NCERT Solutions Class 6 Maths Chapter 3 Number Play

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 6 · M AT H S

NCERT Solutions

Chapter 3: Number Play

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

55 – 73 20 78 English

Solutions, notes, sample papers & more at 66 pages

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

CLASS 6 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 3: Number Play
Complete, step-by-step NCERT Solutions for Class 6 Maths Chapter 3 Number Play from the NCERT textbook
Ganita Prakash. Every Figure it Out question from pages 57 to 73 is solved, along with all the in-text questions
— supercells, number lines, digit sums, palindromes, the Kaprekar constant 6174, the Collatz conjecture,
estimation and winning game strategies.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 6) 55 – 73

SECTIONS QUESTIONS

20 78

MEDIUM

English

In-text Questions — Pages 55 & 56
Section 3.1 Numbers can Tell us Things

MATH TALK

Q1 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.

Numbers are all around us. Five everyday situations in which we use them:

1. Telling time — the school bell at 8:30, a 40-minute period, a train leaving at 6:15.
2. Money — ₹20 for a plate of samosas, counting the change, saving pocket money.
3. Measuring — your height in centimetres, 2 kg of onions, 1 litre of milk.
4. Counting and marks — 42 children in the class, 87 marks out of 100, 6 runs off one ball.
5. Addresses and codes — house number 14, PIN code 110016, a 10-digit mobile number, bus
route 236.

Math Talk: Notice that the numbers do different jobs. ₹20 and 2 kg measure an
amount, 42 children counts, but a PIN code or a bus route number only names
something — you would never add two PIN codes!

Page 1 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q2 What do you think these numbers mean? (Some children in a park are standing in a
line. Each one says a number: 0, 2, 1, 1, 0, 2, 1, 0)

Each child is telling us how many of the children standing next to them are taller than they
are.

A child says 0 → neither neighbour is taller

A child says 1 → exactly one neighbour is taller

A child says 2 → both neighbours are taller

Look at the picture on page 55 and check it yourself. The first child says 0 because she has only
one neighbour and that neighbour is shorter. The second child (the smallest girl) says 2 because
the children on both her sides are taller. The tall boy in the middle says 0 because both his
neighbours are shorter than him.

Why it happens: a child at either end of the line has only one neighbour, so an end
child can only say 0 or 1 — never 2.

Q3 Did you figure out what these numbers represent? Hint: Could their heights be
playing a role? (The children now rearrange themselves and say 1, 0, 2, 0, 1, 2, 1, 0)

Yes — the numbers again say how many taller neighbours each child has. The children have
only changed places; the rule has not changed.

CHILD (LEFT TO RIGHT) 1ST 2ND 3RD 4TH 5TH 6TH 7TH 8TH

Number said 1 0 2 0 1 2 1 0

The second child is the tallest of the group, so he says 0. The child next to him is shorter than
both his neighbours, so he says 2.

Check it yourself: add up all the numbers said. Here 1 + 0 + 2 + 0 + 1 + 2 + 1 + 0 = 7,
which is one less than the 8 children. This always happens when all the heights are
different — try it!

Page 2 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

In-text Questions — Page 56
Section 3.1 Numbers can Tell us Things

MATH TALK

Q1 Can the children rearrange themselves so that the children standing at the ends
say ‘2’?

No, it is impossible.

Why it happens: a child standing at an end of the line has only one neighbour —
there is nobody on the outer side. Saying ‘2’ means “both my neighbours are taller”,
but an end child does not even have two neighbours. So a child at an end can only
say 0 or 1.

Q2 Can we arrange the children in a line so that all would say only 0s?

Yes — but only if all the children are of the same height.
If everybody is equally tall, then nobody has a taller neighbour, so every child says 0.

Why nothing else works: suppose the heights are not all equal. Take the shortest
child in the line. Somebody standing next to that child is taller, so the shortest child
would have to say 1 or 2 — not 0.

Q3 Can two children standing next to each other say the same number?

Yes. Two neighbours can easily say the same number.
Look at the very first picture on page 55: the children say 0, 2, 1, 1, 0, 2, 1, 0 — the third and the
fourth child both say ‘1’.
This happens when the children are standing in increasing order of height:

Page 3 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
em.
shortest → taller → taller → taller → tallest
m l as
.co
numbers said: 1, 1, 1, 1, 0
m a g
l a se
a g
Here the first four children each have exactly one taller neighbour (the one on their right).

co m
m . ag
l a se
aglast one says ‘0’? Why or why not?
Q4 There are 5 children in a group, all of different heights. Can they stand such that
four of them say ‘1’ and the

co m
m.

m as e
.co
Yes, they can. Make them stand in order of height — shortest to tallest.
a g l
se m
g l a
a
0
a s
. com 1
agl
e m
as
1
1 agl
1
co m
m .
m as e
.co a g l
se m
g l a Children standing shortest to tallest
a
se m
Five children of different heights standing in increasing order. The first four each have exactly one

com
taller neighbour, so they say 1; the tallest says 0.
g l a
m. a
ase
agl
Why it works: for each of the first four children the neighbour on the right is taller
and the neighbour on the left (if any) is shorter — that is exactly one taller

co m
.
neighbour. The tallest child stands at the end and has nobody taller beside him, so
e m
m as
he says 0.

.co a g l
se m
g l a
a c
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
m .
e
Q5

m a s
e m . co agl
as

No, it is not possible.
a g l

co m
m .
m as e
.co


a g l Page 4 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Why it happens: think about the tallest child of the group. Nobody is taller than
that child, so whatever place the tallest child takes in the line, he or she must say 0. A
‘0’ must therefore appear somewhere in the sequence, and so all five children can
never say 1.

Tip: the same reasoning shows that the sequence can never be all 1s or all 2s for any
number of children with different heights — a 0 always appears where the tallest
child stands.

Q6 Is the sequence 0, 1, 2, 1, 0 possible? Why or why not?

Yes, it is possible. Arrange the children so that the two tallest stand at the two ends and the
shortest stands in the middle.

0
0
1
1
2

Tallest at both ends, shortest in the middle
Heights going down towards the middle and up again. The middle child is shorter than both
neighbours, so she says 2.

Say the heights (in cm) are 150, 140, 120, 130, 145 from left to right:

Page 5 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

150 → right neighbour 140 is shorter → says 0

140 → 150 is taller, 120 is shorter → says 1

120 → both 140 and 130 are taller → says 2
130 → 120 is shorter, 145 is taller → says 1

145 → 130 is shorter → says 0

Q7 How would you rearrange the five children so that the maximum number of
children say ‘2’?

At most two children out of five can say ‘2’. Arrange them so that the two shortest children
stand in the 2nd and 4th places, with taller children on both sides of each:

0
0
0

2
2

Short children in the 2nd and 4th places
Heights 145, 120, 150, 125, 140. The 2nd and 4th children are shorter than both their neighbours, so
each says 2 — the sequence is 0, 2, 0, 2, 0.

Why only two: the children at the two ends can never say 2, so only the 2nd, 3rd
and 4th children are possible. But two children saying ‘2’ can never stand next to
each other — if the 3rd child said 2, the 2nd and 4th would both be taller than him
and could not say 2 themselves. So the best we can do is the 2nd and the 4th: 2
children.

Figure it Out — Pages 57 & 58

Page 6 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.2 Supercells

MATH TALK TRY THIS

Q1 Colour or mark the supercells in the table below.

6828 670 9435 3780 3708 7308 8000 5583 52

The nine-cell table printed on page 57.

A cell is a supercell if its number is bigger than the numbers in the cells touching it. Compare
each number with its left and right neighbours:

CELL 6828 670 9435 3780 3708 7308 8000 5583 52

Neighbours 670 6828, 670, 9435, 3780, 3708, 7308, 8000, 5583
9435 3780 3708 7308 8000 5583 52

Supercell? Yes No Yes No No No Yes No No

Answer: 6828, 9435 and 8000 are the supercells.

Why 52 is not a supercell: it has only one neighbour, 5583, but 52 is smaller than
5583. Compare it with 6828 at the other end — 6828 also has one neighbour (670)
but 6828 is bigger, so 6828 is a supercell.

Page 7 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q2 Fill the table below with only 4-digit numbers such that the supercells are exactly
the coloured cells.

5346 1258 9635

The row printed on page 57 — the 2nd, 4th and 9th cells are coloured, and three
numbers are already filled in.

The coloured cells are the 2nd, 4th and 9th. So exactly those three must beat their neighbours,
and no other cell may.

CELL 1 2 ✓ 3 4 ✓ 5 6 7 8 9 ✓
Number 5346 5678 1000 1258 1100 1200 1300 9635 9800

Check the three coloured cells:

5678 > 5346 and 5678 > 1000 ✓

1258 > 1000 and 1258 > 1100 ✓

9800 > 9635 (its only neighbour) ✓

And no other cell wins: 1300 loses to 9635, 9635 loses to 9800, 1200 loses to 1300, and so on.

Tip: the last cell is coloured and touches only 9635, so the number you write there
must be bigger than 9635 — any 4-digit number from 9636 to 9999 will do.

Page 8 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Fill the table below such that we get as many supercells as possible. Use numbers
e
Q3

m l as
.co
between 100 and 1000 without repetitions.

a g
se m
g l a
a

co m
ag
The empty nine-cell table printed on page 57.

m .
as e
a g l

co m
m.
The table has 9 cells, so put the big numbers in the 1st, 3rd, 5th, 7th and 9th cells and small
numbers in between:
m as e
.co a g l
se m ✓ ✓ ✓ ✓ ✓
a
CELL 1 2 3 4 5 6 7 8 9

g l
a Number 500 100 600 200 700 300 800 400 900

m a s
m.co agl
500 > 100 ✓ 600 > 100, 200 ✓
l a se
700 > 200, 300 ✓ 800 > 300, 400 ✓ 900 > 400 ✓
a g
All nine numbers are different and each lies between 100 and 1000, so the conditions are

co m
satisfied. This gives 5 supercells, the most that 9 cells can have.
m .
m ase
.co a g l
a s em Out of the 9 numbers, how many supercells are there in the table above? ___________
agl Q4

se m
com g l a
m . a
ase
agl
The nine-cell table of Question 3 — the one you filled with numbers between 100 and
1000.

co m
m .
m as e
.co
a g l
a s em5 supercells — the numbers written in the 1st, 3rd, 5th, 7th and 9th cells.
agl
.c
Why 5 is the maximum: two supercells can never be side by side (of two
s e m
com a
. agl
neighbours, only the bigger one can win). So the supercells must be separated by at
m of 9 cells, the most you can mark with a gap in
least one ordinary cell. In aerow
s
a
between is 1, 3, 5, 7, 9gl— that is 5 cells.
a

co m
m .
m as e
.co


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Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q5 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.

Fill the biggest numbers in the 1st, 3rd, 5th, … cells (alternate cells starting from the first) and
the small numbers in between. Then:

NUMBER OF CELLS 2 3 4 5 6 7 8 9 10

Most supercells 1 2 2 3 3 4 4 5 5

The pattern:

If the number of cells n is even → most supercells = n ÷ 2
If the number of cells n is odd → most supercells = (n + 1) ÷ 2

Why this is the best possible: no two neighbouring cells can both be supercells. So
between any two supercells there must be at least one other cell. Marking cells 1, 3,
5, 7 … packs them as tightly as the rule allows.

Math Talk: Try it with 6 cells — 600, 100, 700, 200, 800, 300 gives supercells at cells
1, 3 and 5, that is 6 ÷ 2 = 3 supercells.

Q6 Can you fill a supercell table without repeating numbers such that there are no
supercells? Why or why not?

No — it is impossible. There will always be at least one supercell.

Why it happens: look at the largest number you have written anywhere in the
table. Since no number is repeated, every neighbour of that cell holds a smaller
number. So that cell is greater than all its neighbours — it is a supercell, wherever
you place it.

Page 10 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Try This: write any 5 different numbers in a row and hunt for the biggest one. It is a
supercell every single time.

Q7 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?

The largest number is always a supercell. The smallest number can never be a supercell
(as long as the table has more than one cell).

Largest: every other number in the table is smaller, so in particular all its neighbours are
smaller. Being greater than all its neighbours is exactly what a supercell means.
Smallest: every neighbour holds a bigger number, so the cell loses the comparison at once.

Tip: this is why question 6 has the answer “no” — a table always contains its largest
number, and that cell is always a supercell.

Q8 Fill a table such that the cell having the second largest number is not a supercell.

Simply write the second largest number right next to the largest number.

CELL 1 2 3 4 5 6 7 8 9

Number 110 120 130 140 150 160 170 190 180

Largest = 190 (cell 8) → supercell
Second largest = 180 (cell 9), and its only neighbour is 190

180 < 190, so 180 is not a supercell ✓

Page 11 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q9 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?

Yes, it is possible. Put the second largest number beside the largest, and put the second
smallest number beside the smallest one at the far end.

CELL 1 2 3 4 5 6 7 8 9

Number 120 110 130 140 150 160 170 190 180

Second smallest = 120 (cell 1). Its only neighbour is 110, and 120 > 110 → supercell ✓
Second largest = 180 (cell 9). Its only neighbour is 190, and 180 < 190 → not a supercell ✓

The trick: a cell at the end of the row has only one neighbour, so it is very easy to
control. Keep the smallest number at one end and the largest number next to the
other end.

Q10 Make other variations of this puzzle and challenge your classmates.

Here are some puzzles you can set for your friends — all of them use a row of 9 cells:

Fill the table with 3-digit numbers so that there are exactly 2 supercells.
Fill the table so that the middle cell (5th) is the only supercell.
Fill the table so that both end cells are supercells but no cell in between is.
Can you get 6 supercells out of 9 cells? (Answer: no — and asking your friend to explain why
is the real puzzle.)
Use only even numbers between 100 and 1000 and get the maximum number of supercells.

Try This: make the same puzzles for a table with 2 rows and 4 columns. Remember,
there the neighbours are the cells to the left, right, above and below.

In-text Questions — Pages 58 & 59

Page 12 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.2 Supercells — Table 2

TRY THIS

Q1 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.
Once you have filled the table above, put commas appropriately after the
thousands digit.

96,301 36,109

13,609 60,319 19,306

60,193

10,963

Table 2 on page 58 — printed with seven numbers already filled in and five cells
coloured.

Every number must use the digits 1, 0, 6, 3, 9 exactly once. The five coloured cells are the 1st
and 4th of row 1, the 3rd of row 2, and the 2nd and 4th of row 4. Here is one correct filling (the
numbers already printed in the book are shown in grey):

96,310 96,301 36,109 39,160

96,103 13,609 60,319 19,306

13,906 10,396 60,193 60,931

10,369 10,963 10,936 69,031

Check the five coloured cells (orange):

96,310 > 96,301 and 96,103 ✓ (it is the biggest number you can make with these digits)

39,160 > 36,109 and 19,306 ✓

60,319 > 36,109, 13,609, 19,306 and 60,193 ✓

10,963 > 10,396, 10,369 and 10,936 ✓

69,031 > 60,931 and 10,936 ✓

Page 13 of 66

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Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

No other cell beats all its neighbours — for example 96,103 loses to 96,310 above it, and 60,193
co m
loses to 60,319 above it.
e m.
m l as
.co a g
s em
Tip: in the Indian system the comma goes after the thousands digit, so 96310 is
a
gl 96,310 and read as “ninety-six thousand three hundred ten”.
awritten

co m
e m . ag
Q2
l as
The biggest number in the table is ____________ .
g
a

co m
m.
96,301 36,109

m as e
.co
13,609 60,319
a g l 19,306

se m
g l a 60,193

a 10,963

m a s
m .co
Table 2 on page 58 — printed with seven numbers already filled in and five cells
agl
l a se coloured.

a g

co m
m .
as e
comorder: g l
The biggest number you can make from the digits 1, 0, 6, 3, 9 is got by writing them in
. a
em
decreasing
a s
agl
m
9 > 6 > 3 > 1 > 0 → 96,310
a se
. com a g l
m
ase
Answer: 96,310. It sits in the top-left coloured cell of Table 2.

agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 14 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q3 The smallest even number in the table is ____________.

96,301 36,109

13,609 60,319 19,306

60,193

10,963

Table 2 on page 58 — printed with seven numbers already filled in and five cells
coloured.

A number is even when its ones digit is 0, 6 (or 2, 4, 8 — not available here). So the number
must end in 0 or 6.
Even numbers in the completed table: 96,310 39,160 19,306 13,906 10,396 10,936.

Smallest of these = 10,396

Answer: 10,396.

Why not 10,369? Because 10,369 ends in 9, which is odd. Among all the
arrangements of 1, 0, 6, 3, 9, the smallest even one is 10,396.

Page 15 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q4 The smallest number greater than 50,000 in the table is ____________.

96,301 36,109

13,609 60,319 19,306

60,193

10,963

Table 2 on page 58 — printed with seven numbers already filled in and five cells
coloured.

A number is greater than 50,000 only if its first digit is 6 or 9 (5, 7, 8 are not among our digits).
Such numbers in the table: 96,310 96,301 96,103 69,031 60,931 60,319 60,193.

Smallest of these = 60,193

Answer: 60,193.

In-text Questions — Page 59

Page 16 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.3 Patterns of Numbers on the Number Line

Q1 We are quite familiar with number lines now. Let’s see if we can place some
numbers in their appropriate positions on the number line. Here are the numbers:
2180, 2754, 1500, 3600, 9950, 9590, 1050, 3050, 5030, 5300 and 8400.

2180

1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000

2754

The number line printed on page 59, with 2180 and 2754 already marked on it.

First decide which pair of thousands each number lies between, then place it nearer the left or
the right mark:

NUMBER LIES BETWEEN WHERE EXACTLY

1050 1000 and 2000 just after 1000

1500 1000 and 2000 exactly halfway

2180 2000 and 3000 a little after 2000

2754 2000 and 3000 closer to 3000

3050 3000 and 4000 just after 3000

3600 3000 and 4000 a bit past halfway

5030 5000 and 6000 just after 5000

5300 5000 and 6000 nearer 5000

8400 8000 and 9000 a little before halfway

9590 9000 and 10,000 just past halfway

9950 9000 and 10,000 almost at 10,000

Page 17 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

1500 2754 3600 5300 9590
1050 2180 3050 5030 8400 9950

1000 2000 3000 4000 5000 6000 7000 8000 9000
10,000

All eleven numbers marked on the number line. Numbers like 1050, 3050 and 5030 sit very close to a
thousands mark; 1500 is exactly in the middle of 1000 and 2000.

Tip: to place a number, first look at its thousands digit — that fixes the pair of
marks. Then look at the hundreds digit to see how far along to go: 3600 has 6
hundreds, so it goes a little past the middle of 3000 and 4000.

Figure it Out — Page 59
Section 3.3 Patterns of Numbers on the Number Line

Q1 Identify the numbers marked on the number lines below, and label the remaining
positions. a. …, 2010, …, 2020, … b. …, 9996, 9997, … c. 15,077, 15,078, …, 15,083, … d.
…, 86,705, 87,705, …

First find the jump between two neighbouring marks, then count forwards and backwards.
a. 2010 and 2020 are two marks apart, so 20 ÷ 2 gives a jump of 5:

1990 1995 2000 2005 2010 2015 2020 2025 2030 2035

b. 9996 and 9997 are next to each other, so the jump is 1:

9993 9994 9995 9996 9997 9998 9999 10,000 10,001 10,002

c. 15,077 and 15,078 are next to each other, so the jump is 1:

15,077 15,078 15,079 15,080 15,081 15,082 15,083 15,084 15,085 15,086

d. 86,705 and 87,705 are next to each other, so the jump is 1000:

83,705 84,705 85,705 86,705 87,705 88,705 89,705 90,705 91,705 92,705

Page 18 of 66

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Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
The method in one line: difference between two known marks ÷ number of steps

m as e
l
between them = the jump. In (a) the difference is 2020 − 2010 = 10 over 2 steps, so
each step.ciso5. a g
a s em
agl

co m
ag
Put a circle around the smallest number and a box around the largest number in
.
Q2

e m
as
each of the sequences above.

a g l

In every one of these number lines the numbers grow from left to right. So the first mark is
co m
always the smallest and the last mark is always the largest.
e m.
c o m g l as
. a
a s em
SEQUENCE SMALLEST (CIRCLE IT) LARGEST (BOX IT)

agla. 1990 2035

m a s
.co agl
b. 9993 10,002

se m
c. 15,077
g l a 15,086

a
d. 83,705 92,705

co m
m .
mwhere a 4-digit number becomes a 5-digit number.glas
Did you notice? In (b) the sequence crosses a big landmark — 9999 is followed by e
. c o a
e m
10,000,
as
agl
se m
In-text Questions — Page 60
com g l a
m . a
ase
Section 3.4 Playing with Digits

agl
We start writing numbers from 1, 2, 3 … and so on. There are nine 1-digit numbers.
m
Q1
Find out how many numbers have two digits, three digits, four digits, and five
. co
e m
as
digits.
m l
.co a g
a s emANSWER
agl Count from the first such number to the last one:
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 19 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

TYPE FROM – TO WORKING HOW MANY

1-digit 1–9 9−1+1 9

2-digit 10 – 99 99 − 10 + 1 90

3-digit 100 – 999 999 − 100 + 1 900

4-digit 1000 – 9999 9999 − 1000 + 1 9000

5-digit 10,000 – 99,999 99,999 − 10,000 + 1 90,000

Why the count is 9, 90, 900, 9000 …: for a 3-digit number the first digit can be filled
in 9 ways (1 to 9, not 0) and each of the other two digits in 10 ways (0 to 9). So 9 × 10
× 10 = 900. Every extra digit multiplies the count by 10.

Figure it Out — Page 60
Section 3.4 Digit sums of numbers

MATH TALK

Q1 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?

a. Any number whose digits add up to 14 will do:

59 → 5 + 9 = 14 77 → 7 + 7 = 14 86 → 8 + 6 = 14 95 → 9 + 5 = 14

158 → 1 + 5 + 8 = 14 356 → 3 + 5 + 6 = 14 806 → 8 + 0 + 6 = 14

1247 → 1 + 2 + 4 + 7 = 14 3407 → 3 + 4 + 0 + 7 = 14 22226 → 2+2+2+2+6 = 14

b. The smallest number is 59.

Why: a 1-digit number can have a digit sum of at most 9, so we need at least 2
digits. To make the number small, keep the tens digit as small as possible: 5 in the
tens place leaves 9 for the ones place, and 5 + 9 = 14. A tens digit of 4 would need 10
in the ones place, which is impossible.

Page 20 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

c. The largest 5-digit number is 95,000.

Put the biggest digit first: 9

14 − 9 = 5, so the next digit is 5

Nothing is left, so fill the rest with zeros → 95,000

Check: 9 + 5 + 0 + 0 + 0 = 14 ✓

d. There is no biggest such number — you can always make a bigger one.

95 → 9005 → 900005 → 9000000005 → 9000000000000005 …

Each time, push the 5 further to the right and fill the gap with zeros. Zeros add nothing to the
digit sum, so the sum stays 14 while the number keeps growing. Hence you can always make
an even bigger number.

Q2 Find out the digit sums of all the numbers from 40 to 70. Share your observations
with the class.

NUMBER 40 41 42 43 44 45 46 47 48 49

Digit sum 4 5 6 7 8 9 10 11 12 13

NUMBER 50 51 52 53 54 55 56 57 58 59

Digit sum 5 6 7 8 9 10 11 12 13 14

NUMBER 60 61 62 63 64 65 66 67 68 69

Digit sum 6 7 8 9 10 11 12 13 14 15

And 70 → 7 + 0 = 7.
Observations

Inside one decade (40–49, 50–59, 60–69) the digit sum goes up by 1 each time.
When we cross into a new decade the digit sum drops by 8: 49 → 13 but 50 → 5, and 59 →
14 but 60 → 6.
The largest digit sum in this range is 15 (at 69) and the smallest is 4 (at 40).

Page 21 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Some digit sums repeat — 5, 6, 7, 8, 9 appear in more than one decade. For example 41, 50
both have digit sum 5.

Why the drop is 8: going from 49 to 50 the ones digit falls from 9 to 0 (−9) while the
tens digit rises from 4 to 5 (+1). Altogether the digit sum changes by −9 + 1 = −8.

Q3 Calculate the digit sums of 3-digit numbers whose digits are consecutive (for
example, 345). Do you see a pattern? Will this pattern continue?

NUMBER DIGIT SUM VALUE 3 × MIDDLE DIGIT

123 1+2+3 6 3×2=6

234 2+3+4 9 3×3=9

345 3+4+5 12 3 × 4 = 12

456 4+5+6 15 3 × 5 = 15

567 5+6+7 18 3 × 6 = 18

678 6+7+8 21 3 × 7 = 21

789 7+8+9 24 3 × 8 = 24

The pattern: every digit sum is a multiple of 3, and the sums go up in steps of 3 — 6, 9, 12, 15,
18, 21, 24.

Why it happens: if the middle digit is m, the three consecutive digits are (m − 1), m,
(m + 1). Adding them:

(m − 1) + m + (m + 1) = 3 × m

The −1 and the +1 cancel out, so the digit sum is always exactly three times the
middle digit — and therefore always a multiple of 3.

Will the pattern continue? No — it has to stop. The next number would need the digits 8, 9,
10, but 10 is not a single digit. So 789 is the last one, and there are only these seven such
numbers.

Page 22 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Try This: the same seven sums appear if you write the digits the other way round —
321, 432, 543 … Check that 987 also has digit sum 24.

In-text Questions — Pages 61 & 62
Digit Detectives & Section 3.5 Pretty Palindromic Patterns

MATH TALK

Q1 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?

From 1 to 100 → the digit 7 occurs 20 times. Count place by place:

In the ones place: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97 → 10 times

In the tens place: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79 → 10 times

Total = 10 + 10 = 20

Note that 77 is counted twice — and correctly so, because Dinesh writes the digit 7 twice while
writing 77.
From 1 to 1000 → the digit 7 occurs 300 times.

Think of every number from 000 to 999 written with 3 digits.

That is 1000 numbers × 3 digit-places = 3000 digit-places.

Each of the ten digits 0–9 fills the same share: 3000 ÷ 10 = 300

(1000 itself has no 7, so it changes nothing.)

Another way: in the ones place a 7 appears 100 times, in the tens place 100 times
and in the hundreds place 100 times (700–799). Total 100 + 100 + 100 = 300.

Page 23 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
The numbers 121, 313, 222 are some examples of palindromes using the digits ‘1’, ‘2’,
e
Q2

m l as
.co
‘3’. Write all possible 3-digit palindromes using these digits.

a g
se m
g l a
a

A 3-digit palindrome looks like aba — the first and last digits must be the same. Here a and b

m
can each be 1, 2 or 3.
. co ag
em
g l as
a
3 choices for a × 3 choices for b = 9 palindromes

co m
m.
FIRST & LAST DIGIT MIDDLE DIGIT 1 MIDDLE DIGIT 2 MIDDLE DIGIT 3

m a e
s131
1
.co
111 121
ag l
se m
g l a
a
2 212 222 232

s
3 313 323 333

om a
e
. c
m 323, 333 — nine in all. agl
s
Answer: 111, 121, 131, 212, 222, 232, 313,
a
agl
Q3
. c om
Try the same procedure for some other numbers, and perform the same steps. Stop

s e
if you get a palindrome. Are there numbers for which you do notm reach a
. om
cpalindrome at all?
a gla
sem
aANSWER
a g l
Try the reverse-and-add steps yourself:
se m
com g l a
m . a
34 + 43 = 77 (1 step)
gl ase
29 + 92 = 121 (1 step) a

co m
.
48 + 84 = 132, 132 + 231 = 363 (2 steps)
em
m l as
.co
76 + 67 = 143, 143 + 341 = 484 (2 steps)
a g
s e m 59 + 95 = 154, 154 + 451 = 605, 605 + 506 = 1111 (3 steps)
agla
.c
For 2-digit starting numbers you always reach a palindrome — some just take longer. The
s e m
m a
. co agl
hardest one is 89, which needs 24 steps before it becomes the palindrome 8813200023188!

e m
g l as
a

com
m .
m ase
.co


a g l Page 24 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Did you know? For 3-digit numbers nobody knows the answer. Mathematicians
suspect that starting with 196 you never reach a palindrome, even after millions of
steps — but this has never been proved.

Q4 Will reversing and adding numbers repeatedly, starting with a 2-digit number,
always give a palindrome? Explore and find out.

Yes. Every 2-digit number reaches a palindrome after a few reverse-and-add steps.

START STEPS PALINDROME REACHED

12 12 + 21 33

47 47 + 74 121

68 68 + 86 = 154; 154 + 451 = 605; 605 + 506 1111

86 86 + 68 = 154; then as above 1111

89 89 + 98 = 187; … 24 steps in all 8813200023188

A short cut: if the two digits of the number add up to 9 or less (like 34 → 3 + 4 = 7),
the very first addition already gives a palindrome. That is because there is no
carrying: 34 + 43 = 77.

Math Talk: the two numbers 68 and 86 reach the same palindrome, and by the
same route. Can you see why? Reversing a number and adding gives the same result
whichever of the two you start from.

Q5 Puzzle time. 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? _________________ Write the
number in words:

Write the number as tth th h t u. Because it is a palindrome, the digits must mirror each other:

Page 25 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

tth = u and th = t

So the number looks like u t h t u

Now use the clues one by one:

The number is odd, so the ones digit u is 1, 3, 5, 7 or 9.
t = 2 × u and h = 2 × t = 4 × u. Since h must stay a single digit, 4 × u ≤ 9.
Only u = 1 works: u = 3 would give h = 12, which is not a digit.

u=1 → t=2×1=2 → h=2×2=4

Number = 1 2 4 2 1 = 12,421

TTH TH H T U

1 2 4 2 1

Answer: 12,421. In words — twelve thousand four hundred twenty-one.

Check it yourself: 12421 read backwards is 12421 ✓ palindrome; it ends in 1 ✓ odd;
t = 2 = 2 × 1 ✓; h = 4 = 2 × 2 ✓.

In-text Questions — Page 63
Section 3.6 The Magic Number of Kaprekar

Q1 Take different 4-digit numbers and try carrying out these steps. Find out what
happens. Check with your friends what they got. (Also complete the last column: A =
___, B = ___, C = ___)

Whatever 4-digit number you start with (as long as at least two of its digits are different), you
always land on 6174 — the Kaprekar constant — and then it repeats for ever.
The chain started in the book with 6382 finishes like this:

Page 26 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

ROUND 1 2 3 4

A (largest) 8632 6642 7641 7641

B (smallest) 2368 2466 1467 1467

C=A−B 6264 4176 6174 6174

So the last column is A = 7641, B = 1467, C = 7641 − 1467 = 6174. Once you reach 6174 the steps
only give 6174 again and again.
Try a fresh number, say 3524:

5432 − 2345 = 3087

8730 − 3078 = 5652

6552 − 2556 = 3996

9963 − 3699 = 6264

6642 − 2466 = 4176

7641 − 1467 = 6174 ✓

Try This: ask four friends to each pick a different 4-digit number. Everyone will end at
6174 — only the number of rounds will differ. (A number like 3333, with all four digits
the same, gives 0 and is not allowed.)

Q2 Carry out these same steps with a few 3-digit numbers. What number will start
repeating?

With 3-digit numbers you always reach 495, and then it repeats.
Start with 321:

Page 27 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

321 − 123 = 198

981 − 189 = 792

972 − 279 = 693
963 − 369 = 594

954 − 459 = 495

954 − 459 = 495 ← it repeats

Start with 517:

751 − 157 = 594

954 − 459 = 495 ✓

Answer: the number 495 starts repeating. It is the Kaprekar constant for 3-digit numbers, just
as 6174 is for 4-digit numbers.

Tip: the digits of the starting number must not be all the same (111, 222 … give 0).

In-text Questions — Page 64
Section 3.7 Clock and Calendar Numbers

TRY THIS

Q1 On the usual 12-hour clock, there are timings with different patterns. For example,
4:44, 10:10, 12:21. Try and find out all possible times on a 12-hour clock of each of
these types.

There are three different patterns here. Take them one at a time.
Type 1 — the same digit all through (like 4:44)

1:11 2:22 3:33 4:44 5:55 11:11

6:66 and beyond are impossible, because minutes can only go up to 59.
Type 2 — the minutes repeat the hour (like 10:10)

Page 28 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
01:01 02:02 03:03 04:04 05:05 06:06
m l as
.co
07:07 08:08 09:09 10:10 11:11 12:12
m a g
l a se
a g
That is 12 such times.

m
Type 3 — palindromic times, reading the same both ways (like 12:21)

. co ag
e m
01:10 02:20 03:30 04:40 05:50
g l as
10:01 11:11 12:21
a

co m
em.
m as
06:60 is impossible, so the list stops at 05:50.

.co a g l
a s emyou notice? 11:11 belongs to all three lists — it is the only time that is repeated,
gl palindromic and made of a single digit.
Did
a
m a s
m .co agl
l a se
g
Manish has his birthday on 20/12/2012 where the digits ‘2’, ‘0’, ‘1’, and ‘2’ repeat in
a
Q2
that order. Find some other dates of this form from the past.

co m
m .
e

m l as
.co join them, and that is the year. g
In such a date the year simply repeats the day and the month: 20/12/2012. So pick any day
andmmonth, a
l a se
a g DATE DAY & MONTH YEAR FORMED

se m
com g l a
. a
20/01/2001 20 and 01 2001

a s em
agl
20/04/2004 20 and 04 2004

20/06/2006 20 and 06 2006

co m
m .
as e
20/11/2011 20 and 11 2011

com
.19/11/1911 a g l
sem
19 and 11 1911

a
agl 18/12/1812 18 and 12 1812
c
m .
m a s e
.co agl
17/09/1709 17 and 09 1709

se m
g l a
a

co m
m .
m ase
.co


a g l Page 29 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

The rule: for a date in the 2000s the day must be 20 (so that the year starts with 20),
and the month can be anything from 01 to 12. For older centuries the day gives the
century — 19 gives 19xx, 18 gives 18xx, and so on.

Q3 His sister, Meghana, has her birthday on 11/02/2011 where the digits read the same
from left to right and from right to left. Find all possible dates of this form from the
past.

Write the date as eight digits: 1 1 0 2 2 0 1 1. For a palindrome the year must be the day and
month written backwards.
In our own century the year has to start with 2 and then 0, so the month must be 02 (February).
That gives all the palindromic dates of the 2000s:

DATE WRITTEN OUT READ BACKWARDS

10/02/2001 10022001 10022001 ✓

20/02/2002 20022002 20022002 ✓

01/02/2010 01022010 01022010 ✓

11/02/2011 11022011 11022011 ✓

21/02/2012 21022012 21022012 ✓

02/02/2020 02022020 02022020 ✓

12/02/2021 12022021 12022021 ✓

22/02/2022 22022022 22022022 ✓

These eight are all the palindromic dates so far in the 21st century — and every one of
them falls in February!

Did you know? Older examples exist too: 11/11/1111 and 12/11/1121. The next
palindromic date after 22/02/2022 is 03/02/2030.

Page 30 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q4 Jeevan was looking at this year’s calendar. He started wondering, “Why should we
change the calendar every year? Can we not reuse a calendar?” But, will any year’s
calendar repeat again after some years? Will all dates and days in a year match
exactly with that of another year?

Yes — calendars do repeat, and an old calendar can be reused.
An ordinary year has 365 days = 52 weeks + 1 extra day, so the same date shifts forward by one
weekday next year. A leap year has 366 days = 52 weeks + 2 extra days, so it shifts by two.

1 January 2014 was a Wednesday → 1 January 2025 was also a Wednesday

Both years are ordinary years, so the whole calendar of 2014 matches that of 2025.

An ordinary year’s calendar usually returns after 6, 11 or 28 years — for example 2018 and
2029 have exactly the same calendar.
A leap year’s calendar returns after 28 years — 2024’s calendar will be reusable in 2052.

Two things must match: (1) the two years must begin on the same weekday, and
(2) both must be leap years, or both ordinary years. Only then will every date fall on
the same day.

Try This: find an old calendar at home and check which future year it can be used
for.

Figure it Out — Pages 64 & 65

Page 31 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.7 Clock and Calendar Numbers

Q1 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.

Pick four digits, arrange them in decreasing order for the largest number and in increasing
order for the smallest.
a. Difference greater than 5085 → choose digits that are far apart, such as 9, 8, 1, 0.

Largest = 9810, Smallest = 1089
9810 − 1089 = 8721 and 8721 > 5085 ✓

b. Difference less than 5085 → choose digits that are close together, such as 5, 4, 3, 2.

Largest = 5432, Smallest = 2345

5432 − 2345 = 3087 and 3087 < 5085 ✓

c. Sum greater than 9779 → choose big digits, such as 9, 8, 7, 6.

Largest = 9876, Smallest = 6789

9876 + 6789 = 16,665 and 16,665 > 9779 ✓

d. Sum less than 9779 → choose small digits, such as 4, 3, 2, 1.

Largest = 4321, Smallest = 1234

4321 + 1234 = 5555 and 5555 < 9779 ✓

Page 32 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Why this works: the difference depends on how spread out the digits are — 9 and
0 together give a big difference. The sum depends on how big the digits themselves
are — four large digits give a large sum.

Q2 What is the sum of the smallest and largest 5-digit palindrome? What is their
difference?

The smallest 5-digit palindrome is 10001 (it must start with 1 and end with 1, with zeros in
between) and the largest is 99999.

Sum = 10,001 + 99,999 = 1,10,000

Difference = 99,999 − 10,001 = 89,998

Check it yourself: 10001 read backwards is 10001 ✓ and 99999 backwards is 99999
✓ — both are palindromes.

Q3 The time now is 10:01. How many minutes until the clock shows the next
palindromic time? What about the one after that?

10:01 is itself a palindrome (1001 read backwards is 1001). The palindromic times that come
next are 11:11 and then 12:21.

From 10:01 to 11:11 → 1 hour 10 minutes = 60 + 10 = 70 minutes

From 11:11 to 12:21 → 1 hour 10 minutes = 70 minutes

From 10:01 to 12:21 → 2 hours 20 minutes = 120 + 20 = 140 minutes

Answer: 70 minutes to the next palindromic time (11:11), and the one after that (12:21)
comes 140 minutes after 10:01.

Page 33 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Why nothing in between: between 10:01 and 11:11 the clock shows 10:02, 10:03 …

m l a se
11:10. For a 4-digit time to be a palindrome the minutes must be the hour written
o g does that.
.c — 10:01, 11:11, 12:21 — and no other reading in thisastretch
m
backwards

l a se
ag

co m
ag
How many rounds does the number 5683 take to reach the Kaprekar constant?
.
Q4

em
g l as
a
Each round means: make the largest number A, make the smallest number B, and work out C =

co m
m.
A − B.

m l a se C = A − B
ROUND
m.co LARGEST (A) SMALLEST (B)
ag
l a se
a g 1 8653 3568 5085

a s
2
m
8550 5058 3492

m .co 2349 agl
ase
3 9432 7083

4 8730 agl 3078 5652

m
.co
5 6552 2556 3996

se m 6264
6
com
9963 3699
g l a
m . a
ase7
agl
6642 2466 4176

m
8

se
7641 1467 6174

com g l a
m . a
ase
Answer: 8 rounds.

agl
Tip: in round 2 the digits are 5, 0, 8, 5. The smallest 4-digit number from them is

m
5058, because a number cannot begin with 0. (If you do allow 0558, you reach 6174

. co
m
in 7 rounds instead — either way you always get there.)

m as e
.co a g l
a s em
agl In-text Questions — Pages 65 & 66
.c
s e m
m a
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 34 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.8 Mental Math

MATH TALK

Q1 Observe the figure below. What can you say about the numbers and the lines
drawn? Draw arrows from the middle to the numbers on the sides to obtain the
desired sums.

25,000
38,800 3,400

400

28,000 63,000

13,000

61,600 19,500
1,500

31,000 60,000 20,900

The figure on page 65. The arrows already drawn are the two worked examples, for
38,800 and for 3,400.

Each number on the side is built by adding numbers from the middle column, using any of
them as many times as you like. The lines show which middle numbers were used.

Page 35 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

SIDE HOW TO MAKE IT FROM THE MIDDLE CHECK
NUMBER COLUMN

38,800 25,000 + 400 × 2 + 13,000 25,000 + 800 + 13,000 = 38,800
✓

28,000 25,000 + 1,500 × 2 25,000 + 3,000 = 28,000 ✓

61,600 60,000 + 400 × 4 60,000 + 1,600 = 61,600 ✓

31,000 25,000 + 1,500 × 4 25,000 + 6,000 = 31,000 ✓

3,400 1,500 + 1,500 + 400 3,000 + 400 = 3,400 ✓

63,000 60,000 + 1,500 × 2 60,000 + 3,000 = 63,000 ✓

19,500 13,000 + 1,500 × 3 + 400 × 5 13,000 + 4,500 + 2,000 =
19,500 ✓

20,900 13,000 + 1,500 × 5 + 400 13,000 + 7,500 + 400 = 20,900
✓

Tip: start from the biggest middle number that is not more than your target, then
fill the gap. For 61,600 begin with 60,000; only 1,600 is left, and 1,600 = 400 × 4.

Q2 Can we make 1,000 using the numbers in the middle? Why not? What about 14,000,
15,000 and 16,000? Yes, it is possible. Explore how. What thousands cannot be made?

1,000 cannot be made.

Why not: the only middle number smaller than 1,000 is 400, and adding 400s gives
400, 800, 1200, 1600 … — it skips right over 1000. Every other middle number (1,500,
13,000, 25,000, 60,000) is already bigger than 1,000.

14,000, 15,000 and 16,000 can all be made:

14,000 = 1,500 × 8 + 400 × 5 = 12,000 + 2,000 ✓

15,000 = 13,000 + 400 × 5 = 13,000 + 2,000 ✓

16,000 = 13,000 + 1,500 × 2 = 13,000 + 3,000 ✓

Page 36 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Which thousands cannot be made? Only 1,000. Every other whole thousand can be built:

TARGET ONE WAY TO MAKE IT TARGET ONE WAY TO MAKE IT

2,000 400 × 5 7,000 1,500 × 2 + 400 × 10

3,000 1,500 × 2 8,000 400 × 20

4,000 400 × 10 9,000 1,500 × 6

5,000 1,500 × 2 + 400 × 5 10,000 1,500 × 4 + 400 × 10

6,000 1,500 × 4 11,000 1,500 × 6 + 400 × 5

Math Talk: notice that 400 × 5 = 2,000 and 1,500 × 2 = 3,000. Once you can make
2,000 and 3,000, you can make every thousand from 2,000 upwards just by
combining them.

Page 37 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q3 Adding and Subtracting. Here, using the numbers in the boxes, we are allowed to
use both addition and subtraction to get the required number. An example is
shown: 39,800 = 40,000 – 800 + 300 + 300. Find 45,000 = , 5,900 = , 17,500 = , 21,400 =

40,000 7,000

300 1,500

12,000 800

The six boxes printed with this question on page 66.

Now subtraction is allowed too, which makes the target much easier to hit.

TARGET ONE CORRECT WAY CHECK

45,000 40,000 + 12,000 − 7,000 52,000 − 7,000 = 45,000 ✓

5,900 7,000 − 800 − 300 6,200 − 300 = 5,900 ✓

17,500 12,000 + 7,000 − 1,500 19,000 − 1,500 = 17,500 ✓

21,400 40,000 − 12,000 − 7,000 + 1,500 − 300 − 800 21,000 + 1,500 − 1,100 = 21,400 ✓

The last one uses every box exactly once. Step by step:

Page 38 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
40,000 − 12,000 = 28,000
m l as
.co
28,000 − 7,000 = 21,000
m a g
l a se
g
21,000 + 1,500 = 22,500
a22,500 − 300 = 22,200

co m
. ag
22,200 − 800 = 21,400 ✓
e m
g l as
a
Tip: look at the last digits first. To end in 400 you need 1,500 − 300 − 800 = 400. The
thousands then take care of themselves.
co m
em.
m l as
m .co a g
ase it Out — Pages 66 & 67
agl
Figure

m a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.8 Digits and Operations

MATH TALK

Q1 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. Could you find
examples for all the cases? If not, think and discuss what could be the reason.

SCENARIO EXAMPLE POSSIBLE?

5-digit + 5-digit → 5-digit sum more than 45,000 + 45,500 = 90,500 Yes
90,250

5-digit + 3-digit → 6-digit sum 99,999 + 999 = 1,00,998 Yes

4-digit + 4-digit → 6-digit sum largest possible is 9999 + 9999 = 19,998 No

5-digit + 5-digit → 6-digit sum 60,000 + 40,000 = 1,00,000 Yes

5-digit + 5-digit → 18,500 smallest possible is 10,000 + 10,000 = No
20,000

5-digit − 5-digit → difference less than 56,503 80,000 − 50,000 = 30,000 Yes

5-digit − 3-digit → 4-digit difference 10,000 − 999 = 9,001 Yes

5-digit − 4-digit → 4-digit difference 12,000 − 2,500 = 9,500 Yes

5-digit − 5-digit → 3-digit difference 50,999 − 50,000 = 999 Yes

5-digit − 5-digit → 91,500 largest possible is 99,999 − 10,000 = 89,999 No

Three of the ten cases are impossible. Here is the reason for each:

4-digit + 4-digit = 6-digit: even the two biggest 4-digit numbers give 9999 + 9999 = 19,998,
which has only 5 digits. To reach 1,00,000 you would need much larger numbers.
5-digit + 5-digit = 18,500: the smallest 5-digit number is 10,000, so the smallest possible
sum is 20,000 — already bigger than 18,500.
5-digit − 5-digit = 91,500: the biggest difference two 5-digit numbers can have is 99,999 −
10,000 = 89,999, which is less than 91,500.

Page 40 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Math Talk: make your own such questions and let your friends decide. Two good
ones — “4-digit + 4-digit to give 2,900” (possible: 1,900 + 1,000 = 2,900) and “5-digit +
5-digit to give a 7-digit sum” (impossible: the biggest sum is 99,999 + 99,999 =
1,99,998, only 6 digits).

Q2 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? 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

STATEMENT VERDICT REASON WITH AN EXAMPLE

a. 5-digit + 5-digit = 5- Only sometimes 10,000 + 10,000 = 20,000 ✓ but 60,000 + 50,000 = 1,10,000
digit true ✗

b. 4-digit + 2-digit = 4- Only sometimes 1,000 + 10 = 1,010 ✓ but 9,999 + 99 = 10,098 ✗
digit true

c. 4-digit + 2-digit = 6- Never true the biggest possible sum is 9,999 + 99 = 10,098 — only 5
digit digits

d. 5-digit − 5-digit = 5- Only sometimes 99,999 − 10,000 = 89,999 ✓ but 12,000 − 10,000 = 2,000 ✗
digit true

e. 5-digit − 2-digit = 3- Never true the smallest possible answer is 10,000 − 99 = 9,901 — still 4
digit digits

How to decide quickly: test the two extreme cases. If the smallest possible answer
and the largest possible answer are both outside what the statement asks for, the
statement is never true. If one example works and another does not, it is only
sometimes true.

In-text Questions — Pages 67 & 68

Page 41 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.9 Playing with Number Patterns

MATH TALK

Q1 Here are some numbers arranged in some patterns. 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 one of you
used to solve these questions.

a. b.

40 40 40 40

50 50 50 50 50

40 40 40 40

50 50 50 50 50

40 40 40 40

c. d.

32 32 32 32 32 32 32 32

32 32 32 32 32 32 32 32

32 32 32 32 32 32 32 32

32 32 32 32 32 32 32 32

64 64 64 64

64 64 64 64

64 64 64 64

64 64 64 64

e. f.

Page 42 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

15 15 35 35 25 25
125 125
125
15 15 25 25 125
35 35
25 125
25 15 15
250
125
35 35
25 25 15 15 125 250
15 25 250
35 35 500
125
25 25 15 15
35 35 35 35 35 35
15 15 25 25 125
1000 500 250
250 500
35 35 125
25 15
15 15 25 25
35 35 125
500
15 15 25 250 125
25 250
35 35
25 25 15 15 125
250
15 15 125
25 25 35 35
125
125
125
125

The six number patterns printed on pages 67 and 68. In the book, patterns (e) and (f) are
printed over decorative pictures; the numbers and their arrangement are reproduced
here.

The quick way is always the same: count how many times each number appears, multiply,
then add. That is much faster than adding one by one.

FIGURE WHAT IT CONTAINS QUICK WORKING SUM

a. 12 forties and 10 fifties 12 × 40 = 480; 10 × 50 = 980
500

b. an 8 × 8 grid of squares — 44 squares with 1 dot, 20 squares 44 × 1 = 44; 20 × 5 = 100 144
with 5 dots

c. 32 cells of 32 and 16 cells of 64 32 × 32 = 1024; 16 × 64 = 2048
1024

d. a 5 × 7 board of dice faces — 17 cards show 3 dots, 18 cards 17 × 3 = 51; 18 × 4 = 72 123
show 4 dots

e. 22 fifteens, 22 twenty-fives and 22 thirty-fives 22 × (15 + 25 + 35) = 22 × 1650
75

f. 18 of 125, 8 of 250, 4 of 500 and one 1000 2250 + 2000 + 2000 + 7250
1000

Working shown in full

Page 43 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
a. There are 3 rows of four 40s and 2 rows of five 50s.
m l as
.co
(3 × 4) × 40 + (2 × 5) × 50 = 480 + 500 = 980
m a g
l a se
a g
b. The grid is 8 × 8 = 64 squares. 20 of them are the dark squares with 5 dots each; the other

co m
44 have a single dot.
e m . ag
g l as
a
44 + (20 × 5) = 44 + 100 = 144

co m
em.
c. The top block is 4 rows × 8 columns = 32 cells of 32.
m l as
.co
The lower block is 4 rows × 4 cells = 16 cells of 64.
a g
a s em
gl
(32 × 32) + (16 × 64) = 1024 + 1024 = 2048
a
m a s
d. The board has 5 columns × 7 rows = 35 cards.
m .co agl
l a se
a g
The first and last columns (14 cards) and the three cards of the 6th row show 3 dots → 17 ×

3 = 51

co m
The remaining 18 cards show 4 dots → 18 × 4 = 72
m .
m as e
.co
51 + 72 = 123
a g l
se m
g l a
a
m
e. The design is symmetric and contains each of 15, 25 and 35 exactly 22 times.

a se
22 × 15 + 22 × 25 + 22 × 35 = 22 × 75 = 1650
. com a g l
m
ase
agl
f. 18 × 125 = 2250, 8 × 250 = 2000, 4 × 500 = 2000, 1 × 1000 = 1000

co m
2250 + 2000 + 2000 + 1000 = 7250
m .
m as e
.co a g l
se m Why grouping is smarter: in figure (c) adding 32 one at a time needs 47 additions
g l a
a and invites mistakes. Grouping turns it into two multiplications and one addition —
c
m .
and it also shows a hidden fact: the 32s and the 64s contribute exactly the same
m a s e
. co agl
amount, 1024 each.
e m
g l as
a

co m
m .
m ase
.co


a g l Page 44 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Math Talk: different friends will group differently. In figure (a) some will do (40 × 12)
+ (50 × 10); others will add each row: 160 + 250 + 160 + 250 + 160 = 980. Both are
correct — compare the methods in class.

In-text Questions — Pages 68 & 69
Section 3.10 An Unsolved Mystery — the Collatz Conjecture!

Q1 Look at the sequences below—the same rule is applied in all the sequences: a. 12, 6,
3, 10, 5, 16, 8, 4, 2, 1 b. 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 c. 21, 64, 32, 16, 8, 4, 2, 1
d. 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Do you see how these sequences
were formed?

Yes. There is one rule with two parts, applied again and again:

If the number is even → take half of it

If the number is odd → multiply by 3 and add 1

Check sequence (a) step by step:

12 is even → 12 ÷ 2 = 6

6 is even → 6 ÷ 2 = 3

3 is odd → 3 × 3 + 1 = 10

10 is even → 5

5 is odd → 5 × 3 + 1 = 16

16 → 8 → 4 → 2 → 1

Sequence (c) is the shortest because 21 → 64, and 64 is a power of 2, so it just keeps halving: 64,
32, 16, 8, 4, 2, 1.

Did you notice? Every sequence ends with the same tail … 16, 8, 4, 2, 1. Once a
sequence reaches a power of 2, the rest is only halving.

Page 45 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q2 Make some more Collatz sequences like those above, starting with your favourite
whole numbers. Do you always reach 1?

Yes — every number tried so far reaches 1.

START SEQUENCE STEPS

6 6, 3, 10, 5, 16, 8, 4, 2, 1 8

7 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 16

9 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 19

15 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 17

32 32, 16, 8, 4, 2, 1 5

Why the sequence goes up and down: an odd number is nearly tripled, so the
sequence jumps upward; but the answer 3n + 1 is always even, so the very next step
halves it. Over many steps the halving wins and the numbers come down to 1.

Q3 Do you believe the conjecture of Collatz that all such sequences will eventually
reach 1? Why or why not?

Yes, it certainly looks true — but nobody has been able to prove it, and that is exactly why it is
famous.

Reasons to believe it: computers have checked every starting number up to more than
2,00,00,00,00,00,00,00,00,000 and each one comes down to 1. Also, whenever a number is
odd, the next number 3n + 1 is even and gets halved at once — so halving happens at least
half the time, and halving pulls the numbers down fast.
Why that is not a proof: checking crores of numbers is still not all numbers. There are
infinitely many whole numbers, and mathematics demands a reason that covers every single
one of them. Perhaps some huge number rises for ever, or gets stuck in a loop that never
touches 1.

Page 46 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Did you know? Lothar Collatz proposed this in 1937. Almost ninety years later it is
still open — a problem a Class 6 student can understand but no mathematician in
the world has solved.

Figure it Out — Pages 69 & 70
Section 3.11 Simple Estimation

Q1 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

These answers will be different for every child — the point is the method. One step of a Class 6
student is roughly half a metre, so 2 steps ≈ 1 metre.

DISTANCE ROUGH LENGTH ESTIMATED STEPS

a. Seat to classroom door about 5 m about 10 steps

b. Across the school ground about 100 m about 200 steps

c. Classroom door to school gate about 150 m about 300 steps

d. School to home about 1 km = 1000 m about 2000 steps

Check it yourself: actually count your steps from your seat to the door. Then use
that as your “ruler” — if 10 steps cross the classroom, and the ground is about 20
classrooms long, the ground is about 200 steps.

Q2 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

Count for one minute, then multiply. A child usually blinks about 15 times and breathes about
20 times in a minute.

Page 47 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

PERIOD WORKING BLINKS BREATHS

a. In a minute counted directly about 15 about 20

b. In an hour × 60 about 900 about 1,200

c. In a day × 24 about 21,600 about 28,800

Blinks in a day = 15 × 60 × 24 = 15 × 1440 = 21,600

Breaths in a day = 20 × 60 × 24 = 20 × 1440 = 28,800

A sensible correction: we do not blink while asleep. If you sleep for 8 hours, the
blinks come down to about 15 × 60 × 16 = 14,400 a day. Breathing, of course,
continues all night.

Q3 Name some objects around you that are: a. a few thousand in number b. more than
ten thousand in number

a. A few thousand (roughly 1,000 – 9,000)

Bricks in the wall of your classroom.
Words on ten pages of your textbook.
Grains of rice in a small bowl.
Students in a big school, or people in a small village.
Tiles on the floor of the school building.

b. More than ten thousand

Hairs on your head (about 1,00,000).
Grains of rice in a 1 kg packet (about 40,000).
Leaves on a big banyan tree.
People watching a match in a stadium.
Letters printed in your maths textbook.

Tip: to estimate the bricks in a wall, count the bricks in one row and the number of
rows, then multiply — do not try to count them one by one!

Page 48 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
In-text Questions — Pages 70 & 71
m as e
.co
Section 3.11 Estimate the answer
a g l
a s em
aQ1gl Number of words in your maths textbook: a. More than 5000 b. Less than 5000

co m
. ag

e m
(a) More than 5000.
g l as
a
m
Words on one page ≈ 150
co
em.
as
Number of pages ≈ 250
om× 250 = 37,500 words
Total ≈.c150
a g l
a s em
gl
a Even if we count only 100 words on a page and 100 pages, we already get 10,000 — far more
than 5,000. So the answer is clearly more than 5000.
m a s
m .co agl
l a se
Q2 a g
Number of students in your school who travel to school by bus: a. More than 200 b.

m
Less than 200

. co
e m
c
o m g l as
Form . a
s e most schools the answer is (b) less than 200.

agla
Students in the school ≈ 500
se m
o m g l a
m
Children who come by bus ≈ one out of
.c every four ≈ 500 ÷ 4 = 125
a
l a se
If your school runs 4 buses a g each carries about 40 children, that is 4 × 40 = 160 — again less
and
than 200.

co m
m .
m l a se
Check it yourself: count the buses standing in your school and multiply by the seats
o g
.cin one bus. That is a good estimate, and it takes half aaminute.
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 49 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Q3 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?

No — ₹100 is a little too low. Let us estimate the shopping list for 5 people:

ITEM QUANTITY APPROXIMATE COST

Milk 1 litre ₹60

Bananas 4 ₹25

Apple 2 ₹50

Grapes / papaya 250 g ₹30

Sugar and custard powder a small quantity ₹20

Total about ₹185

So around ₹180 – ₹200 would be needed. Roshan could manage in ₹100 only if he uses half a
litre of milk and buys the cheapest seasonal fruits (banana, papaya, guava) in small quantities.

The idea of estimation: we are not looking for the exact bill. We only want to know
whether ₹100 is roughly right — and here it falls short by about half.

Q4 Estimate the distance between Gandhinagar (in Gujarat) to Kohima (in Nagaland).
Hint: Look at the map of India to locate these cities.

Gandhinagar is in the far west of India and Kohima is in the far east, so the two cities are
almost as far apart as India is wide.

Width of India from west to east ≈ 3000 km

Straight-line distance Gandhinagar → Kohima ≈ 2400 km

Distance by road ≈ 3000 km

Answer: roughly 2,500 km in a straight line (about 3,000 km if you actually travel by road or
rail).

Page 50 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Tip: use the scale printed on the map of India. Measure the gap between the two
cities with a ruler and multiply by the scale.

Q5 Sheetal is in Grade 6 and says she has spent around 13,000 hours in school till date.
Do you agree with her? Why or why not?

No — 13,000 hours is far too high. Let us check with a quick calculation.

School hours in a day ≈ 6

School days in a year ≈ 220

Hours in one year = 6 × 220 = 1320 hours
Sheetal has studied for about 7 years (Class 1 to Class 6, plus one year of nursery)

Total ≈ 1320 × 7 = 9,240 hours

So about 9,000 – 10,000 hours is a sensible figure. To reach 13,000 hours she would have had to
be in school for nearly 10 years, which is not possible in Class 6.

Why estimating helps: a big number can “feel” right but still be wrong. Breaking it
into hours per day × days per year × number of years shows the truth in three steps.

Q6 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.

A person walks about 5 km in one hour, and can comfortably walk about 25 km in a day (5
hours of walking).

Page 51 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

JOURNEY APPROXIMATE TIME ON FOOT
DISTANCE

a. To a favourite place nearby (park, 2 km about 25 minutes
market, temple)

b. To a neighbouring state’s capital about 500 km about 20 days of walking

c. Kanniyakumari to the northern tip of about 3,200 km about 128 days, i.e. more than
India 4 months

For (c): 3200 km ÷ 25 km a day = 128 days

Walking 8 hours a day at 5 km/h you would cover 40 km a day → 3200 ÷ 40 = 80 days

Did you know? This is exactly how pilgrims and traders travelled in ancient India —
a journey from Kanniyakumari to Kashmir took several months.

Q7 Make some estimation questions and challenge your classmates!

Here are estimation questions you can ask your friends. Remember, the fun is in explaining how
you got your figure.

How many bricks were used to build our classroom wall?
How many chapatis does your family eat in one year?
How many words does your teacher speak during one period?
How many litres of water does your family use in a day?
How many buckets of water would fill the school’s overhead tank?
How many heartbeats have you had since you were born?
How many rupees does your family spend on vegetables in a month?

Sample answer — chapatis in a year:

5 people × 3 chapatis a meal × 2 meals a day = 30 a day

30 × 365 = about 11,000 chapatis a year

In-text Questions — Pages 71 & 72

Page 52 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Section 3.12 Games and Winning Strategies

Q1 Rules for Game #1: The first player says 1, 2 or 3. Then the two players take turns
adding 1, 2, or 3 to the previous number said. The first player to reach 21 wins!
Which player can always win if they play correctly? What is the pattern of numbers
that the winning player should say?

The first player can always win. The winning numbers to say are

1, 5, 9, 13, 17, 21

Start by saying 1. After that, whatever your friend adds, you add the amount that makes the
total go up by 4 from your last number.

FRIEND ADDS YOU ADD TOTAL GOES UP BY

1 3 4

2 2 4

3 1 4

Why it works: work backwards from 21. To win you must say 21, so you must leave
your friend at 17, 18, 19 or 20 — that is, you should say 17. Working back in the
same way gives 13, 9, 5 and 1. Whoever says 1 first controls the whole game, and the
first player can always say 1.

Try This: the key number is 4 = 3 + 1, that is, the largest allowed addition plus 1. This
is the secret of the whole family of such games.

Q2 Rules for Game #2: The first player says a number between 1 and 10. Then the two
players take turns adding a number between 1 and 10 to the previous number said.
The first player to reach 99 wins! Which player can always win? What is the pattern
of numbers that the winning player should say this time?

This time the second player always wins. The winning numbers are the multiples of 11:

Page 53 of 66

as e
Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
11, 22, 33, 44, 55, 66, 77, 88, 99
m l as
.co a g
a s em
Here the largest allowed addition is 10, so the key number is 10 + 1 = 11. Whatever the first

a gl says (some number x from 1 to 10), the second player replies with 11 − x, bringing the
player
total to 11.

. com2 ag
FIRST PLAYER SAYS
se
1
m 3 … 9 10

l a
Second player adds
ag 10 9 8 … 2 1

Total 11 11 11 … 11
co m 11

m .
m e
as wins.
. c o a g l
Repeating this, the second player reaches 22, 33, 44 … and finally 99 — and

s e m
a glaWhy the first player cannot win: the first player can only say 1 to 10, so he can
never say 11. He must always hand over a total that is not a multiple of 11, and the

m a s
agl
second player instantly restores the pattern.

m .co
l a se
a g
Q3 Make your own variations of this game — decide how much one can add at each
turn, and what number is the winning number. Then play your game several times,
co m
m .
e
and figure out the winning strategy and which player can always win!
m l as
.co a g
a s emANSWER

a gl The rule for every such game is the same. If a player may add anything from 1 to k, then the
magic step is k + 1.
se m
com g l a
m . a
ase
agl
Winning numbers = the target, then target − (k + 1), then target − 2(k + 1), and so on.

If the first winning number in that backward list can be said on the first move, the first
player wins.
. com
m player wins.
If the list ends exactly at the target being a multiple of (k + 1), theesecond
s
co m gl a
. a
a s em
agl c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

GAME K+1 NUMBERS TO SAY WINNER

Add 1–3, reach 21 4 1, 5, 9, 13, 17, 21 first player

Add 1–10, reach 99 11 11, 22, …, 99 second player

Add 1–3, reach 20 4 4, 8, 12, 16, 20 second player

Add 1–5, reach 31 6 1, 7, 13, 19, 25, 31 first player

Add 1–4, reach 25 5 5, 10, 15, 20, 25 second player

Tip: divide the target by (k + 1). If the remainder is 0, the second player wins;
otherwise the first player wins by saying that remainder on the very first move.

Figure it Out — Pages 72 & 73
Section 3.12 Games and Winning Strategies

TRY THIS

Q1 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.

16,200 39,344 29,765

23,609 62,871 45,306

19,381 50,319 38,408

The grid printed on page 72; the shaded cell is the only supercell in it.

At present only 62,871 (the centre cell) is a supercell — it is bigger than 39,344, 23,609, 45,306
and 50,319. Because this one number towers over the whole grid, it blocks everybody else.
Swap the digits 6 and 1 of 62,871 — that is, the first digit and the last digit — to get 12,876.

Page 55 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

16,200 39,344 ✓ 29,765

23,609 ✓ 12,876 45,306 ✓
19,381 50,319 ✓ 38,408

Now check each of the four:

39,344 > 16,200, 29,765 and 12,876 ✓

23,609 > 16,200, 12,876 and 19,381 ✓
45,306 > 29,765, 12,876 and 38,408 ✓

50,319 > 12,876, 19,381 and 38,408 ✓

Answer: swap the 6 and the 1 in 62,871 to make it 12,876 — the grid then has 4 supercells.

Why it works: every one of the four “middle-edge” cells touches the centre. Once
the centre number becomes small (12,876), each of them wins over all its own
neighbours at the same time.

Q2 How many rounds does your year of birth take to reach the Kaprekar constant?

Take the year 2014 as an example (use your own year in the same way).

ROUND LARGEST SMALLEST DIFFERENCE

1 4210 1024 3186

2 8631 1368 7263

3 7632 2367 5265

4 6552 2556 3996

5 9963 3699 6264

6 6642 2466 4176

7 7641 1467 6174

2014 takes 7 rounds. Two more examples:

Page 56 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

2013: 3210 − 1023 = 2187; 8721 − 1278 = 7443; 7443 − 3447 = 3996; 9963 − 3699 = 6264;

6642 − 2466 = 4176; 7641 − 1467 = 6174 → 6 rounds

1980: 9810 − 1089 = 8721; 8721 − 1278 = 7443; 7443 − 3447 = 3996; 9963 − 3699 = 6264;

6642 − 2466 = 4176; 7641 − 1467 = 6174 → 6 rounds

Tip: when a year contains a 0 (as 2014 does), the smallest 4-digit number cannot
begin with that 0 — for the digits 2, 0, 1, 4 the smallest number is 1024.

Q3 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?

Only the digits 1, 3, 5, 7, 9 may be used, and the number must lie between 35,000 and 75,000.
Largest → 73,999

The first digit can be at most 7 (9 is too big — the number must stay below 75,000).

With 7 in front, the next digit must be less than 5, so it is 3.

The rest can be the biggest odd digit, 9 → 73,999

Smallest → 35,111

The first digit must be at least 3.

With 3 in front, the number must be more than 35,000, so the next digit is 5.

The rest are the smallest odd digit, 1 → 35,111

Closest to 50,000 → 51,111

Below 50,000 the best we can do is 39,999 (4 is not an odd digit!) → 50,000 − 39,999 =

10,001 away

Above 50,000 the smallest is 51,111 → 51,111 − 50,000 = 1,111 away

1,111 < 10,001, so 51,111 is the closest

Page 57 of 66

Class 6 Maths Chapter 3 Number Play AglaSem · NCERT Solutions

Why 4 is the villain: every number in the 40,000s starts with the digit 4, which is
even. So the group has no member at all between 39,999 and 51,111 — a gap of
more than eleven thousand.

Q4 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.

Estimate first — count the big pieces:

TYPE OF HOLIDAY ROUGH COUNT

Sundays 52

Second Saturdays 12

Festivals (Diwali, Holi, Eid, Christmas, Pongal …) about 15

Summer vacation about 40

Winter and other breaks about 15

Estimated total about 134 days

Now the exact count. A school year has about 220 working days, so

365 − 220 = 145 holidays

The estimate 134 is quite close to 145 — off by about 11 days, which is less than a tenth. That is
good estimation!

Check it yourself: some festival holidays fall on a Sunday, so counting them twice is
a common mistake. Cross them off your calendar to avoid it.

Page 58 of 66

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Class 6 Maths Chapter 3 Number Play
a g l AglaSem · NCERT Solutions

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Estimate the number of liters a mug, a bucket and an overhead tank can hold.
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Q5

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aCONTAINER ESTIMATE HOW TO CHECK

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Mug
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about 1 litre a 1-litre milk packet just fills it

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Bucket
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about 15 – 20 litres about 15 mugs fill one bucket

Overhead tank
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about 500 – 1000 litres about 40 – 60 buckets fill it

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1 bucket ≈ 15 mugs ≈ 15 litres
em ≈ 50 buckets ≈ 50 × 15 = 750 litres
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The method: start from something you know for sure — a 1-litre milk packet. Then
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measure the bucket in mugs, andethe
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can estimate very big amounts

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Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
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Choose the two 3-digit numbers first, then work out the 5-digit number by subtraction.
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Let the 3-digit numbers be 670 and 500.
670 + 500 = 1,170 agl
18,670 − 1,170 = 17,500
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Check: 17,500 + 670 + 500 = 18,670 ✓
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a s emAnswer: 17,500 + 670 + 500 = 18,670. Two more correct answers:
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18,000 + 370 + 300 = 18,670 ✓

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16,870 + 900 + 900 = 18,670e✓
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