Playing With Numbers

NCERT Solutions for Class 6 Maths Chapter 3

The chapter “Playing With Numbers” in your 6th-grade math book likely focuses on developing foundational skills with numbers, particularly whole numbers. Here’s a summary of the key concepts you might encounter:

1. The Number System:

• Whole Numbers: These are non-negative integers, including 0 (zero) and all natural numbers (1, 2, 3, …).
• Natural Numbers: These are positive numbers used for counting, starting from 1 and going on infinitely.
• Place Value System: You’ll learn how the position of a digit in a number determines its value (units, tens, hundreds, thousands, etc.).

2. Comparing and Ordering Numbers:

• Techniques for comparing whole numbers are introduced. This might involve considering the number of digits or looking at each place value one by one, starting from the leftmost place (highest value).
• You’ll learn how to arrange numbers in a specific order:
  • Ascending Order: Arranging numbers from smallest to biggest (e.g., 12, 15, 20).
  • Descending Order: Arranging numbers from biggest to smallest (e.g., 20, 15, 12).

3. Basic Operations (depending on your textbook):

• Some textbooks might introduce basic addition, subtraction, and multiplication of whole numbers.

4. Other concepts you might encounter:

• Forming Numbers: Creating different numbers using a given set of digits, making sure the digits don’t repeat.
• Representing Numbers on the Number Line: Visualizing numbers on a line where each point corresponds to a specific number.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

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 : 

Number	Factors
24	1, 2, 3, 4, 6, 8, 12, 24
15	1, 3, 5, 15
21	1, 3, 7, 21
27	1, 3, 9, 27
12	1, 2, 3, 4, 6, 12
20	1, 2, 4, 5, 10, 20
18	1, 2, 3, 6, 9, 18
23	1, 23
36	1, 2, 3, 4, 6, 9, 12, 18, 36

2. Write first five multiples of:

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

Ans : 

(a) 5: 5, 10, 15, 20, 25 

(b) 8: 8, 16, 24, 32, 40 

(c) 9: 9, 18, 27, 36, 45

3. Match the items in column I with the items in column II.

Column I	Column II
35	Multiple of 8
15	Multiple of 7
16	Multiple of 70
20	Factor of 30
25	Factor of 50
	Factor of 20

Ans : 

Column I	Column II
35	Multiple of 7
15	Factor of 30
16	Multiple of 8
20	Factor of 20
25	Factor of 50

4. Find all the multiples of 9 upto 100.

Ans : 

| 9 | | 18 | | 27 | | 36 | | 45 | | 54 | | 63 | | 72 | | 81 | | 90 | | 99 |

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Exercise 3.2

1. What is the sum of any two:

(a) Odd numbers?

(b) Even numbers?

Ans : 

(a) Odd numbers: Always results in an even number.

(b) Even numbers: Always results in an even number.

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 only the even prime number.

(i) All even numbers are composite numbers.

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

Ans : 

(a) False 

(b) True 

(c) True 

(d) False

(e) False 

(f) True

(g) False 

(h) True 

(i) False 

(j) True 

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

Ans : 

13 and 31 (as mentioned)

17 and 71

37 and 73

79 and 97

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

Ans : Here are the prime and composite numbers less than 20 written separately:

Prime numbers:

• 2
• 3
• 5
• 7
• 11
• 13
• 17
• 19

Composite numbers:

• 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 = 13 + 31
(b) 36 = 17 + 19
(c) 24 = 7 + 17
(d) 18 = 7 + 11

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)
2. (5, 7)
3. (11, 13)

8. Which of the following numbers are prime?

(a) 23

(b) 51

(c) 37

(d) 26

Ans : 

(a) 23 – This is a prime number. It has only two factors: 1 and 23.

(b) 51 – This is not a prime number. It has factors 1, 3, 17, and 51.

(c) 37 – This is a prime number. It has only two factors: 1 and 37.

(d) 26 – This is not a prime number. It has factors 1, 2, 13, and 26.

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

Ans : 

 90, 91, 92, 93, 94, 95, and 96. Although they are consecutive and all 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) 21 = 3 + 5 + 13

(b) 31 = 5 + 7 + 19

(c) 53 = 3 + 17 + 33

(d) 61 = 7 + 11 + 43

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

(Hint: 3 + 7 = 10)

Ans : 

(2, 3): 2 + 3 = 5 (divisible by 5)

(2, 13): 2 + 13 = 15 (divisible by 5)

(3, 7): 3 + 7 = 10 (divisible by 5)

(11, 19): 11 + 19 = 30 (divisible by 5)

(13, 17): 13 + 17 = 30 (divisible by 5)

12. Fill in the blanks.

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

(b) A number which 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 which has only two factors is called a prime number. 

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

(c) 1 is neither a prime nor a composite number. 

(d) The smallest prime number is 2. 

(e) The smallest composite number is 4. 

(f) The smallest even number is 2.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Exercise 3.3

1. Using divisibility tests, determine which of the following numbers are divisible by 2, by 3, by 4, by 5, by 6, by 8, by 9, by 10, by 11 (Say, Yes or No)

Ans : 

Number	Divisible by 2	Divisible by 3	Divisible by 4	Divisible by 5	Divisible by 6	Divisible by 8	Divisible by 9	Divisible by 10	Divisible by 11
128	Yes	No	Yes	No	No	Yes	No	No	No
990	Yes	Yes	No	Yes	Yes	No	Yes	Yes	No
1586	Yes	No	No	No	No	No	No	No	No
275	No	No	No	Yes	No	No	No	No	Yes
6686	Yes	Yes	No	No	No	No	No	No	No
639210	Yes	Yes	Yes	No	Yes	No	Yes	No	No
429714	Yes	Yes	No	No	No	No	No	No	No
2856	Yes	No	No	No	No	No	No	No	No
3060	Yes	Yes	Yes	Yes	Yes	No	Yes	Yes	No
406839	Yes	No	No	No	No	No	No	No	No

2. Using divisibility tests, determine which of following numbers are divisible by 4; by 8.

(a) 572

(b) 726352

(c) 5500

(d) 6000

(e) 12159

(f) 14560

(g) 21084

(j) 2150

(h) 31795072

(i) 1700

Ans : 

Number	Divisible by 4?	Divisible by 8?
(a) 572	Yes	No
(b) 726352	Yes	Yes
(c) 5500	Yes	No
(d) 6000	Yes	Yes
(e) 12159	No	No
(f) 14560	Yes	No
(g) 21084	Yes	Yes
(h) 31795072	Yes	No
(i) 1700	Yes	Yes
(j) 2150	No	No

3. Using divisibility tests, determine which of the 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 : 

Number	Divisible by 4?	Divisible by 8?
(a) 572	Yes	No
(b) 726352	Yes	Yes
(c) 5500	Yes	No
(d) 6000	Yes	Yes
(e) 12159	No	No
(f) 14560	Yes	No
(g) 21084	Yes	Yes
(h) 31795072	Yes	No
(i) 1700	Yes	Yes
(j) 2150	No	No

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

(a) 5445

(b) 10824

(c) 7138965

(d) 70169308

(e) 10000001.

Ans : 

(a) 5445 – No

(b) 10824 – No

(c) 7138965 – Yes

(d) 70169308 – No

(e) 10000001 – Yes

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.

Ans : 

a) _____6724

• Smallest digit: Since the sum of the existing digits (6 + 7 + 2 + 4) is 19, which is one less than a multiple of 3 (21), we need to add the smallest digit, which is 1. The number becomes 16724.
• Greatest digit: To maximize the value while maintaining divisibility by 3, we can add the digit that makes the sum the closest multiple of 3 without going over. In this case, adding 8 makes the sum 27 (a multiple of 3). The number becomes 86724.

(b) 4765_____2

• Smallest digit: Similar to (a), the sum of existing digits (4 + 7 + 6 + 5) is 22, two less than a multiple of 3 (24). Adding the smallest digit, 0, makes the number 476502, which is divisible by 3.
• Greatest digit: Following the same logic, adding the largest digit, 9, makes the sum 31 (a multiple of 3). The number becomes 476592.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

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 : 

The common factors of each pair of numbers are:

(a) 20 and 28: 1, 2, 4

• Find the factors of 20: 1, 2, 4, 5, 10, 20
• Find the factors of 28: 1, 2, 4, 7, 14, 28
• The common factors are the numbers that appear in both lists: 1, 2, 4.

(b) 15 and 25: 1, 5

• Find the factors of 15: 1, 3, 5, 15
• Find the factors of 25: 1, 5, 25
• The common factors are the numbers that appear in both lists: 1, 5.

(c) 35 and 50: 1, 5

• Find the factors of 35: 1, 5, 7, 35
• Find the factors of 50: 1, 2, 5, 10, 25, 50
• The common factors are the numbers that appear in both lists: 1, 5.

(d) 56 and 120: 1, 2, 4, 8

• Find the factors of 56: 1, 2, 4, 7, 8, 14, 28, 56
• Find the factors of 120: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
• The common factors are the numbers that appear in both lists: 1, 2, 4, 8

2. Find the common factors of:

(a) 4, 8 and 12

(b) 5, 15 and 25

(6) 8 9484.

Ans : 

Set (a): 4, 8, and 12

1. Factorize each number:
   
   • 4 = 2 x 2
   • 8 = 2 x 2 x 2
   • 12 = 2 x 2 x 3
2. Identify common factors:
   
   • Look for factors that appear in all three factorizations.
   • In this case, the common factor is 2 (appearing twice).

Therefore, the common factors of 4, 8, and 12 are 1 and 2 (2 appears twice).

Set (b): 5, 15, and 25

1. Factorize each number:
   
   • 5 is already a prime number (cannot be further factored).
   • 15 = 3 x 5
   • 25 = 5 x 5
2. Identify common factors:
   
   • The only common factor is 5.

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

Set (c): 8 and 9484 (corrected from previous prompt)

1. Factorize each number:
   
   • 8 = 2 x 2 x 2 (prime factorization)
   • 9484 = 2 x 2 x 2371 (prime factorization)
2. Identify common factors:
   
   • There are no common factors other than 1, since 8 only has factors of 2 and 9484 has the additional prime factor 2371.

3. Find first three multiples of:

(a) 6 and 8

(b) 12 and 18

Ans :  

(a) 6 and 8

• Multiples of 6: 6, 12, 18, …
• First three multiples: 6, 12, 18

(b) 12 and 18

• Multiples of 12: 12, 24, 36, …
• First three multiples: 12, 24, 36

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

Ans : The common multiples of 3 and 4 less than 100 are:

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

Here’s how we can find them:

1. Least Common Multiple (LCM): The lowest number that is a multiple of both 3 and 4 is 12 (3 x 4).
   
2. Multiples of LCM: Since we want common multiples less than 100, we can find multiples of 12 that are less than 100.
   

• 12 x 1 = 12
• 12 x 2 = 24
• 12 x 3 = 36
• 12 x 4 = 48
• 12 x 5 = 60
• 12 x 6 = 72
• 12 x 7 = 84
• 12 x 8 = 96 (This is the last multiple less than 100)

Therefore, all the numbers listed (12, 24, 36, 48, 60, 72, 84, and 96) are common multiples of 3 and 4 that are less than 100.

5. Which of the following numbers are 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 : 

(a) 18 and 35:

• 18 can be factored as 2 x 3 x 3.
• 35 can be factored as 5 x 7.
• They share no common factors other than 1. Thus, 18 and 35 are co-prime.

(b) 15 and 37:

• 15 can be factored as 3 x 5.
• 37 is a prime number.
• 15 and 37 share no common factors other than 1. Thus, 15 and 37 are co-prime.

(c) 30 and 415:

• 30 can be factored as 2 x 3 x 5.
• 415 can be factored as 5 x 83.
• They share a common factor of 5. Therefore, 30 and 415 are not co-prime.

(d) 17 and 68:

• 17 is a prime number.
• 68 can be factored as 2 x 2 x 17.
• They share a common factor of 17. Therefore, 17 and 68 are not co-prime.

(e) 216 and 215:

• 216 can be factored as 2 x 2 x 2 x 3 x 3.
• 215 can be factored as 5 x 43.
• They share no common factors other than 1. Thus, 216 and 215 are co-prime.

(f) 81 and 16:

• 81 can be factored as 3 x 3 x 3 x 3.
• 16 can be factored as 2 x 2 x 2 x 2.
• They share a common factor of 2. Therefore, 81 and 16 are not co-prime.

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

Ans : If a number is divisible by both 5 and 12, it will always be divisible by 60

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

Ans : If a number is divisible by 12, it will be divisible by several other numbers because 12 has several factors. Here’s what it’s divisible by:

• Always divisible by:
  
  • 1: Every number is divisible by 1.
  • 12 (itself): Since the number is divisible by 12 by definition.
  • Factors of 12: These include the prime factors of 12 (2, 2, and 3) and any combination of these factors that multiply to 12. So, it will also be divisible by 2, 3, 4 (2 x 2), and 6 (2 x 3).

A number divisible by 12 is guaranteed to be divisible by 1, 12, 2, 3, 4, and 6. It might also be divisible by higher multiples of these factors depending on its specific makeup.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Exercise 3.5 

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

Ans : 

Factor Tree 1:

    60

    /   \

   15    4

   / \   / \

  3   5  2   2

• In both trees, 60 is the starting point.

Explanation:

• We split 60 into 15 and 4, which are both factors of 60.
• 15 can be further factored into 3 and 5 (prime factors).
• 4 can be further factored into 2 x 2 (prime factors).

Factor Tree 2:

    60

    /   \

   30    2

   / \    

  15    2

  / \

 3   5

Explanation

• We split 60 into 30 and 2, which are both factors of 60.
• 30 can be further factored into 15, which can then be broken down to 3 and 5 (prime factors).
• 2 is already a prime number.

Both trees show the prime factorization of 60 (2 x 2 x 3 x 5) through different paths.

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

Ans : 

1. 1: Every number is divisible by 1, so it’s not a unique characteristic related to the number’s composition. Prime factorization focuses on breaking down a number into its most basic building blocks, which are prime numbers. Since 1 divides everything, it’s not considered a prime factor.
2. The number itself: Similar to 1, the number itself is not a prime factor because prime factorization is about finding the prime numbers that multiply together to create the composite number. The number itself is the product, not a building block.

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

Ans : The greatest 4-digit number is 9999.

Here’s its prime factorization:

9999 = 3 * 3 * 11 * 101

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

Ans : The smallest 5-digit number is 10000.

Here’s its prime factorization:

10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5

5. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relations, if any, between the two consecutive prime factors.

Ans : 1729 can be prime factorized as 7 x 13 x 19.

Relation between consecutive prime factors:

In this case, there’s no specific mathematical relationship between the consecutive prime factors (7, 13, and 19). They are all distinct prime numbers with no common divisibility rule.

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

Ans : 

Example 1: Take three consecutive numbers 12 (divisible by 3), 13 (leaves remainder 1), and 14 (leaves remainder 2).

• Product = 12 * 13 * 14 = 2184, which is divisible by 6 (2184 / 6 = 364).

Example 2: Consider 5 (leaves remainder 2), 6 (divisible by 3), and 7 (leaves remainder 1).

• Product = 5 * 6 * 7 = 210, which is divisible by 6 (210 / 6 = 35).

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

Ans : 

Example 1: Take three consecutive numbers 4 (leaves remainder 1), 5 (divisible by 3), and 6 (leaves remainder 2).

• Product = 4 * 5 * 6 = 120, which is divisible by 6 (120 / 6 = 20).

Example 2: Consider 7 (leaves remainder 1), 8 (divisible by 3), and 9 (leaves remainder 0).

• Product = 7 * 8 * 9 = 504, which is divisible by 6 (504 / 6 = 84).

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

Ans : 

Example 1: Consider two consecutive odd numbers 3 (2 * 1 + 1) and 5 (2 * 1 + 3).

• Sum = 3 + 5 = 8, which is divisible by 4 (8 / 4 = 2).

Example 2: Take 9 (2 * 4 + 1) and 11 (2 * 4 + 3).

• Sum = 9 + 11 = 20, which is also divisible by 4 (20 / 4 = 5).

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

(a) 24 = 2 x 3 x 4

(b) 56 = 7 x 2 x 2 x 2

(c) 70 = 2 x 5 x 7

(d) 54 = 2 x 3 x 9.

Ans : 

(a) 24 = 2 x 3 x 4: Here, 4 is not a prime number (it can be further factored as 2 x 2). So, this is not a correct prime factorization.

(b) 56 = 7 x 2 x 2 x 2: All the factors on the right side (7, 2, 2, 2) are prime numbers, and their product results in 56. This is a correct prime factorization.

(c) 70 = 2 x 5 x 7: Similar to (b), all the factors here are prime, and their product is 70. This is a correct prime factorization.

(d) 54 = 2 x 3 x 9: Here, 9 is not a prime number (it can be factored as 3 x 3). So, this is not a correct prime factorization.

10. Determine if 25110 is divisible by 45.

Ans : 

Divisibility by 5: A number is divisible by 5 if the last digit is either a 0 or a 5. In this case, 25110 ends in a 0, so it’s divisible by 5.

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

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

Ans : No, unfortunately, we cannot definitively say that the number must also be divisible by 4 x 6 = 24. Here’s why:

However, the divisibility rule for 4 and 6 is different.

• Divisibility by 4: A number is divisible by 4 if the last two digits are divisible by 4.
• Divisibility by 6: A number is divisible by 6 if it’s divisible by both 2 and 3 (as explained earlier).

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 210.

Here’s the reasoning:

1. Smallest Prime Numbers: To get the smallest number with four prime factors, we want to use the first four prime numbers. These are 2, 3, 5, and 7.
   
2. Product of Prime Factors: The number we’re looking for is the product of these four prime numbers:
   
   Smallest number = 2 x 3 x 5 x 7
   
3. Calculation: Multiplying these prime factors together gives us:
   
   Smallest number = 210

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

Exercises 3.6

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 : 

(a) 18, 48: HCF = 6

• Prime factorize: 18 = 2 * 3 * 3; 48 = 2 * 2 * 2 * 2 * 3
• HCF = 2 * 3 = 6 (consider only the common factors with the lowest powers)

(b) 30, 42: HCF = 6

• Prime factorize: 30 = 2 * 3 * 5; 42 = 2 * 3 * 7
• HCF = 2 * 3 = 6

(c) 18, 60: HCF = 6

• Prime factorize: 18 = 2 * 3 * 3; 60 = 2 * 2 * 3 * 5
• HCF = 2 * 3 = 6

(d) 27, 63: HCF = 9

• Prime factorize: 27 = 3 * 3 * 3; 63 = 3 * 3 * 7
• HCF = 3 * 3 = 9

(e) 36, 84: HCF = 12

• Prime factorize: 36 = 2 * 2 * 3 * 3; 84 = 2 * 2 * 3 * 7
• HCF = 2 * 2 * 3 = 12

(f) 34, 102: HCF = 2 (Correction: The HCF of 34 and 102 is 2, not 34)

• Prime factorize: 34 = 2 * 17; 102 = 2 * 3 * 17
• HCF = 2 (only the common factor 2 is present)

(g) 70, 105, 175: HCF = 35

• Prime factorize: 70 = 2 * 5 * 7; 105 = 3 * 5 * 7; 175 = 5 * 7 * 5
• HCF = 5 * 7 = 35 (all three numbers share these prime factors)

(h) 91, 112, 49: HCF = 7

• Prime factorize: 91 = 7 * 13; 112 = 2 * 2 * 2 * 7; 49 = 7 * 7
• HCF = 7 (only the common factor 7 is present)

(i) 18, 54, 81: HCF = 9

• Prime factorize: 18 = 2 * 3 * 3; 54 = 2 * 3 * 3 * 3; 81 = 3 * 3 * 3 * 3
• HCF = 3 * 3 = 9

(j) 12, 45, 75: HCF = 3

• Prime factorize: 12 = 2 * 2 * 3; 45 = 3 * 3 * 5; 75 = 3 * 5 * 5
• HCF = 3 (only the common factor 3 is present)

2. What is the HCF of two consecutive

(a) numbers?

(b) even numbers?

(c) odd numbers?

Ans : 

(a) Two consecutive numbers: The HCF of two consecutive numbers is always 1.

(b) Two consecutive even numbers: The HCF of two consecutive even numbers is also 1.

(c) Two consecutive odd numbers: The HCF of two consecutive odd numbers is again 1.

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

4 = 2 x 2 and 15 = 3 x 15. Since there is no common prime factors, so HCF of 4 and 15 is 0.

Is the answer correct? If not, what is the correct HCF?

Ans : No, the answer is incorrect. The HCF (Highest Common Factor) of 4 and 15 is actually 1.

1. Prime Factorization: You’ve correctly identified the prime factorizations: 4 = 2 x 2 and 15 = 3 x 5.
   
2. HCF and Common Factors: The HCF refers to the highest common factor, which in this case is simply 1. Even though 4 and 15 don’t share any common prime factors beyond 1, 1 is still a common factor between any two integers.

Therefore, the HCF of 4 and 15 is 1.

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

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 : 

1. Prime Factorize the Weights:
   • Prime factorize 75: 75 = 3 * 5 * 5
   • Prime factorize 69: 69 = 3 * 23
2. Identify Common Prime Factors:
   • In this case, the only common prime factor is 3.
3. Find the Product of Common Prime Factors Raised to the Lowest Power:
   • We only have one common prime factor (3), and it appears once in each prime factorization (3^1 for 75 and 3^1 for 69).
   • Therefore, the HCF (or the maximum weight) = 3^1 = 3.

Therefore, the maximum weight that can measure the weight of the fertilizer 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 : 

1. Prime factorize each number:

• 63 = 3 * 3 * 7
• 70 = 2 * 5 * 7
• 77 = 7 * 11

2. Identify the highest power of each prime factor that appears in any of the numbers:

• Prime factor | Highest power
• ———- | ——–
• 2 | 1
• 3 | 2
• 5 | 1
• 7 | 1
• 11 | 1

3. Multiply the prime factors raised to their highest powers:

• LCM = 2^1 * 3^2 * 5^1 * 7^1 * 11^1 = 6930
• The least common multiple of 63, 70, and 77 is 6930.

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 : 

Prime factorize each dimension:

• Length (825) = 3 * 5 * 5 * 11
• Breadth (675) = 3 * 3 * 3 * 5 * 5
• Height (450) = 2 * 3 * 3 * 5 * 5

Identify the common prime factors with the lowest exponents present in any of the factorizations:

• Common factors: 3 * 5 * 5 (all three dimensions share these prime factors)

Find the product of the common prime factors raised to their lowest powers:

• HCF = 3 * 5 * 5 = 75

Therefore, the longest tape which can measure the exact dimensions of the room is 75 cm.

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

Ans : The smallest 3-digit number divisible by 6, 8, and 12 is 120.

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

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

• Prime factorize:
  • 8 = 2 x 2 x 2
  • 10 = 2 x 5
  • 12 = 2 x 2 x 3
• LCM = 2 x 2 x 2 x 3 x 5 = 120

2. Largest 3-digit multiple of LCM: Since we want the greatest 3-digit number, we find the largest multiple of 120 less than 1000 (the smallest 4-digit number).
   • Divide 999 (largest 3-digit number) by 120: 999 / 120 = 8.325.
3. Checking divisibility: Take the integer quotient (8 in this case) and multiply it by the LCM (120).
   • 8 x 120 = 960.

Therefore, the greatest 3-digit number 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 : 

1. Prime factorize each time interval:
   • 48 = 2 * 2 * 2 * 2 * 3
   • 72 = 2 * 2 * 2 * 3 * 3
   • 108 = 2 * 2 * 3 * 3 * 3
2. Identify the highest power of each prime factor that appears in any of the factorizations:
   • Prime factor | Highest power
   • ———- | ——–
   • 2 | 4
   • 3 | 3
3. Multiply the prime factors raised to their highest powers:
   • LCM = 2^4 * 3^3 = 144

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 : 

1. Prime factorize each quantity:
   • 403 = 13 * 31
   • 434 = 2 * 7 * 31
   • 465 = 3 * 5 * 31
2. Identify the common prime factors present in all three factorizations:
   • Common prime factor: 31
3. The HCF is the product of the common prime factor raised to its lowest power:
   • HCF = 31 (since 31 appears only once in each factorization)

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

Ans : 

• Prime factorize each number:
  • 6 = 2 * 3
  • 15 = 3 * 5
  • 18 = 2 * 3 * 3
• Identify the highest power of each prime factor present in any of the numbers:
  • Prime factor | Highest power
  • ———- | ——–
  • 2 | 1
  • 3 | 2
  • 5 | 1
• Calculate the LCM by multiplying the prime factors with their highest powers:
  • LCM = 2 * 3^2 * 5 = 90
• Therefore, the least number divisible by 6, 15, and 18 with a remainder of 5 in each case is LCM (90) + Remainder (5) = 90 + 5 = 95.

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

Ans : 

• Prime Factorize the Divisors:
  • 18 = 2 * 3 * 3
  • 24 = 2 * 2 * 2 * 3
  • 32 = 2 * 2 * 2 * 2 * 2
• Identify Highest Power of Each Prime Factor: | Prime Factor | Highest Power | |—|—| | 2 | 5 (from 32) | | 3 | 2 (from 18) |
• Calculate LCM: LCM = 2^5 * 3^2 = 288

2. Check the Smallest 4-Digit Multiple:

• Smallest 4-Digit Number: 1000
• Divisibility Check: 1000 / 288 ≈ 3.47 (not a whole number, so 1000 is not divisible)

3. Find the Next 4-Digit Multiple:

Since the division result isn’t a whole number, we need to find the next multiple of 288 that’s a 4-digit number.

• Calculate Next Multiple: Add 3 multiples of 288 to 1000: 1000 + (3 * 288) = 1000 + 864 = 1864

4. Verification:

Check if 1864 is divisible by all three divisors:

• 1864 / 18 = 104 (divisible)
• 1864 / 24 = 77 (divisible)
• 1864 / 32 = 58 (divisible)

Therefore, the smallest 4-digit number divisible by 18, 24, and 32 is 1864.

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 : 

1. Prime factorize each number:
   • (a) 9 = 3 * 3
   • (b) 12 = 2 * 2 * 3
   • (c) 6 = 2 * 3
   • (d) 15 = 3 * 5
2. Find the LCM for each case:
   • (a) LCM(9, 4) = 2 * 3 * 3 = 36
   • (b) LCM(12, 5) = 2 * 2 * 3 * 5 = 60
   • (c) LCM(6, 5) = 2 * 3 * 5 = 30
   • (d) LCM(15, 4) = 2 * 2 * 3 * 5 = 60

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 : 

We can find the Least Common Multiple (LCM) for the following cases where one number is a factor of the other:

(a) 5 and 20 (5 is a factor of 20)

(b) 6 and 18 (6 is a factor of 18)

(c) 12 and 48 (12 is a factor of 48)

(d) 9 and 45 (9 is a factor of 45)

Results:

(a) LCM(5, 20) = 20

(b) LCM(6, 18) = 18

(c) LCM(12, 48) = 48

(d) LCM(9, 45) = 45

NCERT Solutions for Class 6 Maths Chapter 3 Playing With Numbers

FAQ’s

What topics are covered in Class 6 Maths Chapter 3 Playing With Numbers?

Class 6 Maths Chapter 3, “Playing with Numbers,” covers various topics such as number patterns, divisibility rules, factors, multiples, and more.

Why is it important to learn about playing with numbers?

Learning about playing with numbers helps students develop problem-solving skills, understand number patterns, and apply divisibility rules, which are essential in various mathematical concepts.

How can I improve my skills in playing with numbers?

You can improve your skills in playing with numbers by practicing different types of problems, understanding the divisibility rules thoroughly, and exploring number patterns through regular practice.

Are there any real-life applications of playing with numbers?

Yes, playing with numbers has numerous real-life applications, such as in banking, cryptography, computer programming, and various other fields where number manipulation is required.

How can NCERT Solutions help in understanding Class 6 maths chapter 3 Playing With Numbers?

NCERT Solutions provide detailed explanations and step-by-step solutions to all the problems in Class 6 maths chapter 3 Playing With Numbers, helping students understand the concepts better and improve their problem-solving skills.