 4, 6, 8, 9 10 and 15 etc. are composite numbers.

A composite number can be an odd number. The smallest odd number is 9.

Question 6:  What are twin-primes? Write all pairs of twin-primes between 50 and 100. 

Solution 6:  If there is only one composite number between two primes are called twin primes. For example 3, 5 and 5, 7 are twin primes. Twin primes between 50 and 100 are 59, 61 and 71, 73.

Question 7:  What are co-primes? Give examples of five pairs of co-primes. Are co-primes always prime? 

Solution 7:  If two numbers do not have any common factor other than 1 are known as co-prime.

For example

2 and 3

3 and 4

4 and 5

5 and 7

13 and 17

No, it is not necessary two co-primes are always prime for example (3, 4), (6, 7) and (4, 13).

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 

Solution 8:

(i) Two prime numbers:- Yes, 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:- No, 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:- No, two composite numbers are not always co-primes to each other.

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

Question 9:  Express each of the following numbers as a sum of two or more primes:

(i) 13

(ii) 130

(iii) 180 

Solution 9:

(i) 13 = 11 + 2

(ii) 130 = 59 + 71

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

Question 10:  Express each of the following numbers as the sum of two odd primes:

(i) 36

(ii) 42

(iii) 84 

Solution 10:

(i) 36

7 + 29 or 17 + 19 are the sum of two odd primes. 

(ii) 42  

5 + 37 or 13 + 29 are the sum of two odd primes. 

(iii) 84  

17 + 67 or 23 + 61 are the sum of two odd primes.

Question 11:  Express each of the following numbers as the sum of three odd prime numbers:

(i) 31

(ii) 35

(iii) 49 

Solution 11:

(i) 31  

5 + 7+ 9 + 13 or 11 + 13 + 7 are the sum of three odd prime numbers. 

(ii) 35

5 + 7 + 23 or 17 +13 + 5 are the sum of three odd prime numbers. 

(iii) 49

3 + 5 + 41 or 7 +11 +31 are the sum of three odd prime numbers.

Question 12:  Express each of the following numbers as the sum of twin primes:

(i) 36

(ii) 84

(iii) 120 

Solution 12:

(i) 36

17 + 19 is the sums of twin primes. 

(ii) 84

41 + 43 is the sums of twin primes. 

(iii) 120

59 + 61 is the sums of twin primes.

Question 13:  Find the possible missing twins for the following numbers so that they become twin primes:

(i) 29

(ii) 89

(iii) 101 

Solution 13:

(i) 27 and 31 are the possible of missing twins for 29. Here 27 is not a prime number and 31 is a prime number. Hence, 31 is the missing twin. 

(ii) 87 and 91 are the possible of missing twins for 89. Here 87 and 91 both is not a prime number. Hence, 89 is not twin. 

(iii) 99 and 103 are the possible of missing twins for 101. Here 99 is not a prime and 103 is a prime number. Hence, 101 is the missing twin.

Question 14:  A list consists of the following pairs of numbers:

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

Categorize them as pairs of

(i) Co-primes

(ii) Primes

(iii) Composites 

Solution 14:

(i) Co-Primes: Here all the given pairs of numbers are co-primes.

Two natural numbers are said to be co-primes numbers if they have 1 as their only common factor. 

(ii) Primes: 59, 61 and 71, 73 are pairs of prime numbers.

A Natural numbers which have exactly two factors, i.e., 1 and the number itself are called prime numbers. 

(iii) Composites: 55, 57 and 63, 65 are pairs of composite numbers.

Natural numbers which have more than two factors are called composite numbers.

Question 15:  For a number, greater than 10, to be prime what may be the possible digit in the unit’s place? 

Solution 15:   Prime numbers greater than 10 are 11, 13, 17 and 19. The possible digit in the unit’s place be 1, 3, 7 or 9.

Question 16:   Write seven consecutive composite numbers less than 100 so that there is no prime number between them. 

Solution 16:    90, 91, 92, 93, 94, 95 and 96 are the required seven consecutive composite numbers less than 100.

Question 17:   State true (T) and 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 a prime.

(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. 

Solution 17:

(i) False

Reason:- 2 and 3 are prime numbers. Sum of prime number can be a prime 2 + 3 = 5.

(ii) True.

Reason:- The product of two prime number is always a composite number.

(iii) False

Reason:- 2 is prime number and also even.

(iv) False

Reason:- 2 and 3 are consecutive and are also prime numbers.

(v) False

Reason:- 9 is an odd number but it is composite numbers.

(vi) False

Reason:- 2 and 7 are prime numbers. Sum of 7 and 2 is an odd composite number.

(vii) True

Reason:- A number and its successor have only one common factor. i.e. 1

Question 18:  Fill in the Blank:

(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 _________.

Solution 18:

(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    . 

Exercise 2.4

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 

Solution 1:

(i) 24 = 2 × 3 × 4

Here 4 is not a prime number. Hence, the given expression is not a prime factorization.

(ii) 56 = 1 × 7 × 2 × 2 × 2

Here 1 is not a prime number. Hence, the given expression is not a prime factorization.

(iii) 70 = 2 × 5 × 7

2, 5, and 7 are prime numbers. Here given expression is a prime factorization.

(iv) 54 = 2 × 3 × 9

Here 9 is not a prime number. Hence, the given expression is not a prime factorization.

Question 2:  Determine prime factorization of each of the following numbers: 

Solution 2:

Question 3:  Write the smallest 4-digit number and express it as a product of primes. 

Solution 3:

The smallest 4-digit number is 1000.

1000 = 2 × 2 × 2 × 5 × 5 × 5

Hence, 1000 = 2 × 2 × 2 × 5 × 5 × 5

Question: 4:  Write the largest 4-digit number and express it as product of primes. 

Solution  4:   The largest 4-digit number is 9999.

Hence, 3 × 3 × 11 × 101 is the factors of largest 4-digit number 9999.

Question 5:  Find the prime factors of 1729. Arrange the factors in ascending order, and find the relation between two consecutive prime factors.

Solution 5:

Thus, the prime factors of 1729 are 7 x 13 x19.

Ascending Order of factors is 7, 13, 19.

Relations between the consecutive prime factors of 1729 are 7, 13 and 19.

13 – 7 = 6 and 19 – 13 = 6.

Here, the first factor is 6 more the previous one.

Question 6:  Which factors are not included in the prime factorization of a composite number?

Solution 6:   In prime factorization of a composite number, 1 and the number itself are not included.

Example: 6 is a composite number.

Prime factorization of 6 = 2 × 3.

Question 7: Here are two different factor trees for 60. Write the missing numbers:

Solution 7:

(i) Here, 2 × 3 = 6 and 5 x 2 = 10

Exercise 2.5

Question 1:  Test the divisibility of the following numbers by 2:

(i) 6520

(ii) 984325

(iii) 367314 

Solution 1:

(i) 6250

If the unit digit of a number ends with 0, 2, 4, 6 and 8 then the number is divisible by 2.

Here in the given number unit’s digit ends with 0

Thus, the given number is divisible by 2.

(ii) 984325

If the unit digit of a number ends with 0, 2, 4, 6 and 8 then the number is divisible by 2.

Here in the given number unit’s digit ends with 5

Thus, the given number is not divisible by 2.

(iii) 367314

If the unit digit of a number ends with 0, 2, 4, 6 and 8 then the number is divisible by 2.

Here in the given number unit’s digit ends with 4

Thus, the given number is divisible by 2.

Question 2:  Test the divisibility of the following numbers by 3:

(i) 70335

(ii) 607439

(iii) 9082746 

Solution 2:

(i) 70335

If the sum of the digits of a number is divisible by 3 then the number is also divisible by 3.

Here, the sum of the digits 7 + 0 + 3 + 3 + 5 = 18 which is divisible by 3.

Hence, 70,335 is divisible by 3.

(ii) 607439

If the sum of the digits of a number is divisible by 3 then the number is also divisible by 3.

Here, the sum of the digits 6 + 0 + 7 + 4 + 3 + 9 = 19 which is not divisible by 3.

Hence, 607439 is not divisible by 3.

(iii) 9082746

If the sum of the digits of a number is divisible by 3 then the number is also divisible by 3.

Here, the sum of the digits 9 + 0 + 8 + 2 + 7 + 4 + 6 = 36 which is divisible by 3.

Hence, 9082746 is divisible by 3.

Question 3:  Test the divisibility of the following numbers by 6:

(i) 7020

(ii) 56423

(iii) 732510 

Solution 3:

(i) 7020

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

Here, the unit digit is ‘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. Thus, the given number is divisible by 3.

Hence, 7,020 is divisible by 6.

(ii) 56423

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

Here, the unit digit is ‘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. Thus, the given number is not divisible by 3.

Hence, 56423 is not divisible by 6.

(iii) 732510

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

Here, the unit digit is ‘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. Thus, the given number is not divisible by 3.

Hence, 732519 is divisible by 6.

Question 4:  Test the divisibility of the following numbers by 4:

(i) 786532

(ii) 1020531

(iii) 9801523 

Solution 4: 

(i) 786532

In the given number last two digits are ‘32’ which is divisible by 4.

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

(ii) 1020531

In the given number last two digits are ‘31’ which is not divisible by 4.

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

(iii) 9801523

In the given number last two digits are ‘23’ which is not divisible by 4.

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

Question 5:  Test the divisibility of the following numbers by 8:

(i) 8364

(ii) 7314

(iii) 36712 

Solution 5:

(i) 8364

In the given number last three digits are ‘364’ which is not divisible by 8.

Hence, 8364 is not divisible by 8.

(ii) 7314

In the given number last three digits are ‘314’ which is not divisible by 8.

Hence, 7314 is not divisible by 8.

(iii) 36712

In the given number last three digits are ‘712’ which is divisible by 8.

Hence, 36712 is also divisible by 8.

Question 6:  Test the divisibility of the following numbers by 9:

(i) 187245

(ii) 3478

(iii) 547218 

Solution 6:

(i) 187245

If the sum of the digits of a number is divisible by 9 then the number is also divisible by 9.

Here, the sum of the digits 1 + 8 + 7 + 2 + 4 + 5 = 27 which is divisible by 9.

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

(ii) 3478

If the sum of the digits of a number is divisible by 9 then the number is also divisible by 9.

Here, the sum of the digits 3 + 4 + 7 + 8 = 22 which is not divisible by 9.

Hence, 3,478 is divisible by 9.

(iii) 547218

If the sum of the digits of a number is divisible by 9 then the number is also divisible by 9.

Here, the sum of the digits 5 + 4 + 7 + 2 + 1 + 8 = 27 which is not divisible by 9.

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

Question 7:  Test the divisibility of the following numbers by 11:

(i) 5,335

(ii) 7,01,69,803

(iii) 1,00,00,001

Solution 7:

(i) 5,335.

The sum of odd places digits = 5 + 3 = 8

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

Difference = 8 – 8 = 0

Hence, 5,335 is divisible by 11.

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

The sum of odd places digits = 7 + 1 + 9 + 0 = 17

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

Their difference = 17 – 17 = 0

Hence, 7,01,69,803 is divisible by 11.

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

The sum of odd places digits = 1 + 0 + 0 + 0 = 1

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

Their difference = 1 – 1 = 0

Hence, 1,00,00,001 is divisible by 11.

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 

Solution 8:

(i) 75*5

75*5 = 7515

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

(ii) 35*64

35*64 = 35064

3 + 5 + 0 + 6 + 4 = 18, it is divisible by 3.

Hence, the missing number is 0.

(iii) 18 * 71

18 * 71= 18171

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

Hence, the missing number is 1.

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 

Solution 9:

(i) 67 *19

Sum of the digits = 6 + 7 + 1 + 9 = 23

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

Hence, the smallest required number = 27 – 23 = 4

(ii) 66784 *

Sum of the digits = 6 + 6 + 7 + 8 + 4 = 31

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

Hence, the smallest required number = 36 – 31 = 5

(iii) 538 * 8

Sum of the digits = 5 + 3 + 8 + 8 = 24

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

Hence, the smallest required number = 27 – 24 = 3

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 

Solution 10:

(i) 86*72

Sum of odd places digits = 8 + missing value + 2 = 10 + missing value

Sum of even places digits = 6 + 7 = 13

Difference = 10 + missing value – 13

Difference = missing value – 3 

3 = Missing Value

Hence, the missing value = 3

(ii) 467 * 91

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

Sum of even places digits = 6 + Missing Value + 1 = Missing Value + 7

Difference = 20 – [Missing value + 7]

Difference = 20 – Missing value – 7

Missing Value = 13

Missing value is a single digit number.

Missing number = 13 – 11

Missing number = 2

Hence, the missing value = 2

(iii) 9 * 8071

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

Sum of even places digits = Missing Value + 0 + 1 = Missing Value + 1

Difference = 24 – [Missing Value + 1]

Difference = 23 – Missing Value

Missing Value = 23 - 22

Hence, the missing value = 1

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 

Solution 11:

(i) “6” is a number which is divisible by 2 but not by 4.

(ii) “9” is a number which is divisible by 3 but not by 6.

(iii) “8” is a number which is divisible by 4 but not by 8.

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

Question 12:  Which of the following statement 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.

Solution 12:

Exercise 2.6 

Question 1:  Find the H.C. F of the following numbers using prime factors 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 

Solution 1:

(i) 144 and 198

Prime factorization of 144 = 2 × 2 × 2 × 3 × 3

Prime factorization of 198 = 2 × 3 × 3 ×11

Hence, HCF = 2 × 2 × 3 = 18

(ii) 81 and 117

Prime factorization of 81 = 3 × 3 × 3 × 3

Prime factorization of 117 = 3 × 3 ×13

Hence, HCF = 3 × 3 = 9

(iii) 84 and 98

Prime factorization of 84 = 2 × 2 × 3 × 7

Prime factorization of 98 = 2 × 7 × 7

Hence, HCF = 2 × 7 = 14

(iv) 225 and 450

Prime factorization of 225 = 3 × 3 × 5 × 5

Prime factorization of 198 = 2 × 3 × 3 × 5 × 5

Hence, HCF = 3 × 3 × 5 × 5 = 225

(v) 170 and 238

Prime factorization of 170 = 2 × 5 × 17

Prime factorization of 238 = 2 × 7 × 17

Hence, HCF = 2 × 17 = 34

(vi) 504 and 980

Prime factorization of 504 = 2 × 2 × 2 × 3 × 3 × 7

Prime factorization of 980 = 2 × 2 × 5 × 7 × 7

Hence, HCF = 2 × 2 × 7 = 28

(vii) 150, 140 and 210

Prime factorization of 150 = 2 × 3 × 5 × 5

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

Prime factorization of 210 = 2 × 3 × 5 × 7

Hence, HCF = 2 × 5 = 10

(viii) 84, 120 and 138

Prime factorization of 84 = 2 × 2 × 3 × 7

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

Prime factorization of 138 = 2 × 3 × 23

Hence, HCF = 2 × 3 = 6

(i×) 106, 159 and 265

Prime factorization of 106 = 2 × 53

Prime factorization of 159 = 2 × 53

Prime factorization of 265 = 5 × 53

Hence, HCF = 53

Question 2:  What is the H.C.F of two consecutive?

(i) Numbers

(ii) even numbers

(iii) odd numbers 

Solution 2:

(i) Common factor of two consecutive numbers is 1.

Hence, HCF of two consecutive numbers = 1

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

Hence, HCF of two consecutive even numbers = 2

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

Hence, HCF of two consecutive odd numbers = 1

Question 3:  H.C.F of co-primes numbers 4 and 15 was found as follows:

4 = 2 × 2 and 15 = 3 × 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? 

Solution 3:

No, as 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.

Prime Factor of 4 = 2 × 2

Prime Factor of 15 = 3 × 5

Hence, HCF of 4 and 15 is 1. 

Exercise 2.7 

Question 1:  Determine the HCF of the following numbers by using Euclid’s algorithm (i – x):

(i) 300, 450

(ii) 399, 437

(iii) 1045, 1520 

Solution 1:

(i) 300, 450

Question 2:  Show that the following pairs are co-prime:

(i) 59, 97

(ii) 875, 1859

(iii) 288, 1375 

Solution 2:

Question 3:  What is the HCF of two consecutive numbers?

Solution 3:

“1” is the HCF of two consecutive numbers.

Example: two consecutive numbers are 5 and 6.

Question 4:  Write true (T) or false (F) for each of the following statements:

(i) The HCF of two distinct prime numbers is 1.

(ii) The HCF of two co-prime number is 1.

(iii) The HCF of an even and an odd number is 1.

(iv) The HCF of two consecutive even numbers is 2.

(v) The HCF of two consecutive odd numbers is 2. 

Solution 4:

(i) The HCF of two distinct prime numbers is 1.                      True

(ii) The HCF of two co-prime number is 1.                              True

(iii) The HCF of an even and an odd number is 1.                   False

(iv) The HCF of two consecutive even numbers is 2.               True

(v) The HCF of two consecutive odd numbers is 2.                 False

Exercise 2.8

Question 1:  Find the largest number which divides 615 and 963 leaving remainder 6 in each case. 

Solution 1:

615 – 6 = 609

963 – 6 = 957

We should find the HCF of 609 and 957

Prime Factorisation of 609 is 3 × 7 × 29

Prime Factorisation of 957 is 3 × 11 × 29

HCF of 609 and 957 is 29 × 3

HCF of 609 and 957 is 87

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

Question 2:  Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively. 

Solution 2:

285 – 9 = 276

1249 – 7 = 1242

We should find the HCF of 276 and 1242

Prime Factorisation of 276 is 2 × 2 × 3 × 23

Prime Factorisation of 1242 is 2 × 3 × 3 × 3 × 23

HCF of 276 and 1242 = 2 × 3 × 23

HCF of 276 and 1242 = 138

Hence, the greatest number which divides 285 and 1249 leaving remainders 9 and 7 is 138.

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

Solution 3:

626 – 1 = 625

3127 – 2 = 3125

15628 – 3 = 15625

We should find the HCF of 625, 3125 and 15625.

Prime Factorisation of 625 = 5 × 5 × 5 × 5

Prime Factorisation of 3125 = 5 × 5 × 5 × 5 × 5

Prime Factorisation of 15625 = 5 × 5 × 5 × 5 × 5 × 5

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

HCF of 625, 3125 and 15625 = 625

Hence, the largest number that divides 626, 3127 and 15628 and leaves remainders 1, 2 and 3 is 625.

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

Solution 4:

The dimensions of room are

Length = 8m 25cm = 825cm

Breadth = 6m 75cm = 675cm

Height = 4m 50cm = 450cm

So the longest rod = HCF of 825, 675 and 450

We know that the prime factorization of 825 = 3 × 5 × 5 × 11

The same way prime factorization of 675 = 3 × 3 × 3 × 5 × 5

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

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

Therefore, the longest rod which can measure the dimensions of the room exactly is 75cm.

Question 5:  A rectangular courtyard is 20m 16cm long and 15m 60cm broad. It is to be paved with square stones of the same size. Find the least possible number of such stones. 

Solution 5:

Length of courtyard = 20 m 16 cm = 2016 cm

Breadth of courtyard = 15 m 60 cm = 1560 cm

Area of Courtyard = 2016 × 1560

Least possible side of square stones used

HCF of 2016 and 1560

Prime factorization of 2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7

Prime factorization of 1560 = 2 × 2 × 2 × 3 × 5 × 13

HCF of 2016 and 1560 = 2 × 2 × 2 × 3 = 24

Area of Square Stone = 24 × 24

Hence, the least possible side of square stones used is 24cm.

Question 6:  Determine the longest tape which can be used to measure exactly the lengths 7m, 3m 85cm and 12m 95cm. 

Solution 6:

The length of First tape is = 7m = 700cm

The length of Second tape is = 3m 85cm = 385cm

The length of Third tape is = 12m 95cm = 1295cm

For find the length of longest tape we should find the HCF of 700, 385 and 1295.

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

Prime factorizations of 385 = 5 × 7 × 11

Prime factorizations of 1295 = 5 × 7 × 37

HCF of 700, 385 and 1295 is 5 × 7

HCF of 700, 385 and 1295 = 35

Thus, the length of longest tape used is 35cm.

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? 

Solution 7:

The largest possible number of animals = HCF of 105, 140 and 175

Prime factorization of 105 is = 3 × 5 × 7

Prime factorization of 140 is = 2 × 2 × 5 × 7

Prime factorization of 175 is = 5 × 5 × 7

HCF of 105, 140 and 175 = 5 × 7

HCF of 105, 140 and 175 = 35

Thus, animals went in each trip is 35.

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? 

Solution 8:

LCM of 15 and 24 is

Prime Factorization of 15 is = 3 × 5

Prime Factorization of 24 is = 2 × 2 × 2 × 3

So the LCM is = 2 × 2 × 2 × 3 × 5 = 120

Hence, chocolate should be purchased of each kind = 120 

Question 11:   A merchant has 120 litres of oil of one kind, 180 litres 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? 

Solution 11:

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

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

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

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

Thus, the greatest capacity of such a tin is 60 litres.

Exercise 2.9

Question 1:  Determine the LCM 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 

Solution 1: 

(i) 48, 60

Prime factorization of 48 = 2 × 2 × 2 × 2 × 3

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

Thus, the LCM of 48 and 46 is 2 × 2 × 2 × 2 × 3 × 5 = 240

(ii) 42, 63

Prime factorization of 42 = 2 × 3 × 7

Prime factorization of 63 = 3 × 3 × 7

Thus, the LCM of 42 and 63 is 2 × 3 × 3 × 7 = 126

(iii) 18, 17

Prime factorization of 18 = 2 × 3 × 3

Prime factorization of 17 = 17

Thus, the LCM of 18 and 17 is 2 × 3 × 3 × 17 = 306

(iv) 15, 30, 90

Prime factorization of 15 = 3 × 5

Prime factorization of 30 = 2 × 3 × 5

Thus, the LCM of 15, 30 and 49 is 2 × 3 × 3 × 5 = 90

(v) 56, 65, 85

Prime factorization of 56 = 2 × 2 × 2 × 7

Prime factorization of 65 = 5 × 13

Prime factorization of 85 = 5 × 17

Thus, the LCM of 56, 65 and 85 is 2 × 2 × 2 × 5 × 7 × 13 × 17 = 61880

(vi) 180, 384, 144

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

Prime factorization of 384 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

Prime factorization of 144 = 2 × 2 × 2 × 2 × 3 × 3

Thus, the LCM of 180, 384 and 144 is 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 5760

(vii) 108, 135, 162

Prime factorization of 108 = 2 × 2 × 3 × 3 × 3

Prime factorization of 135 = 3 × 3 × 3 × 5

Prime factorization of 162 = 2 × 3 × 3 × 3 × 3

Thus, the LCM of 108,135 and 162 is 2 × 2 × 3 × 3 × 3 × 3 × 5 = 1620

(viii) 28, 36, 45, 60

Prime factorization of 28 = 2 × 2 × 7

Prime factorization of 36 = 2 × 2 × 3 × 3

Prime factorization of 45 = 3 × 3 × 5

Thus, the LCM of 28, 36, 45 and 60 is 2 × 2 × 3 × 3 × 5 × 7 = 1260 

Exercise 2.10

Question 1:  What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time? 

Solution 1:

Prime factorization of 24 is = 2 × 2 × 2 × 3

Prime factorization of 36 is = 2 × 2 × 3 × 3

Prime factorization of 54 is = 2 × 3 × 3 × 3

LCM of 24, 36 and 54 is = 2 × 2 × 2 × 3 × 3 × 3 = 216

The Smallest number which is exactly divisible by 24, 36 and 54 is 216

For getting remainder 5, we should increase the LCM by 5.

Smallest Number = 216 + 5 = 221

Thus, 221 is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time.

Question 2:  What is the smallest number that both 33 and 39 divide leaving remainder of 5? 

Solution 2:

Prime factorization of 33 = 3 × 11

Prime factorization of 39 = 3 × 13

LCM of 33 and 39 is = 3 × 11 × 13 = 429

The smallest number which is exactly divisible by 33 and 39 is 429

For getting remainder 5, we should increase the LCM by 5.

Smallest Number = 429 + 5 = 434

Thus, 434 is the smallest number that both 33 and 39 divide leaving remainder of 5.

Question 3:  Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive). 

Solution 3:

Prime Numbers between 1 and 10 are 2, 3, 5 and 7.

Composite Numbers between 1 to 10 are 4, 6, 8, 9 and 10.

Prime factorization of 4 is = 2 × 2

Prime factorization of 6 is = 2 × 3

Prime factorization of 8 is = 2 × 2 × 2

Prime factorization of 9 is = 3 × 3

Prime factorization of 10 is = 2 × 5

LCM of all the numbers between 1 to 10 is = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520

Hence, 2520 is the least number that is divisible by all the numbers between 1 and 10.

Question 4:  What is the smallest number that, when divided by 35, 56 and 91 leaves remainder of 7 in each case? 

Solution 4:

Prime factorization of 35 is = 5 × 7

Prime factorization of 56 is = 2 × 2 × 2 × 7

Prime factorization of 91 is = 7 × 13

LCM of 35, 56 and 91 is = 2 × 2 × 2 × 5 × 7 × 13 = 3640

So, the smallest number exactly divisible by 35, 56 and 91 is 364.

For getting the remainder 7 we should increase the number by 7.

Smallest number = 3640 + 7 = 3647

Thus, the smallest number that when divided by 35, 56 and 91 leaves remainder of 7 in each case is 3647.

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. 

Solution 5:

Prime factorization of 32 is = 2 × 2 × 2 × 2 × 2

Prime factorization of 36 is = 2 × 2 × 3 × 3

LCM of 32 and 36 is = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

Thus, the minimum 288 books required for their class library.

Question 6:  In a morning walk three persons step off together. Their steps measure 80cm, 85cm and 90cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps? 

Solution 6:

Prime factorization of 80 is = 2 × 2 × 2 × 2 × 5

Prime factorization of 85 is = 5 × 17

Prime factorization of 90 is = 2 × 3 × 3 × 5

LCM of 80, 85 and 90 is = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17 = 12240

Thus, the minimum 122m 40 cm distance each should walk.

Question 7:  Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21. 

Solution 7:

In order to determine the number we must find LCM of 8, 15 and 21

Prime factorization of 8 is = 2 × 2 × 2

Prime factorization of 15 is = 3 × 5

Prime factorization of 21 is = 3 × 7

LCM of 8, 15 and 21 is = 2 × 2 × 2 × 3 × 5 × 7 = 840

Here, 840 is not divisible by 100000 but the nearest number which is divisible by 100000 is 800.

The remainder obtained is 840 – 800 = 40

The number greater than 100000 and exactly divisible by 840 = 100000 + 800 = 100800.

Hence, the required number is 100800.

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? 

Solution 8:

The first bus stop at which both of them will stop = LCM of 6th and 8th block

Prime factorization of 6 is = 2 × 3

Prime factorization of 8 is = 2 × 2 × 2

LCM of 6 and 8 is = 2 × 2 × 2 × 3 = 24

Thus, 24^th block is the first bus stop at which both of them will stop.

Question 9:   Telegraph poles occur at equal distances of 220m along a road and heaps of stones are put at equal distances of 300m 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? 

Solution 9:

Prime factorization of 220 is = 2 × 2 × 5 × 11

Prime factorization of 300 is = 2 × 2 × 3 × 5 × 5

LCM of 220 and 300 is = 2 × 2 × 3 × 5 × 5 × 11 = 3300

Hence, the heap which lies at the foot of a pole is 3300m far along the road.

Question 10:  Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively. 

Solution 10:

Prime factorization of 28 is = 2 × 2 × 7

Prime factorization of 32 is = 2 × 2 × 2 × 2 × 2

LCM of 28 and 32 is = 2 × 2 × 2 × 2 × 2 × 7 = 224

Number leaves remainders 8 and 12 when divided by 28 and 32

28 – 8 = 20 and 32 – 12 = 20

Hence, the number is 224 – 20 = 204 

Exercise 2.11 

Question 1:   For each of the following pairs of numbers, verify the property:

Product of the number = Product of their HCF and LCM 

(i) 25, 65

(ii) 117, 221

(iii) 35, 40

(iv) 87, 145

(v) 490, 1155 

Solution 1:

(i) 25, 65

Prime factorization of

Prime factorization of 25 is 5 × 5

Prime factorization of 65 is 5 × 13

HCF of 25 and 65 = 5

LCM of 25 and 65 = 5 × 5 × 13 = 325

Verify:-

We know that Product of number = Product of HCF and LCM

25 × 65 = 5 × 325

1625 = 1625

Thus, the verification is correct product of the number = product of their HCF and LCM.

(ii) 117, 221

Prime factorization of 117 is 3 × 3 × 13

Prime factorization of 221 is 13 × 17

HCF of 117 and 221 = 13

LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1989

Verify:-

We know that Product of number = Product of HCF and LCM

117 × 221 = 13 × 1989

25857 = 25857

Thus, the verification is correct product of the number = product of their HCF and LCM. 

(iii) 35, 40 

Prime factorization of 35 is 5 × 7

Prime factorization of 40 is 2 × 2 × 2 × 5

HCF of 35 and 40 = 5

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

Verify:-

We know that Product of number = Product of HCF and LCM

35 × 40 = 5 × 280

1400 = 1400

Thus, the verification is correct product of the number = product of their HCF and LCM.

(iv) 87, 145

Prime factorization of 87 is 3 × 29

Prime factorization of 145 is 5 × 29

HCF of 87 and 145 = 29

LCM of 87 and 145 = 3 × 5 × 29 = 435

Verify:-

We know that Product of number = Product of HCF and LCM

87 × 145 = 29 × 435

12615 = 12615

Thus, the verification is correct product of the number = product of their HCF and LCM.

(v) 490, 1155

Prime factorization of 490 is 2 × 5 × 7 × 7

Prime factorization of 1155 is 3 × 5 × 7 × 11

HCF of 490 and 1155 = 35

LCM of 490 and 1155 = 2 × 3 × 3 × 5 × 7 × 7 × 11 = 16170

Verify:-

We know that Product of number = Product of HCF and LCM

490 × 1155 = 35 × 16170

565950 = 565950

Thus, the verification is correct product of the number = product of their HCF and LCM.

Question 2:   Find the HCF and LCM of the following pairs of numbers:

(i) 117, 221

(ii) 234, 572

(iii) 145, 232

(iv) 861, 1353 

Solution 2:

(i) 117, 221

Prime factorization of 117 is 3 × 3 × 13 

Prime factorization of 221 is 13 × 17

HCF = 13

LCM = 3 × 3 × 13 × 17 = 1989

Hence, the HCF and LCM of 117 and 221 are 13 and 1989 respectively.

(ii) 234, 572

Prime factorization of 234 is 2 × 3 × 3 × 13

Prime factorization of 572 is 2 × 2 × 11 × 13

HCF = 26

LCM = 2 × 2 × 3 × 3 × 11 × 13 = 5148

Hence, the HCF and LCM of 234 and 572 are 26 and 5148 respectively.

(iii) 145, 232

Prime factorization of 145 is 5 × 29

Prime factorization of 232 is 2 × 2 × 2 × 29

HCF = 29

LCM = 2 × 2 × 2 × 5 × 29 = 1160

Hence, the HCF and LCM of 145 and 232 are 29 and 1160 respectively.

(iv) 861, 1353

Prime factorization of 861 is 3 × 7 × 41

Prime factorization of 1353 is 3 × 11 × 41

HCF = 123

LCM = 3 × 7 × 11 × 41 = 9471

Hence, the HCF and LCM of 861 and 1353 are 123 and 9471 respectively.

Question 3:  The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, find the other number.

Solution 3:

The LCM of two numbers = 180

The HCF of two numbers = 6

One number is = 30

Let the other number be x

Product of two numbers = Product of HCF and LCM

30 × x = 6 × 180

x = (6 × 180)/30

x = 6 × 6

x = 36

Thus, the other number is 36.

Question 4:   The HCF of two numbers is 16 and their product is 3072. Find their LCM.

Solution 4:

The HCF of two numbers = 16

Product of the numbers = 3072

Let the LCM be x

Product of two numbers = Product of HCF and LCM

3072 = 16 × x

3072/16=x

192=x

Thus, the LCM of two numbers is 192.

Question 5:  The HCF of two numbers is 145, their LCM is 2175. If one number is 725, find the other.

Solution 5:

The HCF of two numbers = 145

The LCM of two numbers = 2175

One number is = 725

Let the Other number be x

Product of two numbers = Product of HCF and LCM

725 × x = 145 × 2175

x = (145 × 2175)/725 

x = 435

Thus, the other number is 435.

Question 6:  Can two numbers have 16 as their HCF and 380 as their LCM? Give reason. 

Solution 6:

We know that the HCF of two numbers exactly divides their LCM.

380/16 = 23.75

Here, LCM is not exactly divisible by HCF

Hence, we can say that two numbers cannot exist with 16 as HCF and 380 as LCM.

Objective Type Questions :->

Question 1:  Which of the following numbers is a perfect number?

(a) 4

(b) 12

(c) 8

(d) 6 

Solution 1: (d)

Reason:-

Here 6 is a perfect number.

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

And the sum of the factors of 6 = 1 + 2 + 3 + 6 = 12

Question 2:  Which of the following are not twin-primes?

(a) 3, 5

(b) 5, 7

(c) 11, 13

(d) 17, 23 

Solution 2: (d)

Reason:-

Difference between 17 and 23 is 6

Thus, 17 and 23 are not twin-primes.

Question 3:  Which of the following are co-primes?

(a) 8,10

(b) 9,10

(c) 6,8

(d) 15,18 

Solution 3: (b)

Reason:-

Prime factorisation of 9 = 3 ×3 × 1

Prime factorisation of 10 = 2 × 5 × 1

9 and 10 are composite numbers with common factor 1.

Hence, 9 and 10 are co-primes

Question 4:  Which of the following is a prime number?

(a) 263

(b) 361

(c) 323

(d) 324 

Solution 4: (a)

Reason:-

Prime factorisation of 263 = 1 × 263

263 has only two factors, 1 and 263

Hence, it is a prime number

Question 5:  The number of primes between 90 and 100 is

(a) 0

(b) 1

(c) 2

(d) 3 

Solution 5: (b)

Reason:-

Prime Number between 90 and 100 is only 97.

Hence, the number of primes between 90 and 100 is only 1.

Question 6:  Which of the following numbers is a perfect number? 

(a) 16

(b) 8

(c) 24

(d) 28 

Solution 6: (d)

Reason:-

The Prime factor of 28 are 1, 2, 4, 7, 14, and 28

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

The Double of the number is also 56 = 2 × 28

Hence, 28 is a perfect number.

Question 7:  Which of the following is a prime number? 

(a) 203

(b) 139

(c) 115

(d) 161 

Solution 7: (b)

Question 8:  The total number of even prime numbers is

(a) 0

(b) 1

(c) 2

(d) unlimited 

Solution 8: (c)

Reason:-  2 is the only even prime number.

Question 9:  Which one of the following is a prime number?

(a) 161

(b) 221

(c) 373

(d) 437 

Solution 9: (c)

Question 10:  The least prime is

(a) 1

(b) 2

(c) 3

(d) 5 

Solution 10: (b)

Question 11:  Which one of the following numbers is divisible by 3?

(a) 27326

(b) 42356

(c) 73545

(d) 45326 

Solution 11: (c)

Reason:-

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

24 is divisible by 3

Hence, 73545 is divisible by 3.

Question 12:  Which of the following numbers is divisible by 4?

(a) 8675231

(b) 9843212

(c) 1234567

(d) 543123 

Solution 12: (b)

Reason:-

In 9843212

Last two digits are 12. Which is divisible by 4.

Hance, 9843212 is also divisible by 4.

Question 13:  Which of the following numbers is divisible by 6?

(a) 7908432

(b) 68719402

(c) 45982014 

Solution 13: (a)

Reason:-

In digit 79,08,432

Sum of the digits = 7 + 9 + 0 + 8 + 4 + 3 + 2 = 33

33 is a divisible by 3.

Last digit is even so 79,08,432 is also divisible by 2.

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

Question 14:  Which of the following numbers is divisible by 8?

(a) 87653234

(b) 78956042

(c) 64298602

(d) 98741032 

Solution 14: (d)

Reason:-

In 98741032

Last three digits are 032. Which is divisible by 8.

Hance, 98741032 is also divisible by 8.

Question 15:  Which of the following numbers is divisible by 9?

(a) 9076185

(b) 92106345

(c) 10349576

(d) 95103476 

Solution 15: (a)

Reason:-

In 90,76,185

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

Here, 36 is divisible by 9

Hence, 9076185 is divisible by 9.

Question 16:  Which of the following numbers is divisible by 11?

(a) 1111111

(b) 22222222

(c) 3333333

(d) 4444444 

Solution 16: (b)

Reason:-

In 2,22,22,222,

Sum of odd places digits = 2 + 2 + 2 + 2 = 8

Sum of even places digits = 2 + 2 + 2 + 2 = 8

Difference = 8 – 8 = 0

Therefore, the number is divisible by 11.

Question 17:  If 1*548 is divisible by 3, then * can take the value

(a) 0

(b) 2

(c) 7

(d) 8 

Solution 17: (a)

Reason:-

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

Here, 18 is a multiple of 3

Hence, the required digit is 0.

Question 18:  5*2 is a three digits 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 

Solution 18: (a)

Reason:-

Sum of the given digits = 5 + 2 = 7

Multiple of 3 is 9 which is nearest 7.

Difference 9 − 7 = 2

Thus, the digit is 2.

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 

Solution 19: (c)

Reason:-

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

Multiple of 3 is 18 which is nearest 16.

Difference = 18 − 16 = 2

Thus, the smallest digit is 2.

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 

Solution 20: (b)

Reason:-

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

Multiple of 9 is 27 which is nearest 23.

Difference = 27 − 23 = 4

Thus, the smallest digit is 4.

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 

Solution 21: (a)

Reason:-

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

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

Difference = 13 − 12 - *

Difference = 1 - *

Thus, * is a digit, so it must be 1.

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 

Solution 22: (c)

Reason:-

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

Last three digits are 6*2. Here * is replaced by 3 because 632 is divisible by 8.

Thus, the least value of * will be 3.

Question 23: Which one of the following numbers is exactly divisible by 11?

(a) 235641

(b) 245642

(c) 315624

(d) 415624 

Solution 23: (d)

Reason:-

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

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

Difference = 11 − 11 = 0

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

Question 24:  If 1*548 is divisible by 3, which of the following digits can replace *? 

(a) 0

(b) 2

(c) 7

(d) 9 

Solution 24: (a)

Reason:-

The sum of the = 1 + 5 + 4 + 8 = 18

The digit replaces * is 0.

Question 25:  The sum of the prime numbers between 60 and 75 is

(a) 199

(b) 201

(c) 211

(d) 272 

Solution 25: (d)

Reason:-

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

Their sum is 61 + 67 + 71 + 73 = 272

Question 26:  The HCF of two consecutive natural numbers is

(a) 0

(b) 1

(c) 2

(d) non-existent 

Solution 26: (b)

Reason:-  The HCF of two consecutive natural numbers is 1.

Question 27:  The HCF of two consecutive even numbers is

(a) 1

(b) 2

(c) 0

(d) non-existent 

Solution 27: (b)

Reason:-  The HCF of two consecutive even numbers is 2.

Question 28:  The HCF of two consecutive odd numbers is

(a) 1

(b) 2

(c) 0

(d) non-existent 

Solution 28: (a)

Reason:-  We know that 1 is the HCF of two consecutive odd number.

Question 29:  The HCF of an even number and an odd number is

(a) 1

(b) 2

(c) 0

(d) non-existent 

Solution 29: (a)

Reason:-  1 is the HCF of an even number and an odd number.

Question 30:  The LCM of 24, 36 and 40 is

(a) 4

(b) 90

(c) 360

(d) 720 

Solution 30: (c)

Reason:-

Prime factorization of 24 = 2 × 2 × 2 × 3 = 23 × 3

Prime factorization of 36 = 2 × 2 × 3 × 3 = 22 × 32

Prime factorization of 40 = 2 × 2 × 2 × 5 = 23 × 5

LCM of 24, 36, and 40 = 23 × 32 × 5 = 8 × 9 × 5 = 360

Question 31:  If x and y are two co-primes, then their LCM is

(a) xy

(b) x + y

(c) x/y

(d) 1 

Solution 31: (a)

Reason:-

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

Hence, LCM of ‘x’ and ‘y’ will be xy.

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 

Solution 32: (b)

Reason:-

Product of HCF and LCM = Product of two numbers

16 × LCM = 3,072

Thus, we get LCM = 3,072/16 = 192

Question 33:  The least number divisible by 15,20,24,32 and 36 is

(a) 1440

(b) 1660

(c) 2880

(d) None of these 

Solution 33: (a)

Reason:-

Prime Factorisation of 15 is = 3 × 5

Prime Factorisation of 20 is = 2 × 2 × 5

Prime Factorisation of 24 is = 2 × 2 × 2 × 3

Prime Factorisation of 32 is = 2 × 2 × 2 × 2 × 2

Prime Factorisation of 36 is = 2 × 2 × 3 × 3

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

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

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 

Solution 34: (d)

Reason:-

LCM of 11, 28, 36 and 45 = 13860

The smallest number diminished by 3 = 13,860 + 3 = 13,863

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 

Solution 35: (c) 

Reason:-

First Number = 1× HCF

First Number = 1 × 6 = 6

Second Number = 2 × HCF

Second Number = 2 × 6 = 12

Third Number = 3 × HCF

Third Number = 3 × 6 = 18

Hence, the numbers are 6, 12 and 18.

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 

Solution 36: (d)

Reason:-

First Number = 3 × HCF

First Number = 3 × 4 = 12

Second Number =  4 × HCF

Second Number = 4 × 4 = 16

Hence, the LCM of 12 and 16 = 48