NCERT Solutions for Class 6 Maths Chapter 3 Number Play

NCERT Class 6 Maths Chapter 3 solutions are a complete guide for students to learn the basic ideas behind "Number Play." They cover important topics including factors, multiples, prime and composite numbers, divisibility rules, and HCF/LCM. These solutions give step-by-step instructions that are meant to help sixth graders do their homework correctly and create a firm foundation in math.

NCERT Class 6 Maths Chapter 3 Solutions are designed to explain each concept in a clear and simple way. The step-by-step answers help students understand how to apply rules correctly and avoid common mistakes. These NCERT solutions for Class 6 Maths Chapter 3 strictly follow the latest NCERT syllabus, making them ideal for school exams and homework practice.

The Number Play Class 6 questions and answers include a variety of problems, from basic to slightly challenging, which improve logical thinking and calculation skills. Practicing Class 6 Math Chapter 3 regularly helps students gain confidence in divisibility tests, prime factorization, and finding HCF and LCM.

Overall, Class 6 Ch 3 Maths is a scoring chapter if practiced well. With reliable NCERT solutions for Class 6 Maths Chapter 3, students can revise concepts easily, strengthen fundamentals, and perform better in exams.

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Class 6 Maths Chapter 3 Number Play Questions and Answers

Below are the detailed solutions for Ganita Prakash Class 6 Maths Chapter 3 Number Play. These solutions are designed to help students understand the basic concepts of number patterns, logical thinking, and how numbers can be used to create fun mathematical arrangements.

3.1 Numbers can Tell us Things 3.2 Supercells Figure it Out (Page No. 57-58)

Question 1.
Colour or mark the supercells in the table below.

Solution:

Question 2.
Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.

Solution:

Question 3.
Fill in the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.

Solution:
Maximum number of supercells that we can get is 5.

Question 4.
Out of the 9 numbers, how many supercells are there in the table above?
Solution:
Out of 9 numbers, there are 5 supercells in the above table.

Question 5. 

Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution:
No, if we fill with numbers increasing by 1, we get at least one supercell, i.e., the last number.

Question 6.
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution:
Yes, the cell with the largest number in a table will always be a supercell. Since the largest number is greater than any other number in the table, it will naturally be greater than its adjacent numbers. Therefore, it will always be a supercell.

No, the cell with the smallest number in a table cannot be a supercell. Since the smallest number in the table is less than or equal to all other numbers, it cannot be greater than its adjacent numbers. Thus, it cannot be a supercell.

Question 7.
Fill a table such that the cell having the second largest number is not a supercell.
Solution:

Question 8.
Make other variations of this puzzle and challenge your classmates.
Solution:
Fill a table such that the cell having the second smallest number is not a supercell.

ntext Question

Question 1.
Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours.

The biggest number in the table is _____________.
The smallest even number in the table is _____________.
The smallest number greater than 50,000 in the table is _____________
Solution:
Once you have filled the table above, put commas appropriately after the thousands digit.

The biggest number in the table is 96,310.
The smallest even number in the table is 10,396.
The smallest number greater than 50,000 in the table is 60,193. (Answer may vary)

3.4 Playing with Digits (Page No. 60)

Question 1.
Digit sum 14
(a) Write other numbers whose digits add up to 14.
(b) What is the smallest number whose digit sum is 14?
(c) What is the largest 5-digit whose digit sum is 14?
(d) How big a number can you form having the digit sum 14? Can you make an even biggermumber?
Solution:
(a) 590, 770, 248, 680 etc.
(b) 59, (∵ 5 + 9 = 14)
(c) 95000,(∵ 9 + 5 + 0 + 0 + 0 = 14)
(d) Infinite numbers can be formed having the digit sum 14.
Yes, we can make an bigger even number.

Question 2.
Find out the digit sums of all the numbers from 40 to 70. Share your observations with the class.
Solution:

Observation:
From 40 to 49: The digit sum increased from 4 to 13
From 50 to 59: The digit sum increased from 5 to 14
From 60 to 69: The digit sum increased from 6 to 15
From 70 to 79: The digit sum will increase from 7 to 16

Question 3.
Calculate the digit sums of 3-digit numbers whose digits are consecutive (for example, 345). Do you see a pattern? Will this pattern continue?
Solution:
1 + 2 + 3 = 6,
2 + 3 + 4 = 9,
3 + 4 + 5= 12,
4 + 5 + 6 = 15,
5 + 6 + 7 = 18,
6 + 7 + 8 = 21,
7 + 8 + 9 = 24.
Yes, the pattern is a table of 3 from 3 × 2 to 3 × 8.
We cannot continue this pattern after 789.

Intext Question

Question 1.
Among the numbers 1-100, how many times will the digit ‘7’occur?
Solution:
The total count of 7 that we get is 20.

Question 2.
Among the numbers 1-1000, how many times will the digit ‘7’occur?
Solution:
The number of times 7 will be written when listing the numbers from 1 to 1000 is 300.

Read More: Addition and Subtraction of Time

3.5 Pretty Palindromic Patterns 3.6 The Magic Number of Kaprekar 3.7 Clock and Calendar Numbers Figure it Out (Page No. 64-65)

Question 1.
Pratibha uses the digits ‘4’, ‘7’, ‘3’ and ‘2’, and makes the smallest and largest 4-digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432 – 2347 = 5085. The sum of these two numbers is 9779. Choose 4-digits to make:
(a) the difference between the largest and smallest numbers greater than 5085.
(b) the difference between the largest and smallest numbers less than 5085.
(c) the sum of the largest and smallest numbers greater than 9779.
(d) the sum of the largest and smallest numbers less than 9779.
Solution:
(a) 5, 8, 3, 2; largest 4-digit number: 8532, smallest 4- digit number: 2358 Difference: 8532 – 2358 = 6174 > 5085
(b) 4, 6, 3, 2; largest 4-digit number: 6432, smallest 4- digit number = 2346 Difference: 6432 – 2346 = 4086 < 5085
(c) 5, 8, 3, 2 → 8532 + 2358 = 10890 > 9779
(d) 4, 6, 3, 2 → 6432 + 2346 = 8778 < 9779
(Answers may vary)

Question 2.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Smallest 5-digit palindrome = 10001
Largest 5-digit palindrome = 99999
Sum of the smallest and largest 5-digit palindrome is 10001 + 99999 = 110000
The difference between the largest and smallest 5-digit palindrome is 99999 – 10001 = 89998

Question 3.
The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?
Solution:
Prime time = 10:01
Next palindromic time after 10:01 = 11:11
Difference = 11:11-10:01 = 70 minutes
Thus, the next palindromic time shows after 70 minutes.
Next palindromic time after 11 : 11=12:21
Difference = 12:21 – 11:11 = 70 minutes
= 1 hour 10 minutes
Thus, the next palindromic time shows after 1 hour 10 minutes and next one is 12 : 21 which comes total 2 hours 20 minutes later.

Question 4.
How many rounds does the number 5683 take to reach the Kaprekar constant?
Solution:
1st round,
A = 8653
B = 3568
C =8653 – 3568 = 5085

2nd round,
A = 8550
B = 0558
C = 8550 – 0558 = 7992

3rd round,
A = 9972
B = 2799
C = 9972 – 2799 = 7173

4th round,
A = 7731
B = 1377
C= 7731 – 1377 = 6354

5th round,
A = 6543
B = 3456
C= 6543 – 3456 = 3087

6th round,
A = 8730
B = 0378
C = 8730 – 0378 = 8352

7th round,
A = 8532
B = 2358
C = 8532 – 2358 = 6174
Thus, the number 5683 reaches the Kaprekar’s constant 6174 after 7 rounds.

Intext Questions

Question 1.
Write all possible 3-digit palindromes using these digits 1,2,3. (Page 61)
Solution:

• 111

• 121

• 131

• 212

• 222

• 232

• 313

• 323

• 333

Question 2.
Will reversing and adding numbers repeatedly, starting with a 2-digit number, always give a palindrome? Explore and find out. (Page 62)

Solution:

All two-digit numbers eventually become palindromes after repeated reversal and addition. About 80% of all numbers under 10,000 resolves into a palindrome ip four or fewer steps; about 90% of those resolve in seven steps or fewer.

Example 1: Number 12
1. Initial Number: 12
2. Reverse: 21

3. Add: 12 + 21 = 33

4. Palindromp

Check: 33 is a palindrome.

• Result: 33 is a palindrome.

Example 2: Number 89
1. Initial Number: 89
2. Reverse: 98
3. Add: 89 + 98 = 187
4. Palindrome Check: 187 is not a palindrome.
5. Reverse 187: 781
6. Add: 187 + 781 =968
7. Palindrome Check: 968 is not a palindrome.

Puzzle time (Page 62)

I am a 5-digit palindrome.
I am an odd number.
My ‘t’ digit is double of my ‘u’ digit.
My ‘h’ digit is double of my ‘t’ digit. Who am I?
Solution: 12421

 Read More: CBSE Class 8 Maths Formulas

Explore (Page 63)
Take different 4-digit numbers and try carrying out these steps. Find out what happens.
Solution:
Selected a 4-digit number 1234

• Starting number: 1234

• Descending order: 4321

• Ascending order: 1234

• Subtract: 4321 – 1234 = 3087

Repeat:

• Descending order of 3087: 8730

• Ascending order of 3087: 0378

• Subtract: 8730 – 0378 = 8352

Repeat:

• Descending order of 8352: 8532

• Ascending order of 8352: 2358

• Subtract: 8532 – 2358 = 6174

Result: 6174 (Kaprekar constant)

Question 1.
Carry out these same steps with a few 3-digit numbers. What number will start repeating? (Page 63)
Solution:
We will do this with help of two examples:
1. Number 123

• Starting number: 123

• Descending order: 321

• Ascending order: 123

• Subtract: 321 – 123 = 198

Repeat:

• Descending order of 198: 981

• Ascending order of 198: 189

• Subtract: 981 – 189 = 792

Repeat:

• Descending order of 792: 972

• Ascending order of 792: 279

• Subtract: 972 – 279 = 693

Repeat:

• Descending order of 693: 963

• Ascending order of 693: 369

• Subtract: 963 – 369 = 594′

Repeat:

• Descending order of 594: 954

• Ascending order of 594: 459

• Subtract: 954 – 459 = 495

Repeat:

• Descending order of 495: 954

• Ascending order of 495: 459

• Subtract: 954 – 459 = 495

Result: The number 495 starts repeating.

2. Now let’s take the number 317

• Starting number: 317

• Descending order: 731

• Ascending order: 137

• Subtract: 731 – 137 = 594

Repeat:

• Descending order of 594: 954

• Ascending order of 594: 459

• Subtract: 954 – 459 = 495 ,

Repeat:

• Descending order of 495: 954

• Ascending order of 495: 459

• Subtract: 954 – 459 = 495

Result: The number 495 starts repeating. When applying Kaprekar’s routine to 3- digit numbers, the number 495 is often reached and starts repeating. This number is known as the Kaprekar constant for 3-digit numbers.

Try and find out all possible times on a 12-hour clock of each of these types. For example, 4:44, 10:10, 12:21. (Page 64)
Solution:
01:10, 02:20, 03:30, 04:40, 05:50, 10:01, 11:11, 12:21

Find some other dates of this form from the past like 20/12/2012 where the digits ‘2’, ‘0’, ‘1 ’, and ‘2 ’ repeat in that order. (Page 64)
Solution:
11/02/2011, 22/02/2022, 01/10/2010, 10/01/2010, 02/02/2020

Read More: Quick Calculation Techniques for School Students 

3.8 Mental Math Figure it Out (Page No. 66 – 67)

Question 1.
Write an example for each of the below scenarios whenever possible.

Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.
Solution:

(Answer may vary)
For further do it yourself.

Question 2.
Always, Sometimes, Never?
Below are some statements. Think, explore and find out if each of the statements is ‘Always true’, ‘Only sometimes true’ or ‘Never true’. Why do you think so? Write your reasoning and discuss this with the class.
(a) 5-digit number + 5-digit number gives a 5-digit number
(b) 4-digit number + 2-digit number gives a 4-digit number
(c) 4-digit number + 2-digit number gives a 6-digit number
(d) 5-digit number – 5-digit number gives a 5-digit number
(e) 5-digit number – 2-digit number gives a 3-digit number
Solution:
(a) Only sometimes true.
The given statement is ‘5-digit number + 5-digit number gives a 5-digit number’. It is only sometimes true.
e.g. 10000 +10000 = 20000 i.e. 5-digit number
and 99999 + 99999 = 199998 i.e. 6-digit number

(b) The given statement is ‘4-digit number + 2-digit number gives a 4-digit number’. It is only sometimes true.
e.g. 1000 +10 = 1010 i.e. 4-digit number
and 9999 +10 = 10009 i.e. 5-digit number

(c) The given statement is ‘4-digit number + 2-digit number gives a 6-digit number’. It is never true.
e.g. 9999 + 99 = 10098 i.e. 5-digit number

(d) The given statement is ‘5-digit number – 5-digit number gives a 5-digit number’.
It is only sometimes true.
e.g. 99999 -10000 = 89999 i.e. 5-digit number
and 98765 – 94321 = 4444 i.e. 4-digit number

(e) The given statement is ‘5-digit number – 2-digit number gives a 3-digit number’. It is never true,
e.g. 10000 – 99 = 9901 i.e. 4-digit number.

3.9 Playing with Number Patterns 3.10 An Unsolved Mystery – the Collatz Conjecture! 3.11 Simple Estimation 3.12 Games and Winning Strategies Figure it Out (Page No. 72 – 73)

Question 1. How many rounds does your year of birth take to reach the Kaprekar constant?
Solution:
Do it yourself.

Question 2. We are the group of 5-digit numbers between 35000 and 75000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50000?
Solution:
The odd numbers between 35000 and 75000 are
35001, 35003, 35005, ……..74999.
Therefore, largest number = 74999,
smallest number = 35001
and closest to 50000 = 49999 or 50001

Question 3. Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then, try to get an exact number and see how close your estimate is.
Solution:
Number of holidays = 170 [estimate]
Number of holidays = 166 [exact]

Question 4. Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Solution:
Mug = 0.35 litres
Bucket = 20 litres
Overhead tank = 2000 litres.

Question 5. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670. ‘
Solution:
5 digit number = 1 8 0 0 0
3 digit number = 6 7 0
Sum = 1 8 000 + 670 = 18670

Question 6. Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number.

Solution:
Sum of No. = 5 × 1 = 5
+ 10 × 3 = 30
+ 15 × 5 = 75
+ 20 × 7 = 140 = 250
which lies between 210 and 390.

Question 7. Recall the sequence of Powers of 2 from Chapter 1, Table 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution:
The square of power of 2 is :
1,2,4, 8, 16, 32, 64

Let’s take the number’ 64 as per Collatz Conjecture

• 64 is even, divide by 2 = 32

• 32 is even, divide by 2 = 16

• 16 is even, divide by 2 = 8

• 8 is even, divide by 2 = 4

• 4 is even, divide by 2 = 2

• 2 is even, divide by 2 = 1

Hence Collatz conjecture is correct in all numbers in the power of 2 sequence.
As it is power of 2, and in Collatz Conjecture even number is divided by 2 in each step.

Question 8. Check if the Collatz Conjecture holds for the starting number 100.
Solution:
As per Collatz Conjecture rule: starts with any number; if the number is even, take half of it; if the number is odd, then multiply it by 3 and add 1; and repeat.
The sequence formed with starting number 100 is as follows:
100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
Hence, the Collatz Conjecture holds for the starting number 100.

Intext Questions

Example: In the following, there is a number pattern on +3 being followed.

Find out the sum of the numbers in each of the below figures. Should we add them one by one or can we use a quicker way? Share and discuss in class the different methods each of you used to solve these questions. (See figures, NCERT TB, Pages 67-68)

Solution:

(a) In figure (a), number 40 is repeated 12 times and number 50 is repeated 10 times

Hence sum of all numbers = 40 × 12 + 50 × 10

= 480 + 500 = 980

(b) In figure (b), 1 dot (•) is 44 times and 5 dots (•) are 20 times
Hence sum of all dots = 1 × 44 + 5 × 20 = 44 + 100 = 144

(c) In figure (c), number 32 is 32 times and number 64 is 16 times
Hence sum of all numbers = 32 × 32 + 64 × 16 = 1024 + 1024 = 2048

(d) In figure (d), 3 dots (•) are 17 times and 4 dots (•) are 18 times
Hence sum of all dots = 17 × 3 + 18 × 4 = 51 + 72 = 123

(e) In figure (e), number 15 is 22 times, number 25 is 22 times and number 35 is 22 times
Hence sum of all numbers = 15 × 22 + 25 × 22 + 35 × 22 = 330 + 550 + 770 = 1650

(f) In figure (f), number 125 is 18 times, number 250 is 8 times and number 500 is 4 times and number 1000 is one time.
Hence sum of all numbers = 125 × 18 + 250 × 8 + 500 × 4 + 1000 = 2250 + 2000 + 2000 +1000 = 7250

Also Read: NCERT Solutions for Class 6 Maths Chapter 5 Prime Time

Mastering Concepts with NCERT Class 6 Maths Chapter 3 Solutions 

To understand numbers, you need to be able to see patterns and connections between distinct values, not just count them. The answers to NCERT Class 6 Math Chapter 3 Ganita Prakash focuses on how the new curriculum teaches "Number Play," which emphasizes logical reasoning over memorization. This chapter teaches pupils the basic math concepts that they need to know to do more advanced math.

Students can use the NCERT Class 6 Maths Chapter 3 answers PDF to get step-by-step help with every problem in the book. Finding factors and multiples is a big part of division and multiplication. A factor of a number is a number that can be divided evenly by that number. A multiple of a number is a number that can be made by multiplying it by a whole number. The numbers that can be multiplied by 6 are 6, 12, 18, and so on.

Important Words in Number Play

Students need to learn specific words used in the NCERT Class 6 Maths Chapter 3 solutions Number Play in order to get through this chapter:

• Perfect Number: A perfect number is one whose elements add up to double the number.

• Prime Numbers: Numbers that only have two factors: 1 and the number itself.

• Composite Numbers: Numbers that have more than two elements.

• Even and Odd Numbers: Even numbers are those that can be divided by 2, whereas odd numbers are those that can't.

Detailed Breakdown of Exercises in Number Play

​​The NCERT Class 6 Maths Chapter 3 solutions are quite thorough, therefore they cover all the ideas. The activities are set up such that you start with simple identification and work your way up to more difficult problem-solving that involves common factors and multiples.

1. Factors and Multiples

In the beginning parts of the chapter, students have to list factors and discover the first few multiples of some numbers. This exercise is necessary for learning how numbers are put together. The answers give you a methodical way to make sure you don't overlook any factors. For example, you can check if a number is divisible by starting with 1 and going up to the square root of the number.

2. Prime and Composite Numbers

Identifying prime numbers is a crucial skill. The solutions explain the Sieve of Eratosthenes method to find all prime numbers between 1 and 100 efficiently. Students learn that '1' is neither prime nor composite, and '2' is the only even prime number.

3. Tests for Divisibility of Numbers

Divisibility rules are "shortcuts" that help determine if a large number is divisible by another without performing long division. The NCERT Class 6 Maths Chapter 3 solutions PDF download includes detailed rules for:

• Divisibility by 2: If the last digit is 0, 2, 4, 6, or 8.

• Divisibility by 3: If the sum of the digits is a multiple of 3.

• Divisibility by 4: If the number formed by the last two digits is divisible by 4.

• Divisibility by 5: If the last digit is 0 or 5.

• Divisibility by 6: If the number is divisible by both 2 and 3.

• Divisibility by 8: If the number formed by the last three digits is divisible by 8.

• Divisibility by 9: If the sum of the digits is divisible by 9.

• Divisibility by 10: If the last digit is 0.

• Divisibility by 11: If the difference between the sum of digits at odd places and even places is either 0 or divisible by 11.

4. Common Factors and Common Multiples

In this section, students learn to find numbers that divide two or more given numbers exactly (common factors) and numbers that are multiples of all given numbers (common multiples). This leads directly into the concepts of HCF and LCM.

Highest Common Factor (HCF) and Lowest Common Multiple (LCM)

The core of the ncert class 6 maths chapter 3 solutions lies in calculating the HCF and LCM. These concepts are applied in various real-life scenarios, such as finding the maximum capacity of a container or the next time several bells will ring together.

Highest Common Factor (HCF)

The HCF of two or more given numbers is the highest of their common factors. It is also known as the Greatest Common Divisor (GCD).

Example: To find the HCF of 20, 28, and 36:

1. List factors of 20: 1, 2, 4, 5, 10, 20

2. List factors of 28: 1, 2, 4, 7, 14, 28

3. List factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36

4. Common factors: 1, 2, 4.

5. The HCF is 4.

Lowest Common Multiple (LCM)

The LCM of two or more given numbers is the lowest of their common multiples. The solutions often use the prime factorization method or the division method to find the LCM efficiently.

Example: To find the LCM of 12 and 18:

1. Multiples of 12: 12, 24, 36, 48...

2. Multiples of 18: 18, 36, 54...

3. The LCM is 36.

Relationship Table: HCF and LCM

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