# Class 6th Chapter 3: Playing with Numbers – HCF, LCM, Factors, and More

June 17, 2023
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## Exercise 3.1

### 1. Write all the factors of the following numbers :

(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36

Ans.

(a) Factors of 24:

1. Divide 24 by 1: 24 ÷ 1 = 24
2. Divide 24 by 2: 24 ÷ 2 = 12
3. Divide 24 by 3: 24 ÷ 3 = 8
4. Divide 24 by 4: 24 ÷ 4 = 6
5. Divide 24 by 6: 24 ÷ 6 = 4
6. Divide 24 by 8: 24 ÷ 8 = 3
7. Divide 24 by 12: 24 ÷ 12 = 2
8. Divide 24 by 24: 24 ÷ 24 = 1

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

(b) Factors of 15:

1. Divide 15 by 1: 15 ÷ 1 = 15
2. Divide 15 by 3: 15 ÷ 3 = 5
3. Divide 15 by 5: 15 ÷ 5 = 3

The factors of 15 are 1, 3, 5, 15.

(c) Factors of 21:

1. Divide 21 by 1: 21 ÷ 1 = 21
2. Divide 21 by 3: 21 ÷ 3 = 7
3. Divide 21 by 7: 21 ÷ 7 = 3

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

(d) Factors of 27:

1. Divide 27 by 1: 27 ÷ 1 = 27
2. Divide 27 by 3: 27 ÷ 3 = 9
3. Divide 27 by 9: 27 ÷ 9 = 3

The factors of 27 are 1, 3, 9, 27.

(e) Factors of 12:

1. Divide 12 by 1: 12 ÷ 1 = 12
2. Divide 12 by 2: 12 ÷ 2 = 6
3. Divide 12 by 3: 12 ÷ 3 = 4
4. Divide 12 by 4: 12 ÷ 4 = 3

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

(f) Factors of 20:

1. Divide 20 by 1: 20 ÷ 1 = 20
2. Divide 20 by 2: 20 ÷ 2 = 10
3. Divide 20 by 4: 20 ÷ 4 = 5
4. Divide 20 by 5: 20 ÷ 5 = 4

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

(g) Factors of 18:

1. Divide 18 by 1: 18 ÷ 1 = 18
2. Divide 18 by 2: 18 ÷ 2 = 9
3. Divide 18 by 3: 18 ÷ 3 = 6
4. Divide 18 by 6: 18 ÷ 6 = 3

The factors of 18 are 1, 2, 3, 6, 9, 18.

(h) Factors of 23:

1. Divide 23 by 1: 23 ÷ 1 = 23
2. Divide 23 by 23: 23 ÷ 23 = 1

The factors of 23 are 1, 23.

(i) Factors of 36:

1. Divide 36 by 1: 36 ÷ 1 = 36
2. Divide 36 by 2: 36 ÷ 2 = 18
3. Divide 36 by 3: 36 ÷ 3 = 12
4. Divide 36 by 4: 36 ÷ 4 = 9
5. Divide 36 by 6: 36 ÷ 6 = 6

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

### 2. Write the first five multiples of :

(a) 5
(b) 8
(c) 9

Ans.

Here are the first five multiples of the given numbers:

(a) Multiples of 5:

1. 5 × 1 = 5
2. 5 × 2 = 10
3. 5 × 3 = 15
4. 5 × 4 = 20
5. 5 × 5 = 25

The first five multiples of 5 are 5, 10, 15, 20, 25.

(b) Multiples of 8:

1. 8 × 1 = 8
2. 8 × 2 = 16
3. 8 × 3 = 24
4. 8 × 4 = 32
5. 8 × 5 = 40

The first five multiples of 8 are 8, 16, 24, 32, 40.

(c) Multiples of 9:

1. 9 × 1 = 9
2. 9 × 2 = 18
3. 9 × 3 = 27
4. 9 × 4 = 36
5. 9 × 5 = 45

The first five multiples of 9 are 9, 18, 27, 36, 45.

Ans.

### 4. Find all the multiples of 9 up to 100.

Ans. To find all the multiples of 9 up to 100 using multiplication, follow these steps:

2. Multiply 9 by 1 to get the first multiple.
3. Continue multiplying 9 by the next whole numbers (2, 3, 4, and so on) to find subsequent multiples.
4. Stop when you reach a multiple that is equal to or exceeds 100.

Using multiplication, we can find the multiples of 9 up to 100:

9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99

Therefore, these are all multiples of 9 up to 100.

## Exercise 3.2

### 1. What is the sum of any two (a) Odd numbers? (b) Even numbers?

Ans.

(a) The sum of any two odd numbers is always an even number. When you add two odd numbers, you are essentially adding an odd quantity (1) to another odd quantity (1). The result is an even number. For example, 3 + 5 = 8, 7 + 9 = 16, and so on.

(b) The sum of any two even numbers is also an even number. When you add two even numbers, you are adding an even quantity (2) to another even quantity (2). The result is always an even number. For example, 4 + 6 = 10, 12 + 8 = 20, and so on.

In both cases, the sum of the two numbers follows a pattern and retains the property of the numbers being added (odd or even).

### 2. State whether the following statements are True or False:

(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.

Ans.

(a) False.
The sum of three odd numbers is always odd.

(b) True.
The sum of two odd numbers and one even number is always even.

(c) True.
The product of three odd numbers is always odd.

(d) False.
If an even number is divided by 2, the quotient is always an even number.

(e) False.
2 is an even prime number, so not all prime numbers are odd.

(f) True.
Prime numbers only have two factors, 1 and themselves.

(g) False.
The sum of two prime numbers can be either even or odd.

(h) True.
2 is the only even prime number.

(i) False.
Not all even numbers are composite numbers. 2 is a prime number.

(j) True.
The product of two even numbers is always even

### 3. The numbers 13 and 31 are prime numbers. Both these numbers have the same digits 1 and 3. Find such pairs of prime numbers up to 100.

Ans.

Here are the pairs of prime numbers up to 100 that have the same digits:

(13, 31) (17, 71) (37, 73) (79, 97)

In each pair, the numbers have the same digits but in different orders. These pairs consist of prime numbers that are permutations of each other.

### 4. Write down separately the prime and composite numbers less than 20.

Ans.

Prime numbers less than 20:
2, 3, 5, 7, 11, 13, 17, 19.

Composite numbers less than 20:
4, 6, 8, 9, 10, 12, 14, 15, 16, 18.

### 5. What is the greatest prime number between 1 and 10?

Ans. The greatest prime number between 1 and 10 is 7.

### 6. Express the following as the sum of two odd primes.

(a) 44
(b) 36
(c) 24
(d) 18

Ans.

(a) 44 cannot be expressed as the sum of two odd primes.
(b) 36 = 17 + 19
(c) 24 cannot be expressed as the sum of two odd primes.
(d) 18 cannot be expressed as the sum of two odd primes.

### 7. Give three pairs of prime numbers whose difference is 2.

[Remark: Two prime numbers whose difference is 2 are called twin primes].

Ans.

1. (3, 5):
The numbers 3 and 5 are both prime numbers, and their difference is 5 – 3 = 2. Therefore, they form a pair of twin primes.
2. (5, 7):
Similarly, the numbers 5 and 7 are both prime numbers, and their difference is 7 – 5 = 2. Hence, they are also twin primes.
3. (11, 13):
In this case, 11 and 13 are both prime numbers, and their difference is 13 – 11 = 2. Therefore, they form another pair of twin primes.

### 8. Which of the following numbers is prime?

(a) 23
(b) 51
(c) 37
(d) 26

Ans.

The prime numbers among the given options are:

(a) 23: Yes, 23 is a prime number because it is only divisible by 1 and itself.
(c) 37: Yes, 37 is also a prime number as it has no divisors other than 1 and itself.

The composite numbers among the given options are:

(b) 51: No, 51 is not a prime number because it has divisors other than 1 and itself (e.g., 3 and 17).
(d) 26: No, 26 is not a prime number as it is divisible by numbers other than 1 and itself (e.g., 2 and 13).

Therefore, the prime numbers in the given options are 23 and 37.

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

Ans.

The following are seven consecutive composite numbers less than 100 without any prime number between them:

91, 92, 93, 94, 95, 96, 97

These numbers are all composite because they have divisors other than 1 and themselves. There are no prime numbers between these consecutive numbers, as all of them are composite.

### 10. Express each of the following numbers as the sum of three odd primes:

(a) 21
(b) 31
(c) 53
(d) 61

Ans.

(a) To express 21 as the sum of three odd primes, we can choose the primes 3, 7, and 11: 21 = 3 + 7 + 11

(b) To express 31 as the sum of three odd primes, we can choose the primes 3, 11, and 17: 31 = 3 + 11 + 17

(c) To express 53 as the sum of three odd primes, we can choose the primes 3, 11, and 39: 53 = 3 + 11 + 39

(d) To express 61 as the sum of three odd primes, we can choose the primes 3, 19, and 39: 61 = 3 + 19 + 39

### 11. Write five pairs of prime numbers less than 20 whose sum is divisible by 5.

(Hint : 3+7 = 10)

Ans.

Here are five pairs of prime numbers less than 20 whose sum is divisible by 5:

1. 3 + 7 = 10
2. 7 + 13 = 20
3. 7 + 17 = 24
4. 13 + 17 = 30
5. 2 + 19 = 21

In each of these pairs, the sum of the primes is divisible by 5.

### 12. Fill in the blanks :

(a) A number that has only two factors is called a _______ .
(b) A number that has more than two factors is called a _______ .
(c) 1 is neither _______ nor _______ .
(d) The smallest prime number is _______ .
(e) The smallest composite number is _______ .
(f) The smallest even number is _______.

Ans.

(a) A number that has only two factors is called a prime number.
(b) A number that has more than two factors is called a composite number.
(c) 1 is neither prime nor composite.
(d) The smallest prime number is 2.
(e) The smallest composite number is 4.
(f) The smallest even number is 2.

## Exercise 3.3

Ans.

(a) 572
(b) 726352
(c) 5500
(d) 6000
(e) 12159
(f) 14560
(g) 21084
(h) 31795072
(i) 1700
(j) 2150

Ans.

### 3. Using divisibility tests, determine which of following numbers are divisible by 6:

(a) 297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 639210
(j) 17852

Ans.

Here is the divisibility analysis for each number:

(a) 297144:
To check divisibility by 6, we need to determine if the number is divisible by both 2 and 3. The number ends with an even digit, so it is divisible by 2. For divisibility by 3, we add the digits: 2 + 9 + 7 + 1 + 4 + 4 = 27, which is divisible by 3. Therefore, 297144 is divisible by 6.

(b) 1258:
The number does not end with an even digit, so it is not divisible by 2. When we sum the digits, 1 + 2 + 5 + 8 = 16, which is not divisible by 3. Thus, 1258 is not divisible by 6.

(c) 4335:
The number ends with an odd digit, so it is not divisible by 2. Summing the digits gives us 4 + 3 + 3 + 5 = 15, which is divisible by 3. Therefore, 4335 is divisible by 6.

(d) 61233: The number ends with an odd digit, so it is not divisible by 2. The sum of the digits is 6 + 1 + 2 + 3 + 3 = 15, which is divisible by 3. Hence, 61233 is divisible by 6.

(e) 901352:
The number ends with an even digit, so it is divisible by 2. Summing the digits, we get 9 + 0 + 1 + 3 + 5 + 2 = 20, which is not divisible by 3. Thus, 901352 is not divisible by 6.

(f) 438750:
The number ends with an even digit, so it is divisible by 2. Adding the digits gives us 4 + 3 + 8 + 7 + 5 + 0 = 27, which is divisible by 3. Therefore, 438750 is divisible by 6.

(g) 1790184:
The number ends with an even digit, so it is divisible by 2. Summing the digits, we get 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30, which is divisible by 3. Hence, 1790184 is divisible by 6.

(h) 12583:
The number does not end with an even digit, so it is not divisible by 2. When we sum the digits, 1 + 2 + 5 + 8 + 3 = 19, which is not divisible by 3. Thus, 12583 is not divisible by 6.

(i) 639210:
The number ends with an even digit, so it is divisible by 2. Adding the digits gives us 6 + 3 + 9 + 2 + 1 + 0 = 21, which is divisible by 3. Therefore, 639210 is divisible by 6.

(j) 17852:
The number ends with an even digit, so it is divisible by 2. Summing the digits, we get 1 + 7 + 8 + 5 + 2 = 23, which is not divisible by 3. Thus, 17852 is not divisible by 6.

Based on the divisibility tests, the numbers that are divisible by 6 are 297144, 4335, 61233, 438750, and 639210.

### 4. Using divisibility tests, determine which of the following numbers are divisible by 11:

(a) 5445
(b) 10824
(c) 7138965
(d) 70169308
(e) 10000001
(f) 901153

Ans.

Here is the divisibility analysis for each number with an explanation:

(a) 5445:
To determine if a number is divisible by 11, we can use the alternating sum of its digits. In this case, (4 + 5) – (5 + 4) = 0, which is divisible by 11. Therefore, 5445 is divisible by 11.

(b) 10824:
To check divisibility by 11, we use the alternating sum of the digits. For 10824, 1 – 0 + 8 – 2 + 4 = 11, which is divisible by 11. Hence, 10824 is divisible by 11.

(c) 7138965:
Applying the alternating sum test, we get 7 – 1 + 3 – 8 + 9 – 6 + 5 = 9, which is not divisible by 11. Therefore, 7138965 is not divisible by 11.

(d) 70169308:
Using the alternating sum, 7 – 0 + 1 – 6 + 9 – 3 + 0 – 8 = 0, which is divisible by 11. Hence, 70169308 is divisible by 11.

(e) 10000001:
The alternating sum of the digits, 1 – 0 + 0 – 0 + 0 – 0 + 0 – 1 = 0, is divisible by 11. Thus, 10000001 is divisible by 11.

(f) 901153:
Applying the alternating sum test, we get 9 – 0 + 1 – 1 + 5 – 3 = 11, which is divisible by 11. Therefore, 901153 is divisible by 11.

Based on the divisibility tests using the alternating sum of digits, we can determine the divisibility of each number by 11.

### 5. Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 :

(a) _ 6724
(b) 4765 _ 2

Ans.

To make each number divisible by 3, we need to determine the smallest and greatest digits that, when added to the given numbers, result in a multiple of 3.

(a) _ 6724: The sum of the digits in this number is 6 + 7 + 2 + 4 = 19. To make it divisible by 3, we need to find the smallest digit that, when added to 19, results in a multiple of 3. The smallest digit that satisfies this condition is 2. The greatest digit that we can choose is 8. Therefore, the number formed is 86724, which is divisible by 3.

(b) 4765 _ 2: The sum of the digits in this number is 4 + 7 + 6 + 5 + 2 = 24. To make it divisible by 3, we need to find the smallest digit that, when added to 24, results in a multiple of 3. The smallest digit that satisfies this condition is 0. The greatest digit that we can choose is 9. Therefore, the number formed is 4765092, which is divisible by 3.

### 6. Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 :

(a) 92 _ 389
(b) 8 _ 9484

Ans.

To make each number divisible by 11, we need to determine the digit that, when placed in the blank space, results in a multiple of 11.

(a) 92 _ 389:

In the number 92 _ 389, the sum of the digits in the odd places is 2 + 3 + 9 = 14, and the sum of the digits in the even places is 9 + _ + 8.

To make the overall sum divisible by 11, we need to find a digit that satisfies the equation 14 + _ – (9 + 8) ≡ 0 (mod 11).

Simplifying the equation, we have 14 + _ – 17 ≡ 0 (mod 11), which can be further simplified to _ ≡ 3 (mod 11).

To find the value of the blank, we need to find a digit that, when added to 14, gives a result congruent to 3 modulo 11. The digit that satisfies this condition is 8.

Therefore, the correct number is 928389, where the digit 8 is placed in the blank space, and it is divisible by 11.

(b) 8 _ 9484:

In the number 8 _ 9484, the sum of the digits in the odd places is _ + 4 + 4, and the sum of the digits in the even places is 8 + 9 + 8.

To make the overall sum divisible by 11, we need to find a digit that satisfies the equation (_ + 4 + 4) – (8 + 9 + 8) = 11.

Simplifying the equation, we have _ + 8 – 25 = 11, which can be further simplified to _ = 17 = 11

From the equation, _ = 17 – 11 = 6, so the missing digit is 6, 869484.

Let’s test it with the alternate sum 8 – 6 + 9 – 4 + 8 – 4 = (8+9+8) – (6+4+4) = 25 – 14 = 11, 0 or 11 outcomes would be divisible by 11

## Exercise 3.4

### 1. Find the common factors of :

(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120

Ans.

To find the common factors of two numbers, we can list the factors of each number and identify the factors that they have in common. Here’s how we can collect the factors for each pair of numbers:

(a) 20 and 28:
Factors of 20: 1, 2, 4, 5, 10, 20
Factors of 28: 1, 2, 4, 7, 14, 28

Common factors: 1, 2, 4

(b) 15 and 25:
Factors of 15: 1, 3, 5, 15
Factors of 25: 1, 5, 25

Common factor: 5

(c) 35 and 50:
Factors of 35: 1, 5, 7, 35
Factors of 50: 1, 2, 5, 10, 25, 50

Common factor: 5

(d) 56 and 120:
Factors of 56: 1, 2, 4, 7, 8, 14, 28, 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, 8

### 2. Find the common factors of :

(a) 4, 8 and 12
(b) 5, 15 and 25

Ans.

To find the common factors of multiple numbers, we can list the factors of each number and identify the factors that they have in common. Here’s how we can collect the factors for each set of numbers:

(a) 4, 8, and 12:
Factors of 4: 1, 2, 4
Factors of 8: 1, 2, 4, 8
Factors of 12: 1, 2, 3, 4, 6, 12

Common factors: 1, 2, 4

(b) 5, 15, and 25:
Factors of 5: 1, 5
Factors of 15: 1, 3, 5, 15
Factors of 25: 1, 5, 25

Common factor: 1, 5

To collect the common factors, we determine the factors of each number and find the numbers that divide all of the given numbers without leaving any remainder. The common factors are the numbers that appear in the factors of all the given numbers.

### 3. Find the first three common multiples of :

(a) 6 and 8
(b) 12 and 18

Ans.

To find the common multiples of two numbers, we can list the multiples of each number and identify the multiples that they have in common. Here are the first three common multiples for each set of numbers:

(a) 6 and 8:
Multiples of 6: 6, 12, 18, 24, 30, 36, …
Multiples of 8: 8, 16, 24, 32, 40, 48, …

Common multiples: 24, 48, 72

(b) 12 and 18:
Multiples of 12: 12, 24, 36, 48, 60, 72, …
Multiples of 18: 18, 36, 54, 72, 90, 108, …

Common multiples: 36, 72, 108

To find the common multiples, we determine the multiples of each number and find the numbers that appear in the multiples of both numbers. The common multiples are the numbers that are divisible by both of the given numbers.

### 4. Write all the numbers less than 100 which are common multiples of 3 and 4.

Ans.

To find the numbers that are common multiples of 3 and 4, we can list the multiples of each number and identify the numbers that they have in common. Here are the numbers less than 100 that are common multiples of 3 and 4:

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

These numbers are obtained by finding the multiples of both 3 and 4 and identifying the numbers that appear in the multiples of both numbers. These numbers are divisible by both 3 and 4.

### 5. Which of the following numbers is co-prime?

(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16

Ans.

Let’s list the factors of each number and identify the co-prime numbers:

(a) Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 35: 1, 5, 7, 35

Co-prime numbers have no common factors other than 1. Looking at the factors, we can see that 18 and 35 do not have any common factors other than 1. Therefore, the pair (18, 35) is co-prime.

(b) Factors of 15: 1, 3, 5, 15
Factors of 37: 1, 37

Again, 15 and 37 do not have any common factors other than 1. So, the pair (15, 37) is co-prime.

(c) Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
Factors of 415: 1, 5, 83, 415

In this case, 30 and 415 have a common factor of 5. Therefore, the pair (30, 415) is not co-prime.

(d) Factors of 17: 1, 17
Factors of 68: 1, 2, 4, 17, 34, 68

The numbers 17 and 68 only share a common factor of 1. So, the pair (17, 68) is co-prime.

(e) Factors of 216: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216
Factors of 215: 1, 5, 43, 215

216 and 215 have a common factor of 1. Therefore, the pair (216, 215) is not co-prime.

(f) Factors of 81: 1, 3, 9, 27, 81
Factors of 16: 1, 2, 4, 8, 16

81 and 16 have a common factor of 1. So, the pair (81, 16) is not co-prime.

To summarize, the co-prime pairs among the given options are: (a) 18 and 35 (b) 15 and 37 (d) 17 and 68

### 6. A number is divisible by both 5 and 12. By which other numbers will that number be always divisible?

Ans.

If a number is divisible by both 5 and 12, it means it is a common multiple of 5 and 12. To determine the numbers by which this number will always be divisible, we need to find the common multiples of 5 and 12.

The common multiples of 5 and 12 can be obtained by finding the multiples of their least common multiple (LCM).

The LCM of 5 and 12 is 60.

Therefore, a number that is divisible by both 5 and 12 will always be divisible by the following numbers: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.

So, any number divisible by both 5 and 12 will also be divisible by these numbers.

### 7. A number is divisible by 12. By what other numbers will that number be divisible?

Ans.

If a number is divisible by 12, it means it is a multiple of 12. To determine the numbers by which this number will be divisible, we need to find its factors.

The factors of 12 are 1, 2, 3, 4, 6, and 12. Therefore, any number divisible by 12 will also be divisible by these numbers.

In addition to these factors, the number will also be divisible by other numbers that are multiples of 12, such as 24, 36, 48, and so on.

So, a number divisible by 12 will be divisible by 1, 2, 3, 4, 6, 12, and any other multiple of 12.

## Exercise 3.5

### 1. Which of the following statements are true?

(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18 if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.

Ans.

Incorrect Statements

(a) This statement is false. A number can be divisible by 3 without being divisible by 9.
For example, 6 is divisible by 3 but not by 9.

(d) This statement is false. A number can be divisible by 9 and 10 but not divisible by 90.
For example, 180 is divisible by 9 and 10, but not divisible by 90.

(f) This statement is false. Numbers divisible by 4 are not necessarily divisible by 8.
For example, 12 is divisible by 4 but not divisible by 8.

Correct Statements

(b) If a number is divisible by 9, it must be divisible by 3. Explanation: This statement is true. Since 9 is a multiple of 3, any number divisible by 9 will also be divisible by 3.

(c) A number is divisible by 18 if it is divisible by both 3 and 6. Explanation: This statement is true. Since 18 is a multiple of both 3 and 6, any number divisible by both 3 and 6 will also be divisible by 18.

(e) If two numbers are co-primes, at least one of them must be prime. Explanation: This statement is true. Co-prime numbers are numbers that have no common factors other than 1. If two numbers have no common factors, it means they cannot share any prime factors as well. Therefore, at least one of the co-prime numbers must be prime.

(g) All numbers which are divisible by 8 must also be divisible by 4. Explanation: This statement is true. Since 8 is a multiple of 4, any number divisible by 8 will also be divisible by 4.

(h) If a number exactly divides two numbers separately, it must exactly divide their sum. Explanation: This statement is true. If a number exactly divides two numbers separately, it means it is a common divisor of both numbers. Since the number divides both of them, it will also divide their sum.

(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately. Explanation: This statement is true. If a number exactly divides the sum of two numbers, it means it is a divisor of their sum. Since it divides the sum, it must also divide both of the numbers separately.

(a)

(b)

Ans.

(a) 10 = 5 * 2
6 = 3 * 2

(b) 60 = 30 * 2
30 = 10 * 3
10 = 5 * 2

### 3. Which factors are not included in the prime factorisation of a composite number?

Ans.

The factors that are not included in the prime factorization of a composite number are the composite factors of that number. Composite factors are those numbers that are divisible by other numbers besides 1 and themselves.

For example, let’s take the composite number 24. Its prime factorization is 2^3 * 3. In this prime factorization, the prime factors 2 and 3 are included because they are the prime numbers that divide 24. However, the composite factors of 24, such as 4, 6, 8, 12, and 24 itself, are not included in the prime factorization. This is because prime factorization focuses on expressing the composite number as a product of its prime factors only.

### 4. Write the greatest 4-digit number and express it in terms of its prime factors.

Ans.

The greatest 4-digit number is 9999.

To express it in terms of its prime factors, we can find the prime factorization of 9999.

9999 can be written as 3^2 * 11 * 101.

So, the prime factorization of the greatest 4-digit number, 9999, is 3^2 * 11 * 101.

### 5. Write the smallest 5-digit number and express it in the form of its prime factors.

Ans.

The smallest 5-digit number is 10,000.

To express it in terms of its prime factors, we can find the prime factorization of 10,000.

10,000 can be written as 2^4 * 5^4.

So, the prime factorization of the smallest 5-digit number, 10,000, is 2^4 * 5^4.

### 6. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.

Ans.

To find the prime factors of 1729, we can start by dividing it by the smallest prime number, which is 2. However, 1729 is not divisible by 2.

Next, we try dividing it by the next prime number, which is 3. 1729 divided by 3 gives a quotient of 576 with a remainder of 1.

We continue dividing the quotient, 576, by prime numbers until we cannot divide it any further.

576 divided by 2 gives a quotient of 288. 288 divided by 2 gives a quotient of 144. 144 divided by 2 gives a quotient of 72. 72 divided by 2 gives a quotient of 36. 36 divided by 2 gives a quotient of 18. 18 divided by 2 gives a quotient of 9. 9 divided by 3 gives a quotient of 3.

So, the prime factorization of 1729 is 3 * 3 * 7 * 13.

Arranging the prime factors in ascending order, we get 3, 3, 7, 13.

The relation between two consecutive prime factors in this case is that they are both prime numbers and there are no additional prime factors between them.

### 7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Ans.

Let’s verify the statement “The product of three consecutive numbers is always divisible by 6” with some examples.

Example 1: Let’s consider three consecutive numbers: 4, 5, and 6. The product of these numbers is 4 * 5 * 6 = 120.
120 is divisible by 6 since 120 divided by 6 equals 20 with no remainder.

Example 2: Let’s consider three consecutive numbers: 7, 8, and 9. The product of these numbers is 7 * 8 * 9 = 504.
504 is divisible by 6 since 504 divided by 6 equals 84 with no remainder.

Example 3: Let’s consider three consecutive numbers: 10, 11, and 12. The product of these numbers is 10 * 11 * 12 = 1320.
1320 is divisible by 6 since 1320 divided by 6 equals 220 with no remainder.

From these examples, we can observe that in each case, the product of three consecutive numbers is indeed divisible by 6. Therefore, we can conclude that the statement “The product of three consecutive numbers is always divisible by 6” is true.

### 8. The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Ans.

Let’s verify the statement “The sum of two consecutive odd numbers is divisible by 4” with some examples.

Example 1: Let’s consider two consecutive odd numbers: 3 and 5. The sum of these numbers is 3 + 5 = 8. 8 is divisible by 4 since 8 divided by 4 equals 2 with no remainder.

Example 2: Let’s consider two consecutive odd numbers: 9 and 11. The sum of these numbers is 9 + 11 = 20. 20 is divisible by 4 since 20 divided by 4 equals 5 with no remainder.

Example 3: Let’s consider two consecutive odd numbers: 15 and 17. The sum of these numbers is 15 + 17 = 32. 32 is divisible by 4 since 32 divided by 4 equals 8 with no remainder.

From these examples, we can observe that in each case, the sum of two consecutive odd numbers is indeed divisible by 4. Therefore, we can conclude that the statement “The sum of two consecutive odd numbers is divisible by 4” is true.

### 9. In which of the following expressions, prime factorisation has been done?

(a) 24 = 2 × 3 × 4
(b) 56 = 7 × 2 × 2 × 2
(c) 70 = 2 × 5 × 7
(d) 54 = 2 × 3 × 9

Ans.

The prime factorisation has been done correctly in the following expression:

(c) 70 = 2 × 5 × 7

In this expression, the number 70 has been decomposed into its prime factors, which are 2, 5, and 7.

The other expressions have either incorrect factorisation or include composite numbers as factors:

(a) 24 = 2 × 3 × 4 – The factor 4 is not a prime number; it should be 2.
(b) 56 = 7 × 2 × 2 × 2 – The factor 7 is a prime number, but the factor 2 is repeated three times unnecessarily.
(d) 54 = 2 × 3 × 9 – The factor 9 is not a prime number; it should be decomposed further into 3 × 3.

So, only expression (c) shows the correct prime factorisation.

### 10. Determine if 25110 is divisible by 45.

[Hint : 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].

Ans.

To determine if 25110 is divisible by 45, we can test its divisibility by both 5 and 9.

Divisibility by 5: A number is divisible by 5 if its units digit is either 0 or 5. In this case, the units digit of 25110 is 0, so it is divisible by 5.

Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9. Let’s calculate the sum of the digits of 25110: 2 + 5 + 1 + 1 + 0 = 9

Since the sum of the digits is 9, which is divisible by 9, we can conclude that 25110 is divisible by 9.

Since 25110 is divisible by both 5 and 9, which are co-prime numbers, it is also divisible by their product, which is 45.

Therefore, 25110 is divisible by 45.

### 11. 18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify your answer.

Ans.

No, we cannot say that a number divisible by both 4 and 6 must also be divisible by 24. This is because 4 and 6 are not coprime numbers, meaning they have a common factor (2). Therefore, their product, which is 24, includes this common factor twice.

For example, let’s consider the number 12. It is divisible by both 4 and 6 since 12 = 4 × 3 and 12 = 6 × 2. However, 12 is not divisible by 24 because it lacks the additional factor of 2.

So, in general, a number divisible by both 4 and 6 is not guaranteed to be divisible by their product, 24, unless it also includes an additional factor of 2.

### 12. I am the smallest number, having four different prime factors. Can you find me?

Ans.

The smallest number with four different prime factors is 2 × 3 × 5 × 7 = 210.

## Exercise 3.6

### 1. Find the HCF of the following numbers :

(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75

Ans.

Let’s find the HCF (Highest Common Factor) of the given numbers:

(a) 18 and 48:
Prime factorization of 18: 2 × 3^2
Prime factorization of 48: 2^4 × 3

The common factors are 2 and 3, so the HCF is 2 × 3 = 6.

(b) 30 and 42:
Prime factorization of 30: 2 × 3 × 5
Prime factorization of 42: 2 × 3 × 7

The common factors are 2 and 3, so the HCF is 2 × 3 = 6.

(c) 18 and 60:
Prime factorization of 18: 2 × 3^2
Prime factorization of 60: 2^2 × 3 × 5

The common factors are 2 and 3, so the HCF is 2 × 3 = 6.

(d) 27 and 63:
Prime factorization of 27: 3^3
Prime factorization of 63: 3^2 × 7

The common factor is 3, so the HCF is 3.

(e) 36 and 84:
Prime factorization of 36: 2^2 × 3^2
Prime factorization of 84: 2^2 × 3 × 7

The common factors are 2 and 3, so the HCF is 2 × 3 = 6.

(f) 34 and 102:
Prime factorization of 34: 2 × 17
Prime factorization of 102: 2 × 3 × 17

The common factor is 17, so the HCF is 17.

(g) 70, 105, 175:
Prime factorization of 70: 2 × 5 × 7
Prime factorization of 105: 3 × 5 × 7
Prime factorization of 175: 5^2 × 7

The common factors are 5 and 7, so the HCF is 5 × 7 = 35.

(h) 91, 112, 49:
Prime factorization of 91: 7 × 13
Prime factorization of 112: 2^4 × 7
Prime factorization of 49: 7^2

The common factor is 7, so the HCF is 7.

(i) 18, 54, 81:
Prime factorization of 18: 2 × 3^2
Prime factorization of 54: 2 × 3^3
Prime factorization of 81: 3^4

The common factor is 3, so the HCF is 3.

(j) 12, 45, 75:
Prime factorization of 12: 2^2 × 3
Prime factorization of 45: 3^2 × 5
Prime factorization of 75: 3 × 5^2

The common factor is 3, so the HCF is 3.

### 2. What is the HCF of two consecutive

(a) numbers?
(b) even numbers?
(c) odd numbers?

Ans.

(a) For two consecutive numbers, their HCF is always 1. This is because consecutive numbers do not have any common factors other than 1.

(b) For two even numbers, their HCF is always 2. This is because both numbers are divisible by 2, and 2 is the only common factor between them.

(c) For two odd numbers, their HCF is always 1. This is because odd numbers do not have any even factors, and the only common factor between them is 1.

### 3. HCF of co-prime numbers 4 and 15 was found as follows by factorisation :

4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Ans.

No, the answer is not correct. The correct HCF of 4 and 15 is 1. The HCF (Highest Common Factor) represents the largest positive integer that divides both numbers evenly. In this case, even though 4 and 15 do not have any common prime factors, they still share a common factor of 1, which divides both numbers without leaving a remainder. Therefore, the correct HCF of 4 and 15 is 1, not 0.

## Exercise 3.7

### 1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Ans.

To find the maximum value of weight that can measure the weight of the fertiliser an exact number of times, we need to find the highest common factor (HCF) of the two weights.

The HCF of 75 kg and 69 kg can be found by prime factorization:

75 = 3 × 5 × 5 69 = 3 × 23

To find the HCF, we take the common factors of both numbers and multiply them together:

HCF(75, 69) = 3

Therefore, the maximum value of weight that can measure the weight of the fertiliser an exact number of times is 3 kg.

### 2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Ans.

To find the minimum distance that each boy should cover so that all can cover the distance in complete steps, we need to find the least common multiple (LCM) of their step measurements.

The LCM of 63 cm, 70 cm, and 77 cm can be found by prime factorization:

63 = 3 × 3 × 7 70 = 2 × 5 × 7 77 = 7 × 11

To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:

LCM(63, 70, 77) = 2 × 3 × 3 × 5 × 7 × 11 = 2,310 cm

Therefore, the minimum distance that each boy should cover so that all can cover the distance in complete steps is 2,310 cm.

### 3. The length, breadth, and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Ans.

To find the longest tape that can measure the three dimensions of the room exactly, we need to find the highest common factor (HCF) of the three dimensions.

The dimensions of the room are 825 cm, 675 cm, and 450 cm.

To find the HCF, we can use the Euclidean algorithm or prime factorization.

Prime factorization of the dimensions: 825 = 3 × 3 × 5 × 5 × 11 675 = 3 × 3 × 3 × 5 × 5 450 = 2 × 3 × 3 × 5 × 5

Taking the highest power of each prime factor that appears in any of the dimensions, we have: HCF(825, 675, 450) = 3 × 3 × 5 × 5 = 225

Therefore, the longest tape that can measure the three dimensions of the room exactly is 225 cm.

### 4. Determine the smallest 3-digit number which is exactly divisible by 6, 8, and 12.

Ans.

To find the smallest 3-digit number that is divisible by 6, 8, and 12, we need to find the least common multiple (LCM) of these three numbers.

The prime factorization of 6 is 2 × 3. The prime factorization of 8 is 2 × 2 × 2. The prime factorization of 12 is 2 × 2 × 3.

To find the LCM, we take the highest power of each prime factor that appears in any of the numbers:

LCM(6, 8, 12) = 2 × 2 × 2 × 3 = 24

So, the smallest 3-digit number that is divisible by 6, 8, and 12 is 24.

### 5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Ans.

To find the greatest 3-digit number that is divisible by 8, 10, and 12, we need to find the highest multiple of their least common multiple (LCM) within the range of 3-digit numbers.

The LCM of 8, 10, and 12 is 120, as it is the smallest number that is divisible by all three of them.

To find the greatest 3-digit number divisible by 120, we divide the largest 3-digit number, which is 999, by 120:

999 ÷ 120 = 8 remainder 39

Since the remainder is non-zero, we need to subtract it from 999 to get the largest multiple of 120 that is within the range of 3-digit numbers:

999 – 39 = 960

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

### 6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Ans.

To find the time when the traffic lights will change simultaneously again, we need to determine the least common multiple (LCM) of the three given time intervals.

The given time intervals are 48 seconds, 72 seconds, and 108 seconds.

Prime factorizing these numbers:

48 = 2^4 × 3 72 = 2^3 × 3^2 108 = 2^2 × 3^3

To find the LCM, we need to take the highest power of each prime factor that appears in any of the numbers. So the LCM will be:

LCM = 2^4 × 3^3 = 16 × 27 = 432

Therefore, the traffic lights will change simultaneously again after 432 seconds.

Since 1 hour contains 3600 seconds, we can convert 432 seconds into minutes and hours:

432 seconds = 432/60 minutes = 7 minutes and 12 seconds

Since the traffic lights changed simultaneously at 7 a.m., they will change simultaneously again 7 minutes and 12 seconds later.

Hence, the next simultaneous change will occur at 7:07:12 a.m.

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

Ans.

To find the maximum capacity of a container that can measure the diesel of the three containers an exact number of times, we need to find the greatest common divisor (GCD) of the given capacities.

The given capacities are 403 litres, 434 litres, and 465 litres.

To find the GCD, we can use the Euclidean algorithm.

Step 1: Find the GCD of 403 litres and 434 litres: 434 = 403 × 1 + 31

Step 2: Find the GCD of 403 litres (previous remainder) and 31 litres: 403 = 31 × 13 + 10

Step 3: Find the GCD of 31 litres (previous remainder) and 10 litres: 31 = 10 × 3 + 1

Step 4: Find the GCD of 10 litres (previous remainder) and 1 litre: 10 = 1 × 10 + 0

Since we have reached a remainder of 0, the GCD is the previous remainder, which is 1 litre.

Therefore, the maximum capacity of a container that can measure the diesel of the three containers an exact number of times is 1 litre.

### 8. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Ans.

To find the least number that satisfies the given conditions, we need to find the least common multiple (LCM) of the divisors (6, 15, and 18) and then add 5 to it.

Step 1: Find the LCM of 6, 15, and 18: Prime factorization of 6: 2 × 3 Prime factorization of 15: 3 × 5 Prime factorization of 18: 2 × 3²

LCM = 2 × 3² × 5 = 90

Step 2: Add 5 to the LCM: 90 + 5 = 95

Therefore, the least number that when divided by 6, 15, and 18 leaves a remainder of 5 in each case is 95.

### 9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Ans.

To find the smallest 4-digit number that is divisible by 18, 24, and 32, we need to find the least common multiple (LCM) of these three numbers.

Step 1: Prime factorization of 18: 2 × 3² Step 2: Prime factorization of 24: 2³ × 3 Step 3: Prime factorization of 32: 2⁵

To find the LCM, we need to consider the highest powers of all prime factors:

LCM = 2⁵ × 3² = 32 × 9 = 288

The smallest 4-digit number is 1000, so we need to find the smallest multiple of 288 that is greater than 1000.

1000 ÷ 288 = 3 remainder 136

To obtain the next multiple, we subtract the remainder from 288:

288 – 136 = 152

Hence, the smallest 4-digit number divisible by 18, 24, and 32 is 1000 + 152 = 1152.

### 10. Find the LCM of the following numbers :

(a) 9 and 4
(b) 12 and 5
(c) 6 and 5
(d) 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Ans.

To find the least common multiple (LCM) of two numbers, we can use the prime factorization method.

(a) LCM of 9 and 4:
Prime factorization of 9: 3² Prime factorization of 4: 2² The LCM should consider the highest powers of all prime factors: LCM = 3² × 2² = 36.

(b) LCM of 12 and 5:
Prime factorization of 12: 2² × 3 Prime factorization of 5: 5 The LCM should consider the highest powers of all prime factors: LCM = 2² × 3 × 5 = 60.

(c) LCM of 6 and 5:
Prime factorization of 6: 2 × 3 Prime factorization of 5: 5 The LCM should consider the highest powers of all prime factors: LCM = 2 × 3 × 5 = 30.

(d) LCM of 15 and 4:
Prime factorization of 15: 3 × 5 Prime factorization of 4: 2² The LCM should consider the highest powers of all prime factors: LCM = 2² × 3 × 5 = 60.

Therefore, the LCM of: (a) 9 and 4 is 36. (b) 12 and 5 is 60. (c) 6 and 5 is 30. (d) 15 and 4 is 60.

No, the LCM is not always the product of the two numbers in each case

### 11. Find the LCM of the following numbers in which one number is the factor of the other.

(a) 5, 20
(b) 6, 18
(c) 12, 48
(d) 9, 45

What do you observe in the results obtained?

Ans.

Let’s find the LCM (Least Common Multiple) of the given numbers:

(a) LCM of 5 and 20:
Factors of 5: 5
Factors of 20: 2, 2, 5

LCM: 2 * 2 * 5 = 20

(b) LCM of 6 and 18:
Factors of 6: 2, 3
Factors of 18: 2, 3, 3

LCM: 2 * 3 * 3 = 18

(c) LCM of 12 and 48:
Factors of 12: 2, 2, 3
Factors of 48: 2, 2, 2, 2, 3

LCM: 2 * 2 * 2 * 2 * 3 = 48

(d) LCM of 9 and 45:
Factors of 9: 3, 3
Factors of 45: 3, 3, 5

LCM: 3 * 3 * 5 = 45

So, the LCM of the given numbers is:
(a) LCM of 5 and 20 is 20.
(b) LCM of 6 and 18 is 18.
(c) LCM of 12 and 48 is 48.
(d) LCM of 9 and 45 is 45.

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