Playing With Numbers

Factors of a number are the numbers that can be multiplied by to get the original number.

To factor a number is to break it into smaller number.

(a) 24

1 × 24 = 24

2 × 12 = 24

3 × 8 = 24

4 × 6 = 24

So,

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

(b) 15

1 × 15 = 15

3 × 5 = 15

So,

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

(c) 21

1 × 21 = 21

3 × 7 =21

So,

Factors of 21 are 1, 3, 7 and 21.

(d) 27

1 × 27 =27

3 × 9 = 27

So,

1, 3, 9 and 27 are the factors of 27.

(e) 12

1 × 12 =12

2 × 6 =12

3 × 4 =12

So,

1, 2, 3, 4, 6 and 12 are the factors of 12.

(f) 20

1 × 20 =20

2 × 10 =20

4 × 5 = 20

So,

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

(g) 18

1 × 18 =18

2 × 9 =18

3 × 6 =18

So,

1, 2, 3, 6, 9, and 18 are the factors of 18.

(h) 23

1 × 23 =23

So,

1 and 23 are the factors of 23.

(i) 36

1 × 36 = 36

2 × 18 = 36

3 × 12 = 36

4 × 9 = 36

6 × 6 = 36

So,

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

A multiple is a number that can be obtained by multiplying a number by another number.

(a) 5

First five multiplies of 5 are;

1 × 5 = 5

2 × 5 = 10

3 × 5 = 15

4 × 5 = 20

5 × 5 = 25

So,

The first five multiplies are 5, 10, 15, 20 and 25.

(b) 8

1 × 8 = 8

2 × 8 = 16

3 × 8 = 24

4 × 8 = 32

5 × 8 = 40

So,

The first five multiplies of 8 are,

8, 16, 24, 32 and 40.

(c) 9

1 × 9 = 9

2 × 9 = 18

3 × 9 = 27

4 × 9 = 36

5 × 9 = 45

So,

First five multiplies are 9, 18, 27, 36 and 45.

1 × 9 = 9

2 × 9 = 18

3 × 9 = 27

4 × 9 = 36

5 × 9 = 45

6 × 9 = 54

7 × 9 = 63

8 × 9 = 72

9 × 9 = 81

10 × 9 = 90

11 × 9 = 99

So,

The multiplies of 9 up to 100 will be,

9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

Odd number is a number that can not be divided by 2.

An even number is a number that can be divede by 2.

(a) The sum of two odd numbers is even;

For example –

3 + 5 = 8

7 + 11 = 18

13 + 19 = 32

(b) The sum of two even numbers is even;

Example-

2+6 = 8

8+10 = 18

16+16 = 32

(a) False.

For example- 1+3+5 = 9, this is an odd number.

(b) True.

Example- 1+3+4 = 8, this is an even number.

(c) True.

Example- 1 × 3 × 5 = 15, it is an odd number.

(d) False.

If an even number is divided by 2, the quotient is not always odd it can be even also.

For example-

8÷2 = 4, it is an even number.

(e) False.

All prime numbers are not odd, 2 is a prime number and it is an even number.

(f) False.

Prime numbers do have factors like 1 and the numbers itself are the factors of the numbers.

(g) False.

Sum of two prime numbers is not always even.

Example- 2+3 = 5 it is an odd number.

(h) True.

2 is the only even prime number.

(i) False.

All even numbers are not composite numbers, 2 is the prime number.

(j) True.

The product of two even numbers is always even.

Example- 4 × 6 = 24, it is an even number.

Prime numbers are the numbers that can be divided by 1 and by the number itself.

Such pairs of prime numbers are:

17 and 71

37 and 73

79 and 97

Prime numbers are the numbers that has only two factors: 1 and number itself.

A composite number has more than two factors.

Prime numbers less than 20 are;

2, 3, 5, 7, 11, 13, 17 and 19.

Composite numbers less than 20 are;

4, 6, 8, 9, 10, 12, 14, 15, 16, and 18.

Note: Every Number that is not a prime number is a composite number.

First look for all the prime numbers between 1 and 10;

2, 3, 5, and 7.

Among these prime numbers 7 is the greatest.

(a) 44 = 39+5

(b) 36 = 29+7

(c) 24 = 17+7

(d) 18 = 13+5

Pairs of prime numbers whose difference is 2 are;

1)5 and 7

2)21 and23

3) 91 and 93

93 - 91 = 2

(a) 23 is a prime number;

23 = 1 × 23

23 = 23 × 1

Prime numbers are the numbers that can be divided by 1 and by number itself. So, 23 has two factors 1 and 23. It is a prime number.

(b) 51 is a composite number,

51 = 1 × 51

51 = 3 × 17

51 = 17 × 3

51 = 51 × 1

So, it has four factors, 1, 3, 17, and 51. Therefore it is a composite number not a prime number.

(c) 37 is a prime number,

37 = 1 × 37

37 = 37 × 1

Prime numbers are the numbers that can be divided by 1 and by number itself. 37 just have two factors 1 and 37. So it is a prime number.

(d) 26

26 = 1 × 26

26 = 13 × 2

26 = 2 × 13

26 = 26 × 1

It has four factors 1, 2, 13 and 26,

So, it’s not a prime number it is a composite number.

Between 89 and 97, both of which are prime numbers, there are 7 composite numbers,

90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90

91 = 1, 7, 13, and 91

92 = 1, 2, 4, 23, 46, and 92

93 = 1, 3, 31 and 93

94 = 1, 2, 47, and 94

95 = 1, 5, 19 and 95

96 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96

Hence, 90,91,92,93,94,95 and 96 are the composite numbers between 89 and 97.

(a) 21 = 5 + 5 + 11

Clearly, 5 and 11 are prime numbers.

(b) 31 = 5 + 7 + 19

Clearly, 3, 7 and 19 are prime numbers.

(c) 53 = 5 + 17 + 31

Clearly, 5, 17 and 31 are prime numbers.

(d) 61 = 13 + 17 + 31

Clearly, 13, 17 and 31 are prime numbers.

These are the pairs of prime numbers below 20 whose sum is divisible by 5;

2+3 = 5

3+7 = 10

7+13 = 20

17+3 = 20

19+11 = 30

(a) A number which has only two factors is called a Prime number

Explanation: A number which has only two whole number factors is called a prime number. It has only two whole number factors i.e. on the number itself and other Example: 2 is a prime number.

(b) A number which has more than two factors is called a Composite number

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

Example: 4 is a prime number.

(c) 1 is neither prime nor composite

Explanation: 1 is neither prime nor composite number. Because 1 has no other whole number factor other than itself. So, it's not prime and therefore not composite as well.

(d) The smallest prime number is 2

Explanation: The smallest prime number is 2 because, it has only two whole number factors, one the number itself and other 1.

(e) The smallest composite number is 4

Explanation: The smallest composite number is 4 because, it has only three whole number factors i.e. 1, 2 and 4.

(f) The smallest even number is 2

Explanation: The smallest even number is 2. Because an integer divisible by 2 is an even number.

In this question, we have to divide the numbers by the given numbers to see whether it is divisible or not.
To check this, we follow the divisibility rules:

For example-

990

= 990 ÷ 2 = 495 so yes it is divisible.

= 990 ÷ 3 = 330 so yes,

= 990 ÷ 4 = 247.5 which is not completely divisible so no its not divisible by 4,

= 990 ÷ 5 = 198 yes

= 990 ÷ 6 = 165 yes

= 990 ÷ 8 = 123.75 (N0)

= 990 ÷ 9 = 110 yes

= 990 ÷ 10 = 99 yes

= 990 ÷ 11 = 90 yes

(a) 572

As per the divisibility rule for 4: if the last two digits of a whole number are divisible by 4, then the entire number is divisible by 4.

The last two digits of the given number are 72, since 72 is divisible by 4, the number is also divisible by 4.

According to the rule of divisibility for 8: if the last three digit of a whole number are divisible by 8, then the entire number is divisible by 8.

The last three digits are 572, as 572 is not divisible by 8, the given number is also not divisible by 8.

(b) 726352

According to the rule of divisibility for 4 the last two digits of the given number are 52, since 52 is divisible by 4, the given number is also divisible by 4.

According to the rule of divisibility for 8 the last three digits are 352, as 352 is divisible by 8, the given number is also divisible by 8.

(c) 5500

According to the rule of divisibility for 4 the last two digits of the given number are 00, the number is divisible by 4.

According to the rule of divisibility for 8, the last three digits are 500, as 500 is not divisible by 8, the given number is also not divisible by 8.

(d) 6000

According to the rule of divisibility for 4 the last two digits of the given number are 00, the number is divisible by 4.

According to the rule of divisibility for 4 the last three digits are 000, the given number is divisible by 8.

(e) 12159

According to the rule of divisibility for 4 the last two digits of the given number are 59, since 59 is not divisible by 4, the number is also not divisible by 4.

According to the rule of divisibility for 8 the last three digits are 159, as 159 is not divisible by 8, the given number is also not divisible by 8.

(f) 14560

According to the rule of divisibility for 4 the last two digits of the given number are 60, since 60 is divisible by 4, the number is also divisible by 4.

According to the rule of divisibility for 8 the last three digits are 560, as 560 is divisible by 8, the given number is also divisible by 8.

(g) 21084

According to the rule of divisibility for 4 the last two digits of the given number are 84, since 84 is divisible by 4, the number is also divisible by 4.

According to the rule of divisibility for 8 the last three digits are 084, as 084 is not divisible by 8, the given number is also not divisible by 8.

(h) 31795072

According to the rule of divisibility for 4 the last two digits of the given number are 72, since 72 is divisible by 4, the number is also divisible by 4.

According to the rule of divisibility for 8 the last three digits are 072, as 072 is divisible by 8, the given number is also divisible by 8.

(i) 1700

According to the rule of divisibility for 4 the last two digits of the given number are 00, the number is also divisible by 4.

According to the rule of divisibility for 8 the last three digits are 700, as 700 is not divisible by 8, the given number is also not divisible by 8.

(j) 2150

According to the rule of divisibility for 4 the last two digits of the given number are 50, since 50 is not divisible by 4, the number is also not divisible by 4.

According to the rule of divisibility for 8 the last three digits are 150, as 150 is not divisible by 8, the given number is also not divisible by 8.

The prime factors of 6 are 2 and 3. So for the number to be divisible by 6, it must also be divisible by 2 and 3. Therefore, we need to check whether the number is divisible by 2 and then the sum of the digits should also be divisible by 3.

(a) 297144

Since the last digit of the number is 4, it is divisible by 2, so, the given number is divisible by 2.

On adding all the digits of the number, the sum obtained is 27 since 27 is divisible by 3, the given number is also divisible by 3.

As per the divisibility rule of 6 the number is divisible by both 2 and 3, so, it is divisible by 6.

(b) 1258

Since the last digit of the number is 8, it is divisible by 2, so, the given number is divisible by 2.

On adding all the digits of the number, the sum obtained is 16 since 16 is not divisible by 3, the given number is also not divisible by 3.

As per the divisibility rule of 6 the number should be divisible by 2 and 3 but the given number is not divisible by both 2 and 3, so, it’s not divisible by 6.

(c) 4335

Since the last digit of the number is 5, it is not divisible by 2, so, the given number is not divisible by 2

On adding all the digits of the number, the sum obtained is 15 since 15 is divisible by 3, the given number is also divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3 the numberbut the given number is not divisible by both 2 and 3, so, it is not divisible by 6.

(d) 61233

Since the last digit of the number is 3, it is not divisible by 2, so, the given number is not divisible by 2

On adding all the digits of the number, the sum obtained is 15 since 15 is divisible by 3, the given number is also divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3 but the given number is not divisible by both 2 and 3, so, it is not divisible by 6.

(e) 901352

Since the last digit of the number is 2, it is divisible by 2, so, the given number is divisible by 2

On adding all the digits of the number, the sum obtained is 20 since 20 is not divisible by 3, the given number is also not divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3 but the given number is not divisible by both 2 and 3, so, it is not divisible by 6.

(f) 438750

Since the last digit of the number is 0, it is divisible by 2, so, the given number is divisible by 2

On adding all the digits of the number, the sum obtained is 27 since 27 is divisible by 3, the given number is also divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3, the given number is divisible by both 2 and 3, so, it is divisible by 6.

(g) 1790184

Since the last digit of the number is 4, it is divisible by 2, so, the given number is divisible by 2

On adding all the digits number the sum obtained is 30 since 30 is divisible by 3, the given number is also divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3 and the given number is divisible by both 2 and 3, so, it is divisible by 6.

(h) 12583

Since the last digit of the number is 3, it is divisible by 2, so, the given number is not divisible by 2

On adding all the digits number the sum obtained is 19 since 19 is not divisible by 3, the given number is also not divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3 but the given number is not divisible by both 2 and 3, so, it is not divisible by 6.

(i) 639210

Since the last digit of the number is 0, it is divisible by 2, so, the given number is divisible by 2

On adding all the digits number the sum obtained is 21 since 21 is divisible by 3, the given number is also divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 & 3 and the given number is divisible by both 2 and 3, so, it is divisible by 6.

(j) 17852

Since the last digit of the number is 2, it is divisible by 2, so, the given number is divisible by 2

On adding all the digits number the sum obtained is 23, since 23 is not divisible by 3, the given number is also not divisible by 3.

As per the divisibility rule of 6 the number should be divisible by both 2 and 3 but the given number is not divisible by both 2 and 3, so, it is not divisible by 6.

Divisibility rule of 11 says that if the difference, of the sum of the digits at odd place and the sum of the digits at even place in the given number, is divisible by 11 then the number is also divisible by 11.

(a) 5445

Calculate the sum of the digits at odd places = 5+4 = 9

Calculate the sum of the digits at even places = 4+5 = 9

Difference = 9-9 = 0

As per the divisibility rule of 11 the difference between the sum of the digits at ood places and the sum of the digits at even places is 0, hence, 5445 is divisible by 11.

(b) 10824

Calculate the sum of the digits at odd places = 4+8+1 = 13

Calculate the sum of the digits at even places = 2+0 = 2

Difference = 13-2 = 11

As per the divisibility rule of 11 the difference between the sum of the digits at ood places and the sum of the digits at even places is 11, which is divisible by 11, therefore 10824 is divisible by 11.

(c) 7138965

Calculate the sum of the digits at odd places = 5+9+3+7 = 24

Calculate the sum of the digits at even places = 6+8+1 = 15

Difference = 24-15 = 9

As the difference between the sum of the digits at ood places and the sum of the digits at even places is 9, which is not divisible by 11, therefore, given number is also not divisible by 11.

(d) 70169308

Calculate the sum of the digits at odd places = 8+3+6+0 = 17

Calculate the sum of the digits at even places = 0+9+1+7 = 17

Difference = 17 – 17 = 0

As per the divisibility rule of 11 the difference between the sum of the digits at ood places and the sum of the digits at even places is 0, hence, 70169308 is divisible by 11.

(e) 10000001

Calculate the sum of the digits at odd places = 1

Calculate the sum of the digits at even places = 1

Difference = 1 – 1 = 0

As per the divisibility rule of 11 the difference between the sum of the digits at ood places and the sum of the digits at even places is 0, hence, 10000001 is divisible by 11.

(f) 901153

Calculate the sum of the digits at odd places = 3+1+0 = 4

Calculate the sum of the digits at even places = 5+1+9 = 15

Difference = 15 – 4 = 11

As per the divisibility rule of 11 the difference between the sum of the digits at ood places and the sum of the digits at even places is 11, which is divible by 11, therefore, 901153 is divisible by 11.

(a) __6724

Add the remaining digits = 6+7+2+4 = 19

To make the number divisible by 3, the sum of its digits should be divisible by 3.

The smallest multiplier of 3 which comes after 19 is 21.

Therefore, smallest number = 21 – 19 = 2

Therefore, the largest number is 8.

(b) 4765__2

Add the remaining digits = 4+7+6+5+2 = 24

To make the number divisible by 3, the sum of its digits should be divisible by 3. As we can see 24 is already divisible by 3,
the smallest number that can be placed here is 0.

if we add 9 then the digit is divisible by 3 ( 24+9=33)
9 is the largest digit

4765 ___2 =4+ 7+6+5+9+2= 33

Therefore, the largest number is 9.

(a) Let suppose missing digit = x

Calculate the sum of the digits at odd places = 9 + 3 + 2 = 14

Calculate the sum of the digits at even places = 8+ x + 9 = 17 + x

Difference = 17 + x – 14 = 3 + x

For a number to be divisible by 11, this difference should be 0 or a multiplier of 11.

If 3 + x = 0, then

x = -3

But the number cannot be negative.

The closest multiplier of 11, which is near to 3 is taken. This is 11 itself.

3 + x = 11

x = 8

Therefore, the required digit is 8.

(b) Let suppose missing digit = x

Calculate the sum of the digits at odd places = 4 + 4 + x = 8 + x

Calculate the sum of the digits at even places = 8 + 9 + 8 = 25

Difference = 25 – (8 + x) = 17 – x

For a number to be divisible by 11, this difference should be 0 or a multiplier of 11.

If 17 - x = 0, then

x = 17

But this is not possible.

The closest multiplier of 11 is taken, taking 11 itself we get,

17 - x = 11

x = 6

Therefore, the required digit is 6.

When we find the factos of tow or more numbers, then some factors may be found in both the numbers, these are called common factors.

(a) 20 and 28

Factors of 20 = 1, 2, 4, 5, 10 and 20

Factors of 28 = 1, 2, 4, 7, 14 and 28

Common factors = 1, 2, 4

(b) 15 and 25

Factors of 15 = 1, 3, 5, 15

Factors of 25 = 1, 5, 25

Common factors = 1 and 5

(c) 35 and 50

Factors of 35 = 1, 5, 7, 35

Factors of 50 = 1, 2, 5, 10, 25, 50

Common factors = 1 and 5

(d) 56 and 120

Factors of 56 = 1, 2, 4, 7, 8, 14, 28 and 56

Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120

Common factors = 1, 2, 4 and 8

(a) 4, 8 and 12

Factors of 4 = 1, 2 and 4

Factors of 8 = 1, 2, 4 and 8

Factors of 12 = 1, 2, 3, 4, 6 and 12

Comman factors = 1, 2, and 4

(b) 5, 15 and 25

Factors of 5 = 1, 5

Factors of 15 = 1, 3, 5 and 15

Factors of 25 = 1, 5 and 25

Common factors = 1 and 5.

(a) 6 and 8

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

Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80

So,

First three common multiples are = 24, 48 and 72

(b) 12 and 18

Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

Multiples of 18 = 18, 36, 54, 72, 90, 108, 126

First three common multiples = 36, 72 and 108

Multiples of 3 less than 100 are;

= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 81, 84, 87, 90, 93, 96, 99.

Multiples of 4 less than 100 are;

= 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96.

Common multiples are;

= 12, 24, 36, 48, 60, 72, 84 and 96

Two numbers are said to be co-prime if they don't have any common factor other than 1

(a) 18 and 35

Factors of 18 = 1, 2, 3, 6, 9, and 18

Factors of 35 = 1, 5, 7 and 35

Common factor = 1

Therefore, the given two numbers are co-prime.

(b) 15 and 37

Factors of 15 = 1, 3, 5, and 15

Factors of 37 = 1, and 37

Common factor = 1

Therefore, the given two numbers are co-prime.

(c) 30 and 415

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, and 30

Factors of 415 = 1, 5, 83 and 415

Common factor = 1 and 5

These two numbers have a factor other than 1, therefore, the given two numbers are not co-prime.

(d) 17 and 68

Factors of 17 = 1, and 17

Factors of 68 = 1, 2, 4, 17, 34 and 68

Common factor = 1, and 17

These two numbers have a factor other than 1, therefore, the given two numbers are not co-prime.

(e) 216 and 215

Factors of 216 = 1, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216

Factors of 215 = 1, 5, 43 and 215

Common factor = 1

Therefore, the given two numbers are co-prime.

(f) 81 and 16

Factors of 81 = 1, 3, 9, 27 and 81

Factors of 16 = 1, 2, 4, 8 and 16

Common factor = 1

Therefore, the given two numbers are co-prime.

As the number is divisible by 12,

So, it will also be divisible by its factors;

That is,

1, 2, 3, 4, 6 and 12.

Therefore,

1, 2, 3, 4 and 6 are the numbers other than 12 by which this number is also divisible.

(a) False.

For example – 6 is divisible by 3, but it is not divisible by 9.

(b) True.

Example- 18, is divisible by 3 as well as 9 so, yes, If a number is divisible by 9, it must be divisible by 3.

(c) False.

If a number is divisible by 3 and 6, it is not neccessory that it is also divisible by 18..

Example – 42 it is divisible by 6 and 3 but not by 18.

(d) True.

As 9 × 10 = 90 So, if a number is divisible by 9 and 10 both, then it must be divisible by 90.

(e) False.

Example – 15 and 32 are co-prime and composite.

(f) False.

Example – 20 is divisible by 4 but not by 8. So, its not neccessory that all numbers which are divisible by 4 must also be divisible by 8.

(g) True.

As we know 8 = 2 × 4 so all numbers which are divisible by 8 must also be divisible by 2 and 4.

(h) True.

Example – 3 divides 6 and 9 separatly and 15, which is the sum of 9 and 6. So, if a number exactly divides two numbers separately, it must exactly divide their sum.

(i) False.

Example- 2 divide 12 but it doesn’t divide 3 and 9. So, if a number exactly divides the sum of two numbers, its not neccessory that it also divide the two numbers separately.

(a) Lets take ‘a’ be the missing number,

So, we have,

2 × a = 6

a = 6/2 = 3

a = 3

So,

2 × 3 = 6

And

5 × a = 10

a = 10/5 = 2

5 × 2 = 10

Missing numbers are 3 and 2.

(b) As we know 60 = 30 × 2

30 = 10 × 3

10 = 5 × 2

Missing numbers are, 2, 5, 3 and 2.

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

Greatest four digit number = 9999

Let’s draw the factor tree-

So the prime factors are;

9999 = 3 × 3 × 11 × 101

Smallest 5 digit number = 10,000

Draw the factor tree,

So the factors are;

10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Calculate the factors first;

So we have;

1729 = 7 × 13 × 19

13 – 7 = 6

19 – 13 = 6

Difference of two consecutive prime factors is 6.

Let’s take,

2 × 3 × 4 = 24, this is divisible by 6

9 × 10 × 11 = 990, this is divisible by 6

20 × 21 × 22 = 9240, this is divisible by 6

31 × 32 × 33 = 32, 736, this is divisible by 6

So, the product of three consecutive numbers is always divisible by 6.

5 + 7 = 12, it is divisible by 4,

15 + 17 = 32, divisible by 4,

17 + 19 = 36, divisible by 4,

21 + 23 = 44, divisible by 4.

Therefore, the sum of two consecutive odd numbers is divisible by 4.

Prime factorization is finding the factors of a number that are all prime.

(a) 24 = 2 × 3 × 4

As we can see in this factorization 4 is a composit number, so prime factorization has not been done.

(b) 56 = 7 × 2 × 2 × 2

Since all the factors are prime factors so, prime factorization has been done.

(c) 70 = 2 × 5 × 7

Since all the factors are prime factors so, prime factorization has been done.

(d) 54 = 2 × 3 × 9

As we can see in this factorization 9 is a composit number, so prime factorization has not been done.

According to the divisibility rule of 45 if the last digit of a number is divisible by 5 and the sum of the digits is divisible by 9, then the number is divisible by 45.

45 = 5 × 9

Factors of 5 = 1 and 5

Factors of 9 = 1, 3 and 9

Therefore, 5 and 9 are co-prime numbers,

As the last digit of the number is 0 so it is divisible by 5.

Sum of the all digits = 2+5+1+1+0 = 9

As the sum of the digits is divisible by 9 so this number is also divisible by 9.

So, the number is divisible by 5 and 9, it is also divisible by 45.

No, it is not necessary because 12 and 36 are divisible by 4 and 6 both but are not divisible by 24.

As it is the smallest number, it will be the product of 4 smallest prime numbers;

2 × 3 × 5 × 7 = 210.

HCF – Highest common factor

The highest common factor of two or more numbers is the largest number that divides saparetly both the numbers.

(a) 18 and 48,

18 = 2 × 3 × 3

48 = 2 × 2 × 2 × 2 × 3

HCF = 2 × 3 = 6

(b) 30, 42

30 = 2 × 3 × 5

42 = 2 × 3 × 7

HCF = 2 × 3 = 6

(c) 18, 60

18 = 2 × 3 × 3

60 = 2 × 2 × 3 × 5

HCF = 2 × 3 =6

(d) 27, 63

27 = 3 × 3 × 3

63 = 3 × 3 × 7

HCF = 3 × 3 = 9

(e) 36, 84

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

HCF = 2 × 2 × 3 = 12

(f) 34, 102

34 = 2 × 17

102 = 2 × 3 × 17

HCF = 2 × 17 = 34

(g) 70, 105, 175

70 = 2 × 5 × 7

105 = 3 × 5 × 7

175 = 5 × 5 × 7

HCF = 5 × 7 = 35

(h) 91, 112, 49

91 = 7 × 13

112 = 2 × 2 × 2 × 2 × 7

49 = 7 × 7

HCF = 7

(i) 18, 54, 81

18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

HCF = 3 × 3 = 9

(j) 12, 45, 75

12 = 2 × 2 × 3

45 = 3 × 3 × 5

HCF = 3

(a) Consecutive numbers

Example- 2 and 3

HCF = 1

(b) Even number

Example – 2 and 4

HCF = 2

(c) Odd numbers

Example – 3 and 5

HCF = 1

No, the answer is not correct. 1 will be the correct HCF.

Explanation: The factorization of 4 and 15 is given as:

4: 1 × 2 × 2

15: 1 × 3 × 5

Here, the highest common factor is 1.

Weight of the two bags = 75 kg and 69 kg

Maximum weight = HCF (75, 69)

75 = 3 × 5 × 5

69 = 3 × 23

HCF = 3

Therefore, the maximum value of weight, which can measure the weight of the fertilizer exact number of time, is 3 kg.

Step measure of first boy = 63 cm

Step measure of second boy = 70 cm

Step measure of third boy = 77cm

Now calculate the LCM of 63, 70 and 77

LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930

Therefore,

The maximum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

Length of the room = 825 cm

Breadth of the room = 675 cm

Height of the room = 450 cm

Longest tape = HCF of 825, 675 and 450

825 = 3 × 5 × 5 × 11

675 = 3 × 3 × 3 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

HCF = Product of common factors = 3 × 5 × 5 = 75 cm

Therefore, the longest tape = 75 cm

Smallest 3-digit number = LCM of 6, 8 and 12

LCM = 2 × 2 × 2 × 3 = 24

LCM of these 3 numbers is 24 so to find out 3-digit number we have to calculate the multiples of 24,

24 × 1 = 24

24 × 2 = 48

24 × 3 = 72

24 × 4 = 96

24 × 5 = 120

24 × 6 = 144

It can be seen that, 120 is the smallest 3-digit multiple of 24 so, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

Let’s take the LCM of 8, 10 and 12

LCM = 2 × 2 × 2 × 3 × 5 = 120

Now to get greatest 3-digit number we have to calculate the multiples of 120

120 × 1 = 120

120 × 2 = 240

120 × 3 = 360

120 × 4 = 480

120 × 5 = 600

120 × 6 = 720

120 × 7 = 840

120 × 8 = 960

120 × 9 = 1080

120 × 10 = 1200

Hence the greatest 3-digit number exactly divisible by 8, 10 and 12 is 960.

Time-period after which these lights will change = LCM of 48, 72 and 108

LCM is the lowest common multiple. LCM of two numbers It is the smallest number that is divisible by both of them.

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

The lights will change together after every 432 seconds.

60 seconds = 1 minute

432 seconds = 432/60 min



= 7 minutes and 12 seconds

Total time will be 7 am + 7 min + 12 sec.

Hence these lights will change simultaneously again at 7:07:12 am.

Maximum capacity of the required tanker = HCF of 403, 434 and 465

403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

HCF = 31

So, the container of capacity 31 litres can measure the diesel of 3 containers exact number of times.

To find the required number we have to calculate LCM of 6, 15 and 18 because LCM is the least number divided by all 6, 15 and 18

LCM = 2 × 3 × 3 × 5 = 90

Since the number required always leaves remainder 5

Required number = 90+5 = 95

LCM 0f 18, 24 and 32

LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

It can be observed that

288 × 3 = 864 and

288 × 4 = 1152

Therefore, the smallest 4-digit number which is divisible by 18, 24 and 32 is 1152.

LCM – Least common multiple

The LCM of two numbers is the smallest number that is a multiple of both the numbers.

(a) 9 and 4

LCM = 2 × 2 × 3 × 3 = 36

(b) 12 and 5

LCM = 2 × 2 × 3 × 5 = 60

(c) 6 and 5

LCM = 2 × 3 × 5 = 30

(d) 15 and 4

LCM = 2 × 2 × 3 × 5 = 60

Yes, it can be seen that in each case, LCM of given numbers is the product of thses numbers. When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.

(a) 5, 20

LCM = 2 × 2 × 5 = 20

(b) 6, 18

LCM = 2 × 3 × 3 = 18

(c) 12, 48

LCM = 2 × 2 × 2 × 2 × 3 = 48

(d) 9, 45

LCM = 3 × 3 × 5 = 45

Yes, it can be seen that in each case the LCM of the given number is the larger number. When one number is a factor of the other, their LCM will be the larger number.