---
title: "NCERT Solutions for Class 6 Maths Chapter 3 Number Play (2026-27)"
url: https://www.swavid.com/maths/class/6/chapter/number-play/ncert-solutions
dateModified: 2026-10-07T14:51:26+00:00
---

# NCERT Solutions for Class 6 Maths Chapter 3 Number Play (2026-27)

This chapter's questions cover various number-based puzzles, patterns, estimation problems, and game strategies designed to build computational thinking.

Free PDF (21 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-6/swavid-ncert-solutions-class-6-maths-chapter-3-number-play-f2cdcd5ec7.pdf

## Math Talk

### Question 1

*3 marks · Short answer*

Think about various situations where we use numbers. List five different situations in which numbers are used. See what your classmates have listed, share, and discuss.

**Solution**

1. Identify five common situations where numbers are used in daily life.
2. List time, calendar, counting objects, measurement of height and weight, and money.
3. Conclude that numbers help us organise our lives in various contexts.

**Answer:** Five different situations in which numbers are used are time, calendar, counting objects, measurement of height and weight, and money.

> Common mistake: Listing situations that do not actually involve numbers.

## Try answering the questions below and share your reasoning.

### Question 1

*3 marks · Short answer*

Can the children rearrange themselves so that the children standing at the ends say '2'?

**Solution**

1. A child at either end of the line has only one neighbour standing next to them.
2. A child says '2' only if both neighbouring cells have taller children standing next to them.
3. Therefore, a child standing at the end cannot have two neighbours, making it impossible to say '2'.

**Answer:** No, because a child at the end has only one neighbour and cannot have two taller neighbours.

> Common mistake: Forgetting that end positions have only one adjacent neighbour instead of two.

### Question 2

*3 marks · Short answer*

Can we arrange the children in a line so that all would say only 0s?

**Solution**

1. A child says '0' if neither of the children standing next to them are taller.
2. If all children in the line are of the exact same height, no child has a taller neighbour.
3. Thus, every child will have zero taller neighbours and will say '0'.

**Answer:** Yes, if all the children in the line are of the same height.

> Common mistake: Assuming all children must be of different heights as given in the main puzzle setup.

### Question 3

*3 marks · Short answer*

Can two children standing next to each other say the same number?

**Solution**

1. Observe the arrangement of children in the figure on page 55 of the textbook.
2. Adjacent children can have the same number of taller neighbours based on their relative heights.
3. Therefore, two children standing next to each other can say the same number.

**Answer:** Yes, two children standing next to each other can say the same number.

> Common mistake: Assuming numbers spoken by adjacent children must always alternate.

### Question 4

*3 marks · Short answer*

There are 5 children in a group, all of different heights. Can they stand such that four of them say '1' and the last one says '0'? Why or why not?

**Solution**

1. Let the 5 children stand in ascending order of height from left to right.
2. Each of the first four children will have exactly one taller neighbour standing to their right, so each says '1'.
3. The last child at the right end has no one taller to their right and no neighbour on the other side is taller, so the last child says '0'.

**Answer:** Yes, they can, if they stand in ascending order of height.

> Common mistake: Not checking the condition for the last child in an ascending arrangement.

### Question 5

*3 marks · Short answer*

For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?

**Solution**

1. A child says '1' if there is only one taller child standing next to them.
2. In a line of 5 children, the tallest child stands at one of the ends and has no one taller next to them.
3. Therefore, the tallest child cannot say '1', making the sequence 1, 1, 1, 1, 1 impossible.

**Answer:** No, because the tallest child at the end cannot say '1'.

> Common mistake: Ignoring the constraint that the tallest person at the extreme end cannot have a taller neighbour.

### Question 6

*3 marks · Short answer*

Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?

**Solution**

1. A sequence of numbers 0, 1, 2, 1, 0 means the heights of the 5 children have a specific pattern.
2. This is possible with an arrangement of heights such as shortest, medium-short, tallest, medium-tall, shortest in terms of neighbours.
3. An arrangement with heights like 2nd shortest, shortest, tallest, 2nd tallest, middle height can yield this sequence.

**Answer:** Yes, it is possible with a proper arrangement of heights.

> Common mistake: Assuming symmetric number sequences require symmetric height patterns.

### Question 7

*3 marks · Short answer*

How would you rearrange the five children so that the maximum number of children say '2'?

**Solution**

1. A child says '2' when both neighbours are taller than that child.
2. In a line of 5 children of different heights, the tallest child cannot say '2' because no one is taller than them.
3. By arranging heights in a wave-like pattern (for example, short, tall, short, tall, medium), at most 2 children can have both neighbours taller than themselves.

**Answer:** At most 2 children can say '2' by arranging them in an alternating height pattern.

> Common mistake: Thinking more than two children can say '2' without violating the neighbour heights.

## Figure it Out

### Question 1

*Activity*

Colour or mark the supercells in the table below.
6828 670 9435 3780 3708 7308 8000 5583 52

**Solution**

1. A cell becomes a supercell if the number in it is larger than its adjacent cells.
2. Compare each cell with its left and right neighbours to find the supercells.

**Answer:** The supercells in the given table are 6828, 9435, 7308, 8000, and 52.

### Question 2

*3 marks · Short answer*

Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.
5346 [ ] [ ] 1258 [ ] [ ] [ ] 9635 [ ]

**Solution**

1. A cell is a supercell if it is greater than its adjacent cells.
2. Choose numbers for the empty cells such that the coloured cells are strictly greater than their neighbours and uncoloured cells are not.
3. One possible filled table is: 5346, 5347, 1000, 1258, 1100, 1200, 1300, 9635, 9636.

**Answer:** 5346, 5347, 1000, 1258, 1100, 1200, 1300, 9635, 9636

> Common mistake: Making an uncoloured cell larger than its adjacent cells.

### Question 3

*3 marks · Short answer*

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

**Solution**

1. To get the maximum number of supercells, start by filling the first cell as a supercell with a large number.
2. Fill the remaining cells alternately with smaller numbers and larger numbers without repetition.
3. One possible arrangement using numbers between 100 and 1000 is: 110, 100, 150, 130, 280, 200, 230, 210, 270.

**Answer:** 110, 100, 150, 130, 280, 200, 230, 210, 270

> Common mistake: Repeating numbers or failing to make alternate cells larger than their neighbours.

### Question 4

*1 mark · Fill in the blank*

Out of the 9 numbers, how many supercells are there in the table above? ___________

**Solution**

1. Count the total number of supercells marked in the table from the earlier activity.

**Answer:** 5

> Common mistake: Counting only the internal supercells and missing the end cells.

### Question 5

*3 marks · Short answer*

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**

1. For an even number of cells ($2, 4, 6, \dots$), the maximum number of supercells is half the total number of cells.
2. For an odd number of cells ($1, 3, 5, 7, \dots$), the maximum number of supercells is $(n+1)/2$.
3. The strategy to achieve this is to start with a supercell in the first position and fill the table by alternating larger and smaller numbers.

**Answer:** For $n$ cells, the maximum number of supercells is $n/2$ when $n$ is even and $(n+1)/2$ when $n$ is odd.

> Common mistake: Confusing the formulas for even and odd numbers of cells.

## Try This

### Question 6

*3 marks · Short answer*

Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?

**Solution**

1. No, it is not possible to fill a supercell table without repeating numbers such that there are no supercells.
2. A cell becomes a supercell if the number in it is greater than all its adjacent neighbouring cells.
3. The cell which contains the greatest number among all the chosen numbers will always be greater than its neighbours and thus will become a supercell, irrespective of its position in the table.

**Answer:** No, because the largest number in the table will always form a supercell.

> Common mistake: Thinking that placing the largest number at a corner or edge can prevent it from being a supercell.

### Question 7

*3 marks · Short answer*

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**

1. Yes, the cell having the largest number in a table will always be a supercell because its value is greater than all other numbers, including its neighbours.
2. No, the cell having the smallest number in a table can never be a supercell.
3. This is because the numbers in all the adjacent neighbouring cells will always be greater than the smallest number.

**Answer:** The largest number is always a supercell, but the smallest number can never be a supercell.

> Common mistake: Assuming the smallest number can be a supercell if it is placed at a corner.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. To ensure the cell having the second largest number is not a supercell, it must have an adjacent neighbour containing the largest number.
2. Consider a table of 9 cells placed in a row with numbers: $1, 2, 3, 4, 5, 6, 7, 9, 8$.
3. Here, the largest number is $9$, and the second largest number is $8$. Since $9$ is adjacent to $8$, the number $8$ is not greater than all its neighbours, so it is not a supercell.

**Answer:** One valid table arrangement is $1, 2, 3, 4, 5, 6, 7, 9, 8$ where $8$ is adjacent to $9$.

> Common mistake: Placing the second largest number away from the largest number, making it a supercell.

### Question 9

*3 marks · Short answer*

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**

1. Yes, it is possible to fill a table such that the second largest number is not a supercell but the second smallest number is a supercell.
2. Consider a 9-cell table arranged as: $2, 1, 3, 4, 5, 6, 7, 9, 8$.
3. In this arrangement, the second smallest number is $2$ (which is a supercell as it is greater than its neighbour $1$), and the second largest number is $8$ (which is not a supercell as it is adjacent to $9$).

**Answer:** Yes, it is possible with the arrangement $2, 1, 3, 4, 5, 6, 7, 9, 8$.

> Common mistake: Forgetting to check the neighbour conditions for both the second smallest and second largest numbers.

### Question 10

*3 marks · Short answer*

Make other variations of this puzzle and challenge your classmates.

**Solution**

1. We can create variations by changing the number of cells or modifying the conditions for supercells.
2. Variation 1: Can you fill a table with $9$ cells such that there are exactly $4$ supercells?
3. Variation 2: Can you fill a table with $9$ cells using consecutive numbers such that all odd-positioned cells are supercells?

**Answer:** Examples of variations include asking for an exact number of supercells or restricting the choice of numbers.

> Common mistake: Creating variations that are mathematically impossible according to supercell rules.

## Figure it Out

### Question 1

*3 marks · Short answer*

Identify the numbers marked on the number lines below, and label the remaining positions.
a. Number line starting from 1990 to 2035 with marks at 2010 and 2020
b. Number line showing 9996 and 9997
c. Number line showing 15,077, 15,078 and 15,083
d. Number line showing 86,705 and 87,705
Put a circle around the smallest number and a box around the largest number in each of the sequences above.

**Part (a) (1 mark)**

1. The number line goes from 1990 to 2035 with intervals of 5 units.
2. Marked numbers from left to right: 1990, 1995, 2000, 2005, 2010, 2015, 2020, 2025, 2030, 2035.
3. Smallest number: (1990); Largest number: [2035].

Answer (a): Sequence: (1990), 1995, 2000, 2005, 2010, 2015, 2020, 2025, 2030, [2035]

**Part (b) (1 mark)**

1. The number line shows consecutive integers with intervals of 1 unit.
2. Marked numbers from left to right: 9993, 9994, 9995, 9996, 9997, 9998, 9999, 10000, 10001, 10002.
3. Smallest number: (9993); Largest number: [10002].

Answer (b): Sequence: (9993), 9994, 9995, 9996, 9997, 9998, 9999, 10000, 10001, [10002]

**Part (c) (0.5 marks)**

1. The number line shows consecutive integers with intervals of 1 unit.
2. Marked numbers from left to right: 15077, 15078, 15079, 15080, 15081, 15082, 15083, 15084, 15085, 15086.
3. Smallest number: (15077); Largest number: [15086].

Answer (c): Sequence: (15077), 15078, 15079, 15080, 15081, 15082, 15083, 15084, 15085, [15086]

**Part (d) (0.5 marks)**

1. The number line goes from 83,705 to 92,705 with intervals of 1,000 units.
2. Marked numbers from left to right: 83705, 84705, 85705, 86705, 87705, 88705, 89705, 90705, 91705, 92705.
3. Smallest number: (83705); Largest number: [92705].

Answer (d): Sequence: (83705), 84705, 85705, 86705, 87705, 88705, 89705, 90705, 91705, [92705]

**Answer:** Identified numbers and marked positions for all four number lines, with smallest numbers circled and largest numbers boxed.

> Common mistake: Misinterpreting the scale or interval value between consecutive tick marks on the number line.

## Figure it Out

### Question 1

*3 marks · Short answer*

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 of 14? Can you make an even bigger number?

**Part a (1 mark)**

1. Find numbers whose individual digits add up to 14.
2. Examples of such numbers include 248, 653, and 356.

Answer a: 248, 653, 356, 815, 833

**Part b (1 mark)**

1. To make the smallest number with a given digit sum, use the fewest digits and place smaller digits at higher place values.
2. Placing 5 in the tens place and 9 in the ones place gives $5 + 9 = 14$.

Answer b: 59

**Part c (1 mark)**

1. To make the largest 5-digit number with digit sum 14, place the largest possible digit (9) in the highest place value (ten thousands place).
2. The remaining sum is $14 - 9 = 5$, which is placed in the thousands place, and zeros are placed in the remaining places to make a 5-digit number.

Answer c: 95,000

**Answer:** Refer to the parts below.

> Common mistake: Confusing smallest number with digit sum 14 by putting 9 first instead of 5.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Find the digit sum of each number from 40 to 70 by adding its digits.
2. For tens numbers like 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, the digit sums are 4, 5, 6, 7, 8, 9, 1, 2, 3, 4.
3. For 50 to 70, the digit sums follow the repeating cyclical pattern of increasing consecutively and resetting when the sum of digits reaches a multiple of 9.

**Answer:** The digit sums range between 4 and 13, showing repeating patterns and sequences.

> Common mistake: Making addition errors while calculating digit sums for numbers containing 9.

### Question 3

*3 marks · Short answer*

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. Write the digit sums for consecutive 3-digit numbers: $123 \rightarrow 1+2+3 = 6$, $234 \rightarrow 2+3+4 = 9$, $345 \rightarrow 3+4+5 = 12$, $456 \rightarrow 4+5+6 = 15$, $567 \rightarrow 5+6+7 = 18$, $678 \rightarrow 6+7+8 = 21$, and $789 \rightarrow 7+8+9 = 24$.
2. Observe the pattern: all the calculated digit sums are consecutive multiples of 3 ($6, 9, 12, 15, 18, 21, 24$), where each sum is 3 more than the previous sum.
3. The pattern will continue as long as the consecutive digits form a valid 3-digit number (up to $789$), after which consecutive 3-digit numbers like $891$ have digits that are no longer strictly consecutive in increasing order.

**Answer:** Yes, there is a pattern where all the digit sums are multiples of 3, but the pattern stops when consecutive digits exceed $789$.

> Common mistake: Thinking that the pattern of consecutive increasing digits can continue past 789 to numbers like 891 where the digits are not consecutive.

## Digit Detectives

### Question 1

*3 marks · Short answer*

Among the numbers 1–100, how many times will the digit '7' occur? Among the numbers 1–1000, how many times will the digit '7' occur?

**Solution**

1. For the numbers 1 to 100, the digit '7' appears in the units place 10 times (7, 17, 27, 37, 47, 57, 67, 77, 87, 97) and in the tens place 10 times (70, 71, 72, 73, 74, 75, 76, 77, 78, 79), making a total of $10 + 10 = 20$ times.
2. For the numbers 1 to 1000, the digit '7' appears 100 times in each of the hundreds blocks (001–100, 101–200, ..., 601–700, 801–900, 901–1000) except for the 700-series (701–800) where it appears $100 + 100 = 200$ times due to the leading digit 7.
3. Thus, the total number of times the digit '7' occurs in numbers from 1 to 1000 is $9 \times 100 + 200 = 300$ times.

**Answer:** The digit '7' occurs 20 times among numbers 1–100, and 300 times among numbers 1–1000.

> Common mistake: Forgetting to count the occurrences where the digit appears in both the tens and units place (like 77) or miscounting the hundreds blocks.

## All palindromes using 1, 2, 3

### Question 1

*3 marks · Short answer*

Write all possible 3-digit palindromes using these digits.

**Solution**

1. A palindrome reads the same from left to right and from right to left.
2. The 3-digit palindromes using digits 1, 2, and 3 are formed by choosing the first digit and last digit to be the same, with the middle digit placed in between.
3. The complete list of all possible 3-digit palindromes using the digits '1', '2', and '3' is: 111, 121, 131, 212, 222, 232, 313, 323, and 333.

**Answer:** 111, 121, 131, 212, 222, 232, 313, 323, 333

> Common mistake: Forgetting combinations like 212 or 232 where the outer and inner digits differ.

## Puzzle time

### Question 1

*3 marks · Short answer*

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**

1. Let the 5-digit number be represented by the digits tth, th, h, t, u.
2. Since it is a palindrome, the first digit (tth) must equal the last digit (u), and the second digit (th) must equal the fourth digit (t).
3. The number is odd, so the last digit (u) must be 1, 3, 5, 7, or 9.
4. Given that the 't' digit is double the 'u' digit, if u = 1, then t = 2. Since t = th, the number looks like 12h21.
5. Given that the 'h' digit is double the 't' digit, h = 2 × 2 = 4.
6. Substituting these values, the number is 12,421.

**Answer:** The number is 12,421.

> Common mistake: Confusing the place value labels (tth, th, h, t, u) or failing to apply the palindrome property correctly.

## Try This

### Question 1

*3 marks · Short answer*

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.

**Part (a) (1 mark)**

1. Form a 4-digit largest number and a 4-digit smallest number using digits 1, 3, 4, 7 such that their difference exceeds 5085.
2. Largest number = 7431, Smallest number = 1347.
3. Difference = 7431 - 1347 = 6084, which is greater than 5085.

Answer (a): 7431 - 1347 = 6084

**Part (b) (1 mark)**

1. Form numbers using digits 2, 3, 4, 7 such that their difference is less than 5085.
2. Largest number = 7433, Smallest number = 3347.
3. Difference = 7433 - 3347 = 4086, which is less than 5085.

Answer (b): 7433 - 3347 = 4086

**Part (c) (0.5 marks)**

1. Form numbers using digits 2, 3, 4, 7 such that their sum is greater than 9779.
2. Largest number = 7433, Smallest number = 3347.
3. Sum = 7433 + 3347 = 10,780, which is greater than 9779.

Answer (c): 7433 + 3347 = 10,780

**Part (d) (0.5 marks)**

1. Form numbers using digits 1, 3, 4, 7 such that their sum is less than 9779.
2. Largest number = 7431, Smallest number = 1347.
3. Sum = 7431 + 1347 = 8778, which is less than 9779.

Answer (d): 7431 + 1347 = 8778

**Answer:** One possible set of answers is given for each part: (a) 7431 - 1347 = 6084, (b) 7433 - 3347 = 4086, (c) 7433 + 3347 = 10,780, (d) 7431 + 1347 = 8,778

> Common mistake: Selecting digits that do not meet the strict inequality conditions for sum or difference.

### Question 2

*3 marks · Short answer*

What is the sum of the smallest and largest 5-digit palindrome? What is their difference?

**Solution**

1. The smallest 5-digit palindrome is 10,001.
2. The largest 5-digit palindrome is 99,999.
3. Sum of the smallest and largest 5-digit palindrome = 10,001 + 99,999 = 1,10,000.
4. Difference between the largest and smallest 5-digit palindrome = 99,999 - 10,001 = 89,998.

**Answer:** Sum = 1,10,000; Difference = 89,998

> Common mistake: Forgetting that a palindrome reads the same backwards and forwards, leading to incorrect number selection.

### Question 3

*3 marks · Short answer*

The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?

**Solution**

1. The current time is 10:01.
2. The next palindromic time on a 12-hour clock is 11:11, which takes 1 hour and 10 minutes (70 minutes).
3. The palindromic time after that is 12:21, which occurs after 2 hours and 20 minutes (140 minutes) from 10:01.

**Answer:** 70 minutes for the next palindromic time; 140 minutes for the one after that

> Common mistake: Calculating time differences incorrectly by treating hours as base-10 instead of base-60.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. Given number: $5683$
2. Round 1: Largest number $= 8653$, Smallest number $= 3568$, Difference $= 8653 - 3568 = 5085$
3. Round 2: Largest number $= 8550$, Smallest number $= 5058$, Difference $= 8550 - 5058 = 3492$
4. Round 3: Largest number $= 9432$, Smallest number $= 2349$, Difference $= 9432 - 2349 = 7083$
5. Round 4: Largest number $= 8730$, Smallest number $= 3078$, Difference $= 8730 - 3078 = 5652$
6. Round 5: Largest number $= 6552$, Smallest number $= 2556$, Difference $= 6552 - 2556 = 3996$
7. Round 6: Largest number $= 9963$, Smallest number $= 3699$, Difference $= 9963 - 3699 = 6264$
8. Round 7: Largest number $= 6642$, Smallest number $= 2466$, Difference $= 6642 - 2466 = 4176$
9. Round 8: Largest number $= 7641$, Smallest number $= 1467$, Difference $= 7641 - 1467 = 6174$

**Answer:** It takes 8 rounds to reach the Kaprekar constant 6174.

> Common mistake: Making subtraction or rearrangement errors while finding the largest and smallest numbers in each round.

## Figure it Out

### Question 1

*3 marks · Short answer*

Write an example for each of the below scenarios whenever possible.
- 5-digit + 5-digit to give a 5-digit sum more than 90,250
- 5-digit + 3-digit to give a 6-digit sum
- 4-digit + 4-digit to give a 6-digit sum
- 5-digit + 5-digit to give a 6-digit sum
- 5-digit + 5-digit to give 18,500
- 5-digit – 5-digit to give a difference less than 56,503
- 5-digit – 3-digit to give a 4-digit difference
- 5-digit – 4-digit to give a 4-digit difference
- 5-digit – 5-digit to give a 3-digit difference
- 5-digit – 5-digit to give 91,500

**Part (a) (0.3 marks)**

1. We need a 5-digit sum greater than 90,250.
2. Example: $45,000 + 45,400 = 90,400$.

Answer (a): $45,000 + 45,400 = 90,400$

**Part (b) (0.3 marks)**

1. We need a 5-digit and a 3-digit number to give a 6-digit sum.
2. Example: $99,999 + 999 = 100,998$.

Answer (b): $99,999 + 999 = 100,998$

**Part (c) (0.3 marks)**

1. We need a 4-digit + 4-digit sum to be 6-digits.
2. This is not possible as the sum of the greatest 4-digit numbers ($9999 + 9999$) is $19,998$, which has 5 digits.

Answer (c): Not possible

**Part (d) (0.3 marks)**

1. We need a 5-digit + 5-digit sum to give a 6-digit sum.
2. Example: $60,000 + 40,000 = 1,00,000$.

Answer (d): $60,000 + 40,000 = 1,00,000$

**Part (e) (0.3 marks)**

1. We need a 5-digit + 5-digit sum to give $18,500$.
2. This is not possible because the smallest 5-digit number is $10,000$, and $10,000 + 10,000 = 20,000$.

Answer (e): Not possible

**Part (f) (0.3 marks)**

1. We need a 5-digit minus 5-digit difference less than $56,503$.
2. Example: $80,000 - 50,000 = 30,000$.

Answer (f): $80,000 - 50,000 = 30,000$

**Part (g) (0.3 marks)**

1. We need a 5-digit minus 3-digit difference to give a 4-digit difference.
2. Example: $10,000 - 999 = 9,001$.

Answer (g): $10,000 - 999 = 9,001$

**Part (h) (0.3 marks)**

1. We need a 5-digit minus 4-digit difference to give a 4-digit difference.
2. Example: $12,000 - 2,500 = 9,500$.

Answer (h): $12,000 - 2,500 = 9,500$

**Part (i) (0.3 marks)**

1. We need a 5-digit minus 5-digit difference to give a 3-digit difference.
2. Example: $50,999 - 50,000 = 999$.

Answer (i): $50,999 - 50,000 = 999$

**Part (j) (0.3 marks)**

1. We need a 5-digit minus 5-digit difference to give $91,500$.
2. This is not possible because the maximum possible difference between two 5-digit numbers is $99,999 - 10,000 = 89,999$.

Answer (j): Not possible

**Answer:** Examples for possible scenarios provided in parts.

> Common mistake: Stating that impossible combinations are possible without checking number digit limits.

### Question 2

*3 marks · Short answer*

Always, Sometimes, Never?
Below are some statements. Think, explore and find out if each of the statement 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

**Part a (0.6 marks)**

1. Consider the addition: $20,000 + 80,000 = 1,00,000$, which is a 6-digit number.
2. Smaller 5-digit numbers give a 5-digit sum, so the statement holds only sometimes.

Answer a: Only sometimes true

**Part b (0.6 marks)**

1. Consider the addition: $9,999 + 99 = 10,098$, which is a 5-digit number.
2. Smaller numbers give a 4-digit sum, so the statement holds only sometimes.

Answer b: Only sometimes true

**Part c (0.6 marks)**

1. The maximum possible sum of a 4-digit number and a 2-digit number is $9,999 + 99 = 10,098$, which is a 5-digit number.
2. It is never possible to reach a 6-digit number.

Answer c: Never true

**Part d (0.6 marks)**

1. Consider the subtraction: $12,000 - 10,000 = 2,000$, which is a 4-digit number.
2. Other subtractions can give 5-digit results, so it is only sometimes true.

Answer d: Only sometimes true

**Part e (0.6 marks)**

1. Subtracting a 2-digit number from the smallest 5-digit number gives $10,000 - 99 = 9,901$, which is a 4-digit number.
2. Even when subtracting from larger 5-digit numbers, the result remains a 4-digit or 5-digit number, never a 3-digit number.

Answer e: Never true

**Answer:** Classifications for statements provided in parts.

> Common mistake: Assuming addition or subtraction always alters the number of digits based only on typical examples.

## Figure it Out

### Question 1

*3 marks · Short answer*

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.
Grid: 16,200 | 39,344 | 29,765 / 23,609 | 62,871 | 45,306 / 19,381 | 50,319 | 38,408

**Solution**

1. Examine the given 3x3 table where a supercell is a number greater than all its immediate horizontal and vertical neighbors.
2. In the given grid, exchange the digits 1 and 6 in the number $62,871$ to get $12,876$.
3. With this new number, the cells containing $39,344$, $23,609$, $12,876$, and $50,319$ become the 4 supercells.

**Answer:** Exchange the digits 1 and 6 in the number 62,871 to make it 12,876.

> Common mistake: Swapping digits in a non-central number that does not increase the number of supercells to four.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. Take a 4-digit year of birth, for example, 1980.
2. Form the largest and smallest 4-digit numbers using these digits: $A = 9810$ and $B = 1089$.
3. Subtract the smallest number from the largest: $9810 - 1089 = 8721$.
4. Repeat this process until reaching the Kaprekar constant 6174, which takes 6 rounds for the year 1980.

**Answer:** It takes 6 rounds to reach the Kaprekar constant for the year 1980.

> Common mistake: Arranging digits in wrong order while forming the largest and smallest numbers.

### Question 3

*3 marks · Short answer*

We are the 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**

1. Consider 5-digit numbers between 35,000 and 75,000 having only odd digits (1, 3, 5, 7, 9).
2. The largest number with non-repeating odd digits is 73,951 (or 73,999 with repeating digits).
3. The smallest number with non-repeating odd digits is 35,179 (or 35,111 with repeating digits).
4. The number closest to 50,000 with non-repeating odd digits is 51,379 (or 51,111 with repeating digits).

**Answer:** Largest number: 73,951; Smallest number: 35,179; Closest to 50,000: 51,379.

> Common mistake: Using even digits like 0, 2, 4, 6, 8 which are not allowed.

### Question 4

*3 marks · Short answer*

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**

1. Estimate the total number of holidays in a year by considering about 52 Sundays, 52 Saturdays (or alternate Saturdays), 10 festival days, and 30 days of vacation, totaling around 110 to 120 days.
2. Calculate the exact number of holidays from a specific calendar by counting all weekends, declared public holidays falling on weekdays, and school vacation days.
3. Compare the estimate with the exact count to see how close the guess is.

**Answer:** Estimated holidays are around 115 days, which is close to the exact calculated count.

> Common mistake: Double counting holidays that fall on weekends.

### Question 5

*3 marks · Short answer*

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

**Solution**

1. Estimate the capacity of a small mug used for drinking or bathing as about 1 liter.
2. Estimate the capacity of a regular household bucket used for bathing as about 20 liters.
3. Estimate the capacity of a rooftop overhead water tank used for a house as about 500 to 1000 liters.

**Answer:** Mug: about 1 liter, Bucket: about 20 liters, Overhead tank: about 1000 liters.

> Common mistake: Confusing milliliters with liters or giving unrealistic large volumes for small containers.

### Question 6

*3 marks · Short answer*

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

**Solution**

1. Choose one 5-digit number, such as 18,000.
2. Choose two 3-digit numbers, such as 300 and 370.
3. Add the chosen numbers: $18,000 + 300 + 370 = 18,670$.

**Answer:** 18,000 + 300 + 370 = 18,670.

> Common mistake: Choosing numbers whose sum does not equal 18,670 or picking incorrect digit counts.

### Question 7

*3 marks · Short answer*

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**

1. Choose a number between 210 and 390, for example, 250.
2. Arrange numbers in a symmetrical grid or pattern similar to Section 3.9 such that their total sum equals 250.
3. Using twenty-five tens or blocks of 50 and 25 gives the required pattern summing up to 250.

**Answer:** A number pattern summing to 250 using numbers like $25$ and $50$.

> Common mistake: Choosing a number outside the given range of 210 and 390.

### Question 8

*3 marks · Short answer*

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**

1. Recall that the powers of 2 are numbers like $2^1, 2^2, 2^3, \dots, 2^n$, which are always even whole numbers.
2. According to the Collatz rule, since these numbers are even, we repeatedly take half of them.
3. Dividing any power of 2 successively by 2 continuously reduces the exponent by 1 until it reaches $2^1 = 2$, which when divided by 2 gives 1.

**Answer:** The Collatz conjecture holds because repeatedly halving any power of 2 will always reach 2 and then 1.

> Common mistake: Confusing powers of 2 with general odd numbers.

### Question 9

*3 marks · Short answer*

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

**Solution**

1. Start with the number 100, which is even, so take half of it to get 50, then 25.
2. Since 25 is odd, multiply by 3 and add 1 to get 76, then continue applying the even-odd rule.
3. Following the steps successively yields the complete sequence ending at 1.

**Answer:** 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.

> Common mistake: Making arithmetic errors while multiplying odd numbers by 3 and adding 1.

### Question 10

*3 marks · Short answer*

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**

1. Analyze the game where players alternate adding numbers between 1 and 3, starting with 0, and the first to reach 22 wins.
2. Since the allowed additions are 1, 2, or 3, the winning multiples or target numbers can be controlled by the first player.
3. The winning strategy is to be the first player and make the first move to secure control.

**Answer:** The winning strategy is to be the first player.

> Common mistake: Forgetting that the game starts at 0 instead of 1.

## Frequently asked questions

### How many total questions are there in NCERT Solutions for Class 6 Maths Chapter 3 Number Play?

This chapter is based on the new NCERT book for the 2026-27 session and contains multiple sections with a total of 42 questions. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### Which topics are covered in the Class 6 Maths Chapter 3 Number Play questions?

The questions cover various engaging concepts like uses of numbers in daily life, taller neighbour rules, supercells, digit sums, palindromic numbers, Kaprekar constant, and the Collatz sequence. SwaVid's free PDF and step-by-step solutions on this page only will help you master all these topics.

### What are the most challenging question types in this chapter and how should we approach them?

The tricky questions involve complex patterns like supercell identification, digit sums of consecutive numbers, and the Kaprekar constant routine. To approach them, carefully read the given rules, test small numbers first, and refer to SwaVid's free PDF and step-by-step solutions available on this page only for clear guidance.

### How can I write answers to score full marks in Class 6 Maths Chapter 3 Number Play?

To secure full marks, write down your reasoning clearly step-by-step, especially for logic and pattern-based questions like the Collatz sequence and estimation tasks. You can check SwaVid's free PDF and step-by-step solutions provided on this page only to understand the ideal presentation format.

### Is the free PDF for Class 6 Maths Chapter 3 Number Play available for download?

Yes, complete solutions aligned with the latest syllabus are readily accessible. You can easily access SwaVid's free PDF and step-by-step solutions on this page only to prepare effectively for your exams.

## Related pages

- [Class 6 Maths chapters](https://www.swavid.com/maths/class/6)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student