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NCERT Solutions for Class 6 Maths Chapter 3 - Number Play

NCERT Solutions for Class 6 Maths Chapter 3 - Number Play

Edited By Komal Miglani | Updated on Jun 18, 2025 03:00 PM IST

We are all surrounded by numbers from all sides. If we carefully look around, we will find that a world of numbers surrounds us in a whole lot of new ways, from our class, height, weight, date of birth, the scores we obtain in any game, our ranks in a race, etc. Numbers organise and simplify our lives in many different ways and contexts. We have already gained the basic knowledge of mathematical operations and performed addition, subtraction, multiplication, and division on them to solve problems related to our daily lives. In this chapter, we continue this journey and deepen our understanding of numbers and their magical applications. We at Careers360 designed this NCERT Solutions for class 6 article to provide comprehensive, step-by-step solutions for the chapter's questions in a student-friendly format.

NCERT Solutions for Class 6 Maths Chapter 3 - Number Play
NCERT Solutions for Class 6 Maths Chapter 3 - Number Play

These NCERT Solutions for Class 6 Maths are trustworthy and reliable as they are created by subject matter experts in Careers360, making it a very important and essential resource that the students can use for exam preparations. Find everything you need—the Latest syllabus, revision notes, and practice PDFs—all in one place in the following link: NCERT.

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NCERT Solutions for Class 6 Maths Chapter 3 Exercise

Page no. 57

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

1749487694487

Solution:

1749487731844

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

1749487756388

Solution:

1749487783017

(Answers may vary)

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

Solution:

1749487814835

4. Out of the 9 numbers, how many supercells are there in the table above?______________

Solution: As we can count from the table above, there are 5 supercells.

5. Find out how many supercells are possible for different numbers of cells.

Do you notice any pattern? What is the method to fill a given table to get the maximum number of supercells? Explore and share your strategy.

Solution:
If there are an odd number of cells (n), then the number of supercells are equal to the n+12.
If there are an even number of cells (n), then the number of supercells are equal to the n2.
For example:

Number of cells (n)Number of Supercells
11
21
32
42


Yes, there is a pattern in the number of supercells as described by the above formula.

Method to fill a given table to get the maximum number of supercells.
Step 1: Make the first cell a supercell. After that, each alternate cell is to be made a supercell.
Step 2: No consecutive cells can be supercell except in the case of 4 cells, because then the first and fourth cells can be supercell.

6. 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 obtain at least one supercell, i.e., the last number.

7. 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 is always a supercell, while the cell with the smallest number will never be a supercell, as its adjacent cell has a bigger number.

8. Fill a table such that the cell having the second largest number is not a supercell.

Solution:

0

1

2

3

4

5

6

7

8


The second-to-last element is not the supercell. (Answers may vary)

9. Fill a table such that the cell having the second-largest number is not a supercell, but the second-smallest number is a supercell. Is it possible?

Solution:

111

100

222

333

444

555

666

888

999


Here, 100 (the second smallest number) is a supercell while 888 (the Second largest) is not a supercell.

it is possible to make such cells.(Answers may vary)

10. Make other variations of this puzzle and challenge your classmates.

Solution: Do It Yourself by taking different scenarios.

Page No. : 59

Identify the numbers marked on the number lines below, and label the remaining positions.

1749488357518

Put a circle around the smallest number and a box around the

largest number in each of the sequences above.

Solution:

(a)

1749488411193

(b) 1749488436583

(c) 1749488456127

(d) 1749488475774

Page No. : 60

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 number whose digit sum is 14?

d. How big a number can you form having the digit sum of 14? Can you make an even bigger number?

Solution:

a. 95,59,86,68,923,.......... etc. (Answers may vary)

b. 59

c. 95000

d. 101010101010101010101010101,11111111111111,.......... so on.

Yes, we can make even bigger numbers.

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

Solution:

NumberDigit SumNumberDigit SumNumberDigit Sum
404516628
415527639
4265386410
4375496511
44855106612
45956116713
461057126814
471158136915
48125914707
4913606

505617


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, it follows the table of 3. We can’t continue this series after 789.

Page no. 64

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 74322347=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. Digits 8,7,3 and 2

Largest Number - 8732

Smallest Number - 2378

Difference =6354

6354>5085

b. Digits 1,2,3 and 4

Largest Number - 4321

Smallest Number - 1234

Difference =3087

c. Digits 9,8,7 and 6

Largest Number - 9876

Smallest Number - 6789

Sum =16665, which is greater than 9779.

d. Digits - 1, 2, 3, and 8

Largest Number 8321

Smallest Number 1238

Sum = 9559, which is less than 9779.

2. What is the sum of the smallest and largest 5-digit palindromes? What is the difference?

Solution:

Smallest 5-digit palindrome number = 10001

Largest 5-digit palindrome number = 99999

Sum =10001+99999=110000

Difference = 9999910001

=89998

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: The Next palindromic time after 10:01 is 11:11, after 70 minutes. After that, the next palindromic time will be 12:21.

4. How many rounds does the number 5683 take to reach the Kaprekar constant?

Solution:

Starting with 5683, arrange the digits in descending and ascending order. Subtract the smaller from the larger repeatedly:
8653 − 3568 = 5085
8550 − 0558 = 7992
9972 − 2799 = 6174.
It takes three rounds.

Page no. 66

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

1749489071851

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: Do it Yourself

{Hint: Answers may vary. See the hint and try some more examples

  1. 50000 + 30000 = 80000

  2. 99900 + 100 = 100000

  3. 9999 + 9999 = 19998. This is the biggest number that can be achieved by adding 4-digit numbers. So, there is no 6-digit number possible by the sum of two 4-digit numbers.

  4. 99999 + 10000 = 109999

  5. 30000 + (-11500) = 18500

  6. 65503 - 10000 = 55503

  7. 10000 - 999 = 9001

  8. 10000 - 1000 = 9000

  9. 99933 - 99000 = 933

  10. 81500 - (-10000) = 91500

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 [Example: 90000 + 10000=100000]

b. Only Sometimes[Example: 9999+10=10009]

c. That is never true, even if we took the largest four-digit number and the largest two-digit number.

Example: 9999 + 99 = 10098

d. Only Sometimes[Example: 99999-99000=999]

e. That is never true, even if we took the smallest five-digit number and the smallest two-digit number.

Example: 10000 - 99 = 9901

Page no: 69

We shall do some simple estimates. It is a fun exercise, and you may find it amusing to know the various numbers around us. Remember, we are not interested in the exact numbers for the following questions.

Share your methods of estimation with the class.

1. Steps you would take to walk:

a. From the place you are sitting to the classroom door

b. Across the school ground from start to end

c. From your classroom door to the school gate

d. From your school to your home

2. Number of times you blink your eyes or number of breaths you take:

a. In a minute

b. In an hour

c. In a day

3. Name some objects around you that are:

a. a few thousand in number

b. more than ten thousand in number

Solution (1,2,3): Do it yourself as per the instructions given.

Page no: 70

Try to guess within 30 seconds. Check your guess with your friends.

1. Number of words in your maths textbook:

a. More than 5000

b. Less than 5000

Solution: a. More than 5000 (Number of words in a line multiplied by the number of lines in a page multiplied by the number of pages in the book)

2. Number of students in your school who travel to school by bus:

a. More than 200

b. Less than 200

Solution: a. More than 200 (Number of seats multiplied by the number of students in one seat)

3. Roshan wants to buy milk and 3 types of fruit to make fruit custard for 5 people. He estimates the cost to be ₹100. Do you agree with him? Why or why not?

Solution: No, because the milk will cost him somewhere between 25-35 Rupees, after that custard for 5 people will cost him more than 150 or more. Therefore, the amount is not sufficient.

4. Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland).

Hint: Look at the map of India to locate these cities.

Solution: Gandhinagar (in Gujarat) is in the west of India, and Kohima (in Nagaland) is in the Northeast of India. Therefore, the approximate distance is 2500 km to 2800 km.

5. Sheetal is in Grade 6 and says she has spent around 13,000 hours in school to date. Do you agree with her? Why or why not?

Solution: Yes, because on average, a student spends 8 hours a day in school. In an academic year, it is approximately equal to 2000 hours in each class. So, till class 6, a student spends more than 13000 hours.

6. Earlier, people used to walk long distances as they had no other means of transport. Suppose you walk at your normal pace. Approximately, how long would it take you to go from:

a. Your current location to one of your favourite places nearby.

b. Your current location to any neighbouring state's capital city.

c. The southernmost point in India to the northernmost point in India.

Solution: Do it yourself

7. Make some estimation questions and challenge your classmates!

Solution: Do it Yourself

Page No: 72

1. There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap.

1749489526669

Solution:

1749489601555

2. How many rounds does your year of birth take to reach the Kaprekar constant?

Solution:
Let yout year of birth is 2009, here's how the process looks:

  1. 2009 -> 9002 - 0029 = 8973
  2. 8973 -> 9873 - 3789 = 6084
  3. 6084 -> 8640 - 0468 = 8172
  4. 8172 -> 8721 - 1278 = 7443
  5. 7443 -> 7443 - 3447 = 3996
  6. 3996 -> 9963 - 3699 = 6264
  7. 6264 -> 6642 - 2466 = 4176
  8. 4176 -> 7614 - 1467 = 6147 (8 rounds)

3. We are a group of 5-digit numbers between 35,000 and 75,000 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 50,000?

Solution: The possible odd digits are: 1,3,5,7 and 9.

  • Largest number: 73,999

  • Smallest number: 35,111

  • Closest to 50,000: 51,111

4. 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: Holidays in a year:

  • Weekends: 104 days

  • Summer vacation: 50 days

  • Winter vacation: 12 days

  • Festival holidays: 12 days

  • Public holidays: 4 days

Total estimated holidays =104+50+12+12+4=182 days.

Now compare this result with the calendar of current year.

5. Estimate the number of litres a mug, a bucket and an overhead tank can hold.

Solution: A mug can hold approximately 0.8 litres of water, whereas a bucket can hold 10 litres of water, and an overhead tank can hold 1000 litres of water.

6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.

Solution: The combination of a 5-digit number and two 3-digit numbers that add up to 18,670:

5-digit number: 12,345

3-digit number 1: 789

3-digit number 2: 536

Sum: 12,345+789+536=18,670

7. 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: We can create a simple addition pattern using a series of repeated numbers.

50+50+50+55+55+55+60=375

The pattern uses 50 repeated 3 times, 55 repeated 3 times and then a 60 to complete the sum.

50+50+50=150

55+55+55=165

150+165+60=375

8. 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 Collatz Conjecture and Powers of 2: The Collatz conjecture suggests that starting with any positive integer, repeatedly applying the following steps will eventually lead to the number 1:

If the number is even, divide it by 2.

If the number is odd, multiply it by 3 and add 1.

For powers of 2 (like 1, 2, 4, 8, 16, etc., as listed in Table 1 of Chapter 1), the Collatz conjecture holds because every power of 2 is even, so It always divide by 2.

Dividing powers of 2 by 2 continuously leads to smaller powers of 2, eventually reaching 1. This explains why the Collatz conjecture is correct for all starting numbers in the sequence of powers of 2.

9. Check if the Collatz Conjecture holds for the starting number 100.

Solution: Checking: Applying the steps of the Collatz Conjecture to 100:

100 is even: 1002=50

50 is even: 502=25

25 is odd: 25×3+1=76

76 is even: 762=38

38 is even: 382=19

19 is odd: 19×3+1=58

58 is even: 582=29

29 is odd: 29×3+1=88

88 is even: 882=44

44 is even: 442=22

22 is even: 222=11

11 is odd: 11×3+1=34

34 is even: 342=17

17 is odd: 17×3+1=52

52 is even: 522=26

26 is even: 262=13

13 is odd: 13×3+1=40

40 is even: 402=20

20 is even: 202=10

10 is even: 102=5

5 is odd: 5×3+1=16

16 is even: 162=8

8 is even: 82=4

4 is even: 42=2

2 is even: 22=1

After these steps, we reach 1, confirming that the Collatz Conjecture holds for the starting number 100.

10. Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?

Solution: Do It Yourself

NCERT Solutions for Class 6 Maths Chapter 3 Number Play - Notes

Arrangement of Numbers

The numbers can be arranged in two ways, the ascending and descending orders. If the numbers are arranged from smallest to largest, then they are in ascending order. If the numbers are arranged from largest to smallest, then it is in descending order.

For example,

Numbers: 3,7,8,9
Ascending order: 3,7,8,9
Descending order: 9,8,7,3

Supercells

A number is a supercell if it is greater than the adjacent numbers in the grid.

Observe the numbers written below. We observe that some numbers are coloured.

200 577 626 345 790 694 109 198

43 79 75 63 10 29 28 34

A cell is coloured if the number in it is larger than its adjacent cells. The number 626 is coloured as it is larger than 577 and 345, whereas 200 is not coloured as it is smaller than 577. The number 198 is coloured as it has only one adjacent cell with 109 in it, and 198 is larger than 109.

Patterns of Numbers on the Number Line

We can also formulate some patterns on the number line.

1749490652620

Digit sums of numbers

In certain numbers, the sum of their digits is the same.

For example, adding the digits of the number 68 will be the same as adding the digits of 176 or 545.

Pretty Palindromic Patterns

Palindromic numbers are numbers that remain the same when written or read from front to back or back to front.

1749490611236

Let us consider few numbers: 33, 848, 575, 454, 22, 88, etc. If we look carefully, we will surely figure out that these numbers read the same from left to right and from right to left.

Clock and Calendar Numbers

According to the chapter number play, on a 12-hour clock, certain times have interesting patterns, like 4:44, 10:10, and 12:21.

Examples

  • Rekha's birthday is on 20/12/2012, where the digits ‘2’, ‘0’, ‘1’, and ‘2’ repeat in that order.
  • Rahul's birthday is on 11/02/2011, where the digits read the same from left to right and from right to left.

An Unsolved Mystery — The Collatz Conjecture

As stated in the chapter on number play, the Collatz Conjecture is a famous unsolved problem in mathematics. The rule is simple:

  1. Start with any number.
  2. If the number is even, take half of it.
  3. If the number is odd, multiply it by 3 and add 1.
  4. Repeat.

Eventually, all sequences will reach the number 1, regardless of the whole number you start with.

For example, 21, 64, 32, 16, 8, 4, 2, 1

Number Play Class 6 Maths Chapter 3 - Topics

  • Supercells
  • Patterns of numbers on the number line
  • Playing with digits
  • Palindromic patterns
  • Clock and calendar numbers
  • Mental health
  • Playing with number patterns
  • Simple estimation
  • Games and winning strategies

NCERT Solutions for Class 6 Maths Chapter 4 Data Handling and Presentation - Points to Remember

Ascending order: The arrangement of numbers from low to high.

Descending order: The arrangement of numbers from high to low.

Supercells: A number is a supercell if it is greater than the adjacent numbers in the grid.

Palindromic numbers: Palindromic numbers are numbers that remain the same when written or read from front to back or back to front.

NCERT Solutions for Class 6 Maths Chapter Wise

NCERT Solutions for Class 6 Subject-wise

Students can refer to other resources of NCERT Class 6, like subject-wise NCERT Solutions, here.

Also read,

Frequently Asked Questions (FAQs)

1. Write the Topics that are covered in NCERT Solution Class 6 Maths chapter 3.

Here are the topics covered in the NCERT Class 6 Maths Chapter 3

  • Supercells
  • Patterns of numbers on the number line
  • Playing with digits
  • Palindromic patterns
  • Clock and calendar numbers 
  • Mental health
  • Playing with number patterns
  • Simple estimation
  • Games and winning strategies

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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