NCERT Solutions Class 7 Maths Chapter 6 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 7 · M AT H S

NCERT Solutions

Chapter 6: Number Play

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

127 - 145 20 75 English

Solutions, notes, sample papers & more at 60 pages

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 6: Number Play
Chapter 6 of Ganita Prakash Grade 7 turns numbers into games. You read a line of children by the numbers
they call out, learn parity (even or odd) and use it to decide what is possible, build magic squares, meet the
Virahāṅka–Fibonacci numbers that came from Indian poetry, and crack cryptarithms where letters hide
digits.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) 127 - 145

SECTIONS QUESTIONS

20 75

MEDIUM

English

In-text Questions — Page 127
Section 6.1 Numbers Tell us Things

Q1 What do the numbers in the figure below tell us?

0
1
0 2
2
2
1

The figure on page 127 — seven children standing in a line, each calling out a number.

Each number tells how many children standing ahead of that child in the line are taller than
the child. Nothing about actual heights is revealed — only a comparison.

Page 1 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

In the first picture the seven children call out, from the left:

0, 1, 0, 2, 2, 2, 1

Read them one by one:

The first child has nobody ahead, so the number must be 0.
The second child has one taller child ahead, so 1.
The third child is the tallest of all, so no one ahead is taller — 0.
The last child has only the tallest child and one more taller child ahead — 1.

Why it happens: the number is a count, not a measurement. Two groups with
completely different heights can give exactly the same list of numbers, as long as
the order of tallness is the same.

Q2 What do you think these numbers mean?

They mean the same thing in the second picture too — the number of taller children standing
in front.
After the children rearrange themselves, the numbers called out from the left are:

0, 0, 1, 2, 2, 5, 6

The tallest child (second from the left) says 0, and the shortest child stands last and says 6,
because every one of the six children ahead is taller.

Tip: the very last number tells you how many children in the whole line are taller
than the last child. If the last child is the shortest, that number is one less than the
number of children.

Page 2 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q3 Could you figure out what these numbers convey? Observe and try to find out.

0
1
2 2
0
5
6

The second figure on page 127 — the same seven children after they rearrange
themselves, each calling out a number.

Yes. The rule is: each child calls out the number of children in front of them who are taller
than them.
Check it on both pictures — every number matches.

POSITION IN LINE 1ST 2ND 3RD 4TH 5TH 6TH 7TH

First picture 0 1 0 2 2 2 1

Second picture 0 0 1 2 2 5 6

Why it happens: the first child can never see anyone ahead, so the first number is
always 0. A child who is the tallest so far also says 0, because nobody ahead is taller.

In-text Questions — Page 128

Page 3 of 60

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Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

Section 6.1 Numbers Tell us Things
co m
e m.
m l as
Q1
.co
Write down the number each child should say based on this rule for the
m a g
l a se
arrangement shown below.

ag

com
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g
The arrangement on page 128 — the speech bubbles are blank in the book.

co m
m .
m as e
.co a g l
a s em
Reading the picture from the left, the order of the children from tallest to shortest is: red, green,

agl blue, pink, dark-yellow, orange, pale-yellow.
Now count, for each child, how many taller children stand ahead:
se m
com g l a
m . a
ase
CHILD (FROM THE LEFT) TALLER CHILDREN IN FRONT NUMBER CALLED OUT

1st — pale yellow (shortest) agl none 0

none (only the shortest is ahead)
co m
.
2nd — light blue 0

e1 m
m l as
.co g
3rd — dark yellow light blue

m 4th — green a
l a se none 0

ag
.c
m
light blue, dark yellow, green
e
5th — orange 3

m a s
m .
none
co agl
6th — red (tallest) 0

l a se light blue, green, red
ag
7th — pink 3

co m
m .
m as e
.co


a g l Page 4 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

So the sequence is 0, 0, 1, 0, 3, 0, 3.

Check it yourself: the numbers 0 appear exactly where a child is taller than
everyone ahead — the pale-yellow child (nobody ahead), the light-blue child, the
green child and the red child.

Figure it Out — Page 128
Section 6.1 Numbers Tell us Things

Q1 Arrange the stick figure cutouts given at the end of the book or draw a height
arrangement such that the sequence reads: (a) 0, 1, 1, 2, 4, 1, 5 (b) 0, 0, 0, 0, 0, 0, 0 (c)
0, 1, 2, 3, 4, 5, 6 (d) 0, 1, 0, 1, 0, 1, 0 (e) 0, 1, 1, 1, 1, 1, 1 (f) 0, 0, 0, 3, 3, 3, 3

Give the seven cutouts heights 1 to 7, where 7 is the tallest and 1 is the shortest. Then arrange
them left to right as shown.

SEQUENCE TO BE READ OUT HEIGHTS FROM LEFT TO RIGHT (7 = TALLEST)

(a) 0, 1, 1, 2, 4, 1, 5 7, 3, 5, 4, 1, 6, 2

(b) 0, 0, 0, 0, 0, 0, 0 1, 2, 3, 4, 5, 6, 7

(c) 0, 1, 2, 3, 4, 5, 6 7, 6, 5, 4, 3, 2, 1

(d) 0, 1, 0, 1, 0, 1, 0 2, 1, 4, 3, 6, 5, 7

(e) 0, 1, 1, 1, 1, 1, 1 7, 1, 2, 3, 4, 5, 6

(f) 0, 0, 0, 3, 3, 3, 3 5, 6, 7, 1, 2, 3, 4

Check (a) child by child:

Page 5 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

7 → nobody ahead = 0

3 → taller ahead: 7 = 1

5 → taller ahead: 7 = 1
4 → taller ahead: 7, 5 = 2

1 → taller ahead: 7, 3, 5, 4 = 4

6 → taller ahead: 7 = 1

2 → taller ahead: 7, 3, 5, 4, 6 = 5

And check (f): heights 5, 6, 7, 1, 2, 3, 4 give 0, 0, 0, 3, 3, 3, 3, because the first three are taller than
each of the last four.

Why it happens: build the line from the last child backwards. If a child must say k,
then exactly k of the children already placed ahead must be taller — so that child is
the (k + 1)th tallest among those in front of it plus itself. Each sequence in this
question fixes the whole order of heights, so there is only one height order for each
part.

Try This: not every list is possible. A list like 0, 2, 0, 0, 0, 0, 0 can never happen,
because the second child has only one child ahead and cannot see two taller people.

Page 6 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q2 For each of the statements given below, think and identify if it is Always True, Only
Sometimes True, or Never True. Share your reasoning. (a) If a person says ‘0’, then
they are the tallest in the group. (b) If a person is the tallest, then their number is
‘0’. (c) The first person’s number is ‘0’. (d) If a person is not first or last in line (i.e., if
they are standing somewhere in between), then they cannot say ‘0’. (e) The person
who calls out the largest number is the shortest. (f) What is the largest number
possible in a group of 8 people?

STATEMENT ANSWER REASON

(a) Says ‘0’ ⇒ tallest in Only Saying 0 only means nobody ahead is taller. The first child
the group Sometimes always says 0 even if he is the shortest of all.
True

(b) Tallest ⇒ number is Always True Nobody in the whole line is taller, so certainly nobody
‘0’ ahead is taller.

(c) First person’s number Always True There is nobody in front of the first person to count.
is ‘0’

(d) A person in between Only A middle child says 0 whenever he is taller than everybody
cannot say ‘0’ Sometimes ahead of him — as the third child does in 5, 6, 7, 1, 2, 3, 4.
True

(e) Largest number ⇒ Only In 7, 3, 5, 4, 1, 6, 2 the largest number 5 is said by the
shortest person Sometimes second-shortest child, not the shortest.
True

(f) The largest number possible in a group of 8 people is 7.

Only the children ahead of you can be counted.

The 8th (last) child has 7 children in front.

At best all 7 are taller → largest number = 7

Why it happens: a child standing at position p can say at most p – 1. So in a group of
n people the biggest number anybody can say is n – 1, and it happens only when the
last child is the shortest.

In-text Questions — Page 129

Page 7 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Section 6.2 Picking Parity

Q1 Can you figure out which 5 cards add to 30? Is it possible? There are many ways of
choosing 5 cards from this collection. Is there a way to find a solution without
checking all possibilities?

9 9 7 7 5 5 3 3
13 13 1 1 11 11
9 9 7 7 5 5 3 3
13 1 11

Kishor’s collection of number cards, page 129.

It is not possible. No choice of 5 cards from that collection adds up to 30.
Look at the cards Kishor has: 13, 9, 1, 7, 11, 5 and 3 — every card carries an odd number.

odd + odd = even

(odd + odd) + odd = even + odd = odd

5 odd numbers = (odd + odd) + (odd + odd) + odd = even + even + odd = odd

30 is even, so the sum can never be 30.

Why it happens: you do not have to test the hundreds of choices. Odd numbers pair
up two at a time to make even sums; with 5 cards one odd number is always left
over, so the total is always odd.

Tip: if the target had been an odd number such as 31, it would be easy — for
example 13 + 9 + 5 + 3 + 1 = 31.

Page 8 of 60

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Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Add a few even numbers together. What kind of number do you get? Does it matter
e
Q2

m l as
.co
how many numbers are added?

a g
se m
g l a
a

The sum is always even, and it does not matter how many even numbers you add.

co m
e m . ag
as
4 + 6 = 10 (even)

4 + 6 + 8 = 18 (even) a g l

m
4 + 6 + 8 + 10 + 12 = 40 (even)
co
se m.
o m l a
g pairs of dots with
Why it.chappens: every even number can be laid out as complete
m a
l a se left over. Put many such collections of pairs together and you still have only
nothing
a g complete pairs — so the total is even again.

m a s
m .co agl
l a se
Q3
a g
Now, add a few odd numbers together. What kind of number do you get? Does it
matter how many odd numbers are added?

co m
m .
e

m l as
.co g
Here it does matter how many you add.
m
eHOW a
a s
agl MANY ODD NUMBERS EXAMPLE PARITY OF THE SUM

se m
com a
2 3+5=8 even

. 3 + 5 + 7 = 15 a g l
e m
as
3 odd

4 agl 3 + 5 + 7 + 9 = 24 even

co m
.
5 3 + 5 + 7 + 9 + 11 = 35 odd

e m
m l as
.co a g
So: an even count of odd numbers gives an even sum, and an odd count of odd numbers gives

a s em an odd sum.

agl
.c
m
Why it happens: each odd number is a set of pairs with one extra dot. Two extra

m a s e
co agl
dots join to make one more pair, so the leftovers cancel two at a time. If the number

m .
e
of odd numbers is even, all leftovers pair up and the sum is even; if it is odd, exactly

g l as
one dot is left over and the sum is odd.
a

co m
m .
m as e
.co


a g l Page 9 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

In-text Questions — Page 130
Section 6.2 Picking Parity

Q1 What about adding 3 odd numbers? Can the resulting sum be arranged in pairs?

No. The sum of 3 odd numbers is always odd, so it can never be arranged in pairs.

odd + odd = even (the two leftover dots join into a pair)

even + odd = odd (one dot is left over again)

So odd + odd + odd = odd

Example: 5 + 7 + 9 = 21, and 21 dots leave one dot alone.

Why it happens: the three leftover dots cannot all be matched — two of them make
a pair and the third has no partner.

Q2 Explore what happens to the sum of (a) 4 odd numbers, (b) 5 odd numbers, and (c) 6
odd numbers.

NUMBER OF ODD NUMBERS EXAMPLE SUM PARITY

(a) 4 1+3+5+7 16 even

(b) 5 1+3+5+7+9 25 odd

(c) 6 1 + 3 + 5 + 7 + 9 + 11 36 even

(a) even (b) odd (c) even

Why it happens: the leftover dots cancel two at a time. With 4 or 6 odd numbers
every leftover finds a partner; with 5 odd numbers one dot is always stranded.

Page 10 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q3 Two siblings, Martin and Maria, were born exactly one year apart. Today they are
celebrating their birthday. Maria exclaims that the sum of their ages is 112. Is this
possible? Why or why not?

No, it is not possible.

Born one year apart ⇒ their ages are consecutive numbers.

In any two consecutive numbers, one is even and the other is odd.

even + odd = odd

But 112 is even. So the sum of their ages can never be 112.

Try a few pairs and see:

55 + 56 = 111 (odd)

56 + 57 = 113 (odd)

51 + 52 = 103 (odd)

Tip: 111 and 113 are the sums just below and just above 112 — the sum jumps by 2
each time and skips every even number.

Figure it Out — Page 131
Section 6.2 Picking Parity

Q1 Using your understanding of the pictorial representation of odd and even numbers,
find out the parity of the following sums: (a) Sum of 2 even numbers and 2 odd
numbers (e.g., even + even + odd + odd) (b) Sum of 2 odd numbers and 3 even
numbers (c) Sum of 5 even numbers (d) Sum of 8 odd numbers

Even numbers bring only complete pairs. Only the odd numbers bring leftover dots, so just
count how many odd numbers there are.

Page 11 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

SUM NUMBER OF ODD NUMBERS LEFTOVER DOTS PARITY

(a) 2 even + 2 odd 2 (even count) pair up fully even

(b) 2 odd + 3 even 2 (even count) pair up fully even

(c) 5 even 0 none even

(d) 8 odd 8 (even count) pair up fully even

All four sums are even.

(a) 4 + 6 + 3 + 5 = 18 even

(b) 3 + 5 + 2 + 4 + 6 = 20 even

(c) 2 + 4 + 6 + 8 + 10 = 30 even

(d) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 even

Why it happens: the even numbers never disturb the pairing. The parity of the
whole sum depends only on how many odd numbers are added — even count →
even sum, odd count → odd sum.

Q2 Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins and an even
number of ₹10 coins in his piggy bank. He calculated the total and got ₹205. Did he
make a mistake? If he did, explain why. If he didn’t, how many coins of each type
could he have?

Yes, Lakpa has made a mistake. With those counts the total can never be ₹205.
Let him have a coins of ₹1 (a odd), b coins of ₹5 (b odd) and c coins of ₹10 (c even).

Value from ₹1 coins = a → odd

Value from ₹5 coins = 5b, and odd × odd = odd

Value from ₹10 coins = 10c → even (10 is even)

Total = odd + odd + even = even + even = even

But ₹205 is odd. So the total is impossible.

Page 12 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Check with an example: 3 coins of ₹1, 5 coins of ₹5, 4 coins of ₹10.

3 × 1 + 5 × 5 + 4 × 10 = 3 + 25 + 40 = ₹68 (even, as expected)

Why it happens: a ₹10 coin can never change the parity, and the ₹1 part and the ₹5
part are each odd. Two odd amounts always add to an even amount, so Lakpa's total
must be an even number of rupees.

Check it yourself: ₹204 or ₹206 would be possible, but no odd total ever is.

Q3 We know that: (a) even + even = even (b) odd + odd = even (c) even + odd = odd.
Similarly, find out the parity for the scenarios below: (d) even – even = ___ (e) odd –
odd = ___ (f) even – odd = ___ (g) odd – even = ___

OPERATION PARITY EXAMPLE

(d) even – even even 10 – 4 = 6

(e) odd – odd even 9–3=6

(f) even – odd odd 10 – 3 = 7

(g) odd – even odd 9–4=5

Why it happens: subtraction behaves exactly like addition here. Taking away
complete pairs from complete pairs leaves complete pairs (even). Taking one leftover
dot away from another leftover dot also leaves complete pairs (even). Only when one
side has a leftover dot and the other does not is a dot stranded — and the answer is
odd.

Tip: a short way to remember all eight facts — the answer is odd exactly when the
two numbers have different parity, for both + and –.

In-text Questions — Page 131

Page 13 of 60

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Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

Small Squares in Grids
co m
e m.
m l as
Q1
.co
Given the dimensions of a grid, can you tell the parity of the number of small
m a g
l a se
squares without calculating the product?

a g

co m
. ag
Yes. Just look at the parity of the two dimensions.
e m
g l as
GRID M × N
a
NUMBER OF SMALL SQUARES EXAMPLE

om
both m and n odd odd 3×3=9

. c
em3 × 4 = 12
m as
one of them even even

. c o a g l
a s em
both even even 4 × 6 = 24

agl
The count of small squares is m × n.
m a s
m .co agl
se
odd × odd = odd

g l a
even × any number = even
a

co m
m .
e
Why it happens: if one side is even, the whole grid can be cut into complete pairs of
m l as
.co g
columns (or rows), so the squares pair up perfectly. If both sides are odd, each row

em a
s
has one square left over, and there is an odd number of rows, so one square is finally
a
l stranded.
ag
se m
com g l a
. a
sem
In-text Questions — Pagea132
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


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Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Parity of Expressions

Q1 Find the parity of the number of small squares in these grids: (a) 27 × 13 (b) 42 × 78
(c) 135 × 654

GRID PARITY OF THE TWO PARITY OF THE ACTUAL
SIDES COUNT PRODUCT

(a) 27 × 13 odd, odd odd 351

(b) 42 × 78 even, even even 3276

(c) 135 × odd, even even 88290
654

Why it happens: a product is odd only when every factor is odd. A single even factor
supplies a 2, and that makes the whole product even.

Check it yourself: the parity was found without multiplying — the products are
shown only to confirm the answers.

Q2 Come up with an expression that always has even parity. Some examples are: 100p
and 48w – 2. Try to find more.

Any expression in which every term is a multiple of 2 always gives an even value.

2n • 6k • 10m • 4a + 8 • 12t – 6 • 100p • 48w – 2

Test one of them:

4a + 8 → a = 1: 12, a = 2: 16, a = 3: 20 — all even

Why it happens: write the expression as 2 × (something). For example 4a + 8 = 2(2a
+ 4). A number with 2 as a factor can always be arranged in pairs, so it is even.

Page 15 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q3 Come up with expressions that always have odd parity.

Take an always-even expression and add or subtract 1 (or any odd number).

2n + 1 • 2m – 1 • 4n + 3 • 6k + 5 • 8a + 3 • 10p – 7

Test one of them:

6k + 5 → k = 1: 11, k = 2: 17, k = 3: 23 — all odd

Why it happens: 6k is a collection of complete pairs; adding 5 leaves one dot
without a partner (since 5 itself is odd). So the value can never be even.

Q4 Come up with other expressions, like 3n + 4, which could have either odd or even
parity.

Use a term whose coefficient is odd, so that the parity flips as n changes.

3n + 4 • 5n + 2 • 7n + 1 • 9n – 4 • n + 3

N 5N + 2 PARITY

1 7 odd

2 12 even

3 17 odd

4 22 even

Why it happens: in 5n + 2 the part 5n is odd when n is odd and even when n is even.
Adding the fixed even number 2 does not change that, so the answer alternates
between odd and even.

Page 16 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q5 The expression 6k + 2 evaluates to 8, 14, 20,... (for k = 1, 2, 3,...) — many even
numbers are missing. Are there expressions using which we can list all the even
numbers? Hint: All even numbers have a factor 2.

Yes — the expression is 2n.

n = 1, 2, 3, 4, 5, …

2n = 2, 4, 6, 8, 10, … — every even number appears, none is skipped.

Compare with 6k + 2:

EXPRESSION VALUES WHAT IS MISSED

6k + 2 8, 14, 20, 26, … 2, 4, 6, 10, 12, 16, …

2n 2, 4, 6, 8, 10, … nothing

Why it happens: 6k + 2 = 2(3k + 1), so it only produces even numbers whose half is
of the form 3k + 1 — it jumps in steps of 6. The expression 2n moves in steps of 2
and therefore catches every even number.

Q6 Are there expressions using which we can list all odd numbers?

Yes — 2n – 1 (with n = 1, 2, 3, …).

n = 1 → 2(1) – 1 = 1

n = 2 → 2(2) – 1 = 3

n = 3 → 2(3) – 1 = 5

n = 4 → 2(4) – 1 = 7 … and so on

The expression 2n + 1 also gives only odd numbers, but starting from n = 1 it gives 3, 5, 7, … and
misses 1.

Page 17 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Why it happens: every odd number is one less than an even number. Since 2n runs
through all even numbers, 2n – 1 runs through all odd numbers.

Q7 What would be the nth term for multiples of 2? Or, what is the nth even number?

The nth even number is 2n.

1st even number = 2 × 1 = 2

2nd even number = 2 × 2 = 4

5th even number = 2 × 5 = 10

23rd even number = 2 × 23 = 46

Why it happens: the even numbers are exactly the multiples of 2, and the nth
multiple of 2 is 2 × n.

Q8 What is the 100th odd number?

The 100th odd number is 199.

100th even number = 2 × 100 = 200

At every position the odd number is one less than the even number:
Even: 2, 4, 6, 8, 10, …

Odd: 1, 3, 5, 7, 9, …

100th odd number = 200 – 1 = 199

Why it happens: the two lists march together, position by position, and the odd list
is always exactly 1 behind — so 2n and 2n – 1 give the nth even and odd numbers.

Page 18 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
In-text Questions — Page 133
m as e
.co
Section 6.3 Some Explorations in Grids
a g l
a s em
aQ1gl What is the 100th even number?

co m
. ag

e m
The 100th even number is 200.
g l as
a
m
nth even number = 2n
co
em.
as
100th even number = 2 × 100 = 200
m l
.co a g
a s em
a gl knowing this immediately gives the 100th odd number as 200 – 1 = 199.
Tip:

m a s
m .co agl
Q2
l a
Write a formula to find the nth odd number.
se
a g
m

. co
The nth odd number is 2n – 1.
e m
m l as
m .co a g
l a se Step 1: find the even number at that position → 2n

ag Step 2: subtract 1 from it → 2n – 1

se m
com g l a
m . a
ase
N 2N 2N – 1

1 2 agl 1

co m
.
2 4 3

em
m l as
.co g
7 14 13

em 50 a
a s
gl
100 99

a c
m .
m a s e
co agl
Why it happens: 2n is always even, so 2n – 1 is always odd, and as n runs through 1,

m .
e
2, 3, … the values 1, 3, 5, 7, … appear in order without a gap.

g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q3 Are you able to see what the circled numbers represent?

Yes — the number in each yellow circle is the sum of the row or column it sits beside.
For the grid shown in the book:

C1 C2 C3 ROW SUM

R1 4 7 5 16

R2 6 1 2 9

R3 3 9 8 20

Column sum 13 17 15 —

4 + 7 + 5 = 16 • 6 + 1 + 2 = 9 • 3 + 9 + 8 = 20
4 + 6 + 3 = 13 • 7 + 1 + 9 = 17 • 5 + 2 + 8 = 15

Check it yourself: 16 + 9 + 20 = 45 and 13 + 17 + 15 = 45 — both equal 1 + 2 + … + 9.

Q4 Fill the grids below based on the rule mentioned above:

Use the numbers 1 – 9, each exactly once, so that every row and every column adds to the
circled number.
First grid — row sums 13, 14, 18 and column sums 24, 9, 12, with 9 in the top-left corner and 5
in the bottom-right corner.

C1 C2 C3 ROW SUM

R1 9 1 3 13

R2 8 2 4 14

R3 7 6 5 18

Column sum 24 9 12 45

Page 20 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

How it was found:

Row 1: 9 + ? + ? = 13 → the other two add to 4 → they are 1 and 3

Column 1: 9 + ? + ? = 24 → the other two add to 15 → they are 7 and 8

Row 3: ? + ? + 5 = 18 → the first two add to 13 → so 7 goes in R3C1 and 6 in R3C2

Column 2: 1 + ? + 6 = 9 → the middle entry is 2

Row 2: 8 + 2 + ? = 14 → the last entry is 4 ✓
Second grid — row sums 24, 15, 6 and column sums 12, 16, 17, with 4 in the middle-left cell and
3 in the bottom-right cell.

C1 C2 C3 ROW SUM

R1 7 8 9 24

R2 4 6 5 15

R3 1 2 3 6

Column sum 12 16 17 45

Row 1 must be 24, and the largest possible sum of three different numbers is 9 + 8 + 7 = 24

→ row 1 is 7, 8, 9 in some order

Row 3 must be 6, the smallest possible sum, and 3 is already there → the other two are 1 and

2

Column 1: ? + 4 + ? = 12 → 7 + 4 + 1 = 12 ✓

Column 2: 8 + 6 + 2 = 16 ✓ Column 3: 9 + 5 + 3 = 17 ✓

Did you know? the second grid has one more answer: 7, 9, 8 / 4, 5, 6 / 1, 2, 3 also fits
every circle. The first grid, however, has exactly one answer.

Why it happens: the three row sums always add to 1 + 2 + … + 9 = 45, and so do the
three column sums. Checking this first (13 + 14 + 18 = 45, 24 + 9 + 12 = 45) tells you
the puzzle is worth attempting.

Page 21 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

In-text Questions — Page 134
Section 6.3 Some Explorations in Grids

Q1 Make a couple of questions like this on your own and challenge your peers.

Start from a filled grid and hide the numbers — then only the circles remain.
Here is one you can set:

C1 C2 C3 ROW SUM

R1 2 9 4 15

R2 7 5 3 15

R3 6 1 8 15

Column sum 15 15 15 45

Give your friend only the six circled numbers 15, 15, 15 (rows) and 15, 15, 15 (columns), plus one
filled cell as a clue.

Tip for setting a good puzzle: always check that your three row sums add to 45 and
your three column sums add to 45, and that no circle is below 6 or above 24.
Otherwise your puzzle will have no answer at all.

Page 22 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q2 You might have realised that it is not possible to find a solution for this grid. Why is
this the case?

5

6 21

19

9 11 26

The grid on page 134 — the circles carry the row sums and the column sums.

Because two of the circled numbers lie outside the possible range.

Smallest sum of three different numbers from 1 – 9 = 1 + 2 + 3 = 6

Largest sum of three different numbers from 1 – 9 = 9 + 8 + 7 = 24

So every circle must satisfy 6 ≤ circle ≤ 24.

This grid shows the circles 5, 21, 19 (rows) and 9, 11, 26 (columns).

Page 23 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
CIRCLE ALLOWED? REASON

m No as e
5
.co a
less than the smallest possible sum 6
g l
se m
g l a more than the largest possible sum 24
a
26 No

21, 19, 9, 11 Yes between 6 and 24

co m
e m . ag
as
So the grid is impossible.

a g l
Why it happens: you did not have to try filling a single cell. Just testing the range of
the circles settles the question at once.
co m
se m.
o m
Check.cit yourself: the row sums here are 5 + 21 + 19 = 45 and l a
gthe column sums are
m a
9s+e 11 + 26 = 46. The column total is not even 45, which is a second reason the grid
g l a
a cannot work.

m a s
m .co agl
l a se
Q3
a g
Why should the row sums and column sums always add to 45?

. c om
m exactly once.
Because adding all three row sums means adding every number in theegrid
s
co m gl a
.
m sums added together a
a s eRow
a gl
m
= (all numbers of row 1) + (all of row 2) + (all of row 3)
a se
=1+2+3+4+5+6+7+8+9
.com a g l
m
ase
agl
= 45

m
The same argument works for columns, because the three columns also cover the whole grid
. co
m
once.

a se
comSOLVED EARLIER
.GRID l
agCOLUMN SUMS
sem
ROW SUMS TOTAL TOTAL

a
agl 475/612/398 16, 9, 20 45 13, 17, 15 45
c
m .
m a s e
.co agl
913/824/765 13, 14, 18 45 24, 9, 12 45

se m24, 15, 6
l a
789/465/123 45 12, 16, 17 45

ag

co m
m .
m as e
.co


a g l Page 24 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Why it happens: the rows do not overlap and together they use up all nine cells. So
their total is fixed at 1 + 2 + … + 9 = 45, whatever arrangement you choose. That also
explains Kishor's observation that all six circles add to 45 + 45 = 90.

Q4 What can the magic sum be? Can it be any number?

No. For a magic square made from the numbers 1 – 9 the magic sum must be 15.

The three row sums always add to 45.

In a magic square all three row sums are equal, say S.

S + S + S = 45

3S = 45

S = 15

Observation 1: In a magic square made using the numbers 1 – 9, the magic sum must be 15.

Why it happens: the magic sum is not free to choose — it is forced by the numbers
you are allowed to use. Use 2 – 10 instead and the total becomes 54, so the magic
sum becomes 54 ÷ 3 = 18.

In-text Questions — Page 135
Section 6.3 Some Explorations in Grids

Q1 What are the possible numbers that could occur at the centre of a magic square?

Only 5 can sit at the centre.
Test 9 at the centre. Then 8 must go somewhere else, and 8 shares a line with the centre or does
not; wherever it goes, some line contains 8 and 9:

Page 25 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

8 + 9 + (another number) = 15

another number = 15 – 17 = –2 — impossible with 1 – 9.

Test 1 at the centre. Then 2 must go somewhere, and some line contains 1 and 2:

2 + 1 + (another number) = 15

another number = 15 – 3 = 12 — impossible, we only have 1 – 9.

So neither 9 nor 1 can be at the centre.

Why it happens: the centre cell belongs to four lines — one row, one column and
both diagonals. So the centre number must pair up with the other eight numbers in
four different pairs, each pair adding to 15 – centre.

Q2 Using such reasoning, find out which other numbers 1 – 9 cannot occur at the
centre.

Every number except 5 fails. Count, for each choice of centre c, how many pairs from the
remaining numbers add up to 15 – c. Four pairs are needed.

Page 26 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

CENTRE PAIRS MUST ADD TO 15 – PAIRS HOW ALLOWED?
C C AVAILABLE MANY

1 14 5+9, 6+8 2 No

2 13 4+9, 5+8, 6+7 3 No

3 12 4+8, 5+7 2 No

4 11 2+9, 3+8, 5+6 3 No

5 10 1+9, 2+8, 3+7, 4+6 4 Yes

6 9 1+8, 2+7, 4+5 3 No

7 8 2+6, 3+5 2 No

8 7 1+6, 2+5, 3+4 3 No

9 6 1+5, 2+4 2 No

Observation 2: The number occurring at the centre of a magic square, filled using 1 – 9, must
be 5.

Why it happens: the centre lies on four lines, so the other eight numbers must split
into four pairs, each adding to 15 – c. Only c = 5 gives exactly four such pairs — and 5
is also the middle number of 1 – 9.

In-text Questions — Page 136
Section 6.3 Some Explorations in Grids

Q1 If yes, then there should exist three ways of adding 1 with two other numbers to
give 15. We have 1 + 5 + 9 = 1 + 6 + 8 = 15. Is any other combination possible?

No, there is no third combination. Only two pairs add to 14.

Page 27 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

1 + ? + ? = 15 → the other two must add to 14

14 = 5 + 9 ✓

14 = 6 + 8 ✓
14 = 7 + 7 ✗ (a number cannot repeat)

So only two lines through 1 are possible.

A corner cell lies on three lines — one row, one column and one diagonal — so it needs three
such combinations. Since 1 has only two, 1 cannot occur in a corner.

Why it happens: a corner is a busy cell. It must join two other numbers three
separate times, all making 15. A small number like 1 needs very large partners, and
there are simply not enough large numbers to do it three times.

Q2 Similarly, can 9 can be placed in a corner position?

No. The same counting rules it out.

9 + ? + ? = 15 → the other two must add to 6

6=1+5 ✓

6=2+4 ✓

6 = 3 + 3 ✗ (repeat not allowed)

Only two lines through 9 are possible, but a corner needs three.

Observation 3: The numbers 1 and 9 cannot occur in any corner, so they should occur in one of
the middle positions.

Why it happens: 9 is so large that its two partners must be very small, and there are
only two such small pairs available.

Page 28 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Can you find the other possible positions for 1 and 9?
e
Q3

m l as
.co a g
a
s em
a gl 9 must sit in middle cells of the edges, and they must be opposite each other, with 5
1 and
between them.

co m
m . ag
1 + 5 + 9 = 15, and 5 is at the centre.
l a se
ag
So 1 and 9 lie at the two ends of a line through the centre.

A corner is ruled out, so that line is a middle row or a middle column.
co m
em.
m l as
.co
The four possibilities are:
a g
a emthe middle of the left edge, 9 in the middle of the right edge
1sin
a gl 9 in the middle of the left edge, 1 in the middle of the right edge
1 in the middle of the top edge, 9 in the middle of the bottom edge
m a s
em
.co
9 in the middle of the top edge, 1 in the middle of the bottom edge
agl
a s
gl at the centre and 1 and 9 are pushed out of the
Why it happens: once 5 isafixed
corners, the only cells left for them are the four edge-middles — and 1, 5, 9 must line

co m
.
up because 1 + 5 + 9 = 15.
e m
m l as
.co a g
a s em
a gl Q4 Now, we have one full row or column of the magic square! Try completing it! [Hint:
m
First fill the row or columns containing 1 and 9]

a se
. com a g l
e m
asrow, then finish the grid.

a g l
Start with 1, 5, 9 filling the middle

m
.co
C1 C2 C3 SUM

sem
com l a
R1 8 3 4 15

.R2 ag
e m
as
1 5 9 15

agl c
15
.
R3 6 7 2

s e m
m15 a
. co agl
Sum 15 15 —

e m
g l as
a

co m
m .
m as e
.co


a g l Page 29 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Diagonals: 8 + 5 + 2 = 15 ✓ and 4 + 5 + 6 = 15 ✓

Column 1: 1 must join two numbers adding to 14 → 8 and 6

Column 3: 9 must join two numbers adding to 6 → 4 and 2
Column 2: 3 + 5 + 7 = 15 ✓

Turning and flipping this grid gives the familiar form of the same magic square:

8 1 6

3 5 7

4 9 2

Check it yourself: rows 15, 15, 15; columns 15, 15, 15; diagonals 8 + 5 + 2 = 15 and 6
+ 5 + 4 = 15.

Figure it Out — Page 136
Section 6.3 Some Explorations in Grids

MATH TALK

Q1 How many different magic squares can be made using the numbers 1 – 9?

There is essentially only one magic square using 1 – 9. Counting turns and mirror images as
different, there are 8.

8 1 6

3 5 7

4 9 2

Rotations: 4 (turn by 0°, 90°, 180°, 270°)

Each rotation also has a mirror image: 4 × 2 = 8 grids

But all 8 are the same square seen differently.

Page 30 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Why it happens: we proved that 5 must be at the centre, and that 1 and 9 must be
opposite edge-middles. Those two facts leave no freedom at all except for turning
the paper or looking at it in a mirror.

Q2 Create a magic square using the numbers 2 – 10. What strategy would you use for
this? Compare it with the magic squares made using 1 – 9.

Strategy: the numbers 2 – 10 are just the numbers 1 – 9 with 1 added to each. So add 1 to
every entry of the 1 – 9 magic square.

MAGIC SQUARE WITH 1 – 9 ADD 1 TO EACH MAGIC SQUARE WITH 2 – 10

8 1 6 → 9 2 7

3 5 7 → 4 6 8

4 9 2 → 5 10 3

Rows: 9 + 2 + 7 = 18, 4 + 6 + 8 = 18, 5 + 10 + 3 = 18

Columns: 9 + 4 + 5 = 18, 2 + 6 + 10 = 18, 7 + 8 + 3 = 18

Diagonals: 9 + 6 + 3 = 18, 7 + 6 + 5 = 18

Magic sum = 15 + 3 × 1 = 18

Comparison: the pattern is exactly the same. The centre is now 6 (the middle
number of 2 – 10), and the magic sum grew by 3, because each of the three numbers
in a line gained 1.

Q3 Take a magic square, and (a) increase each number by 1 (b) double each number. In
each case, is the resulting grid also a magic square? How do the magic sums change
in each case?

Yes — in both cases the result is again a magic square. Start from 8 1 6 / 3 5 7 / 4 9 2 with
magic sum 15.
(a) Increase each number by 1

Page 31 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

9 2 7

4 6 8

5 10 3

Every line gains 1 + 1 + 1 = 3.

New magic sum = 15 + 3 = 18

(b) Double each number

16 2 12

6 10 14

8 18 4

Every line is doubled.

New magic sum = 15 × 2 = 30

Check: 16 + 2 + 12 = 30, 6 + 10 + 14 = 30, 8 + 18 + 4 = 30 ✓

Why it happens: adding the same number k to all nine cells raises every line by 3k;
multiplying all nine cells by k multiplies every line by k. Since all lines change in the
same way, they stay equal — so the grid stays magic.

Page 32 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q4 What other operations can be performed on a magic square to yield another magic
square?

OPERATION STILL MAGIC? NEW MAGIC SUM (IF OLD SUM IS S)

Add a number k to every cell Yes S + 3k

Subtract a number k from every cell Yes S – 3k

Multiply every cell by k Yes k×S

Divide every cell by k (k ≠ 0) Yes S÷k

Rotate the grid by 90°, 180° or 270° Yes S (unchanged)

Reflect the grid in a mirror Yes S (unchanged)

Swap only two numbers Usually no the lines stop being equal

Why it happens: an operation keeps a square magic if it treats every cell in the same
way. Adding, subtracting, multiplying, dividing and turning all do that; changing just
one or two cells does not.

Try This: multiply 8 1 6 / 3 5 7 / 4 9 2 by 3 to get 24 3 18 / 9 15 21 / 12 27 6 with magic
sum 45.

Q5 Discuss ways of creating a magic square using any set of 9 consecutive numbers
(like 2 – 10, 3 – 11, 9 – 17, etc.).

Take the 1 – 9 magic square and add the same number to every cell. The number to add is
(smallest number of the new set) – 1.

Page 33 of 60

ase
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

SET OF 9 CONSECUTIVE ADD TO EACH CELL OF 1 – 9 CENTRE
. om
cSUM
MAGIC

e m
com as
NUMBERS SQUARE

. a g l
m
ase
1–9 0 5 15

a2g– l10 1 6 18

m
.co ag
3 – 11 2 7 21

a sem
agl
9 – 17 8 13 39

For 9 – 17, adding 8 to each entry gives:

co m
em.
m as
16 9 14

.co 15a
g l
m
ase
11 13

agl12 17 10

om a s
Rows: 16 + 9 + 14 = 39, 11 + 13 + 15m
e
. c agl
s
= 39, 12 + 17 + 10 = 39 ✓
a
Columns: 16 + 11 + 12 = 39, a9g+l 13 + 17 = 39, 14 + 15 + 10 = 39 ✓

m
Diagonals: 16 + 13 + 10 = 39, 14 + 13 + 12 = 39 ✓
. co
se m
o m l a
g same spacing pattern
m
Why.c it happens: nine consecutive numbers have exactly athe
aseas 1 – 9 — each one is a fixed amount bigger. The middle number always goes at
agl the centre and the magic sum is 3 × centre.

se m
com g l a
m . a
Math Talk — Page 137 gl ase
Generalising a 3 × 3 Magic Square
a

co m
MATH TALK

m .
o m l a se
m .c Choose any magic square that you have made so faragusing consecutive numbers. If
se
Q1
a
agl
m is the letter-number of the number in the centre, express how other numbers are
related to m, how much more or less than m. [Hint: Remember, how we described a
.c
s e m
m a
2 × 2 grid of a calendar month in the Algebraic Expressions chapter].

e m . co agl
g l as
a
Take the magic square 8 1 6 / 3 5 7 / 4 9 2. Its centre is 5, so write every entry as “5 and
something”.

co m
m .
m ase
.co


a g l Page 34 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

8=5+3 • 1=5–4 • 6=5+1

3=5–2 • 5=5 • 7=5+2

4=5–1 • 9=5+4 • 2=5–3

Replacing 5 by the letter-number m gives the generalised form:

m+3 m–4 m+1

m–2 m m+2

m–1 m+4 m–3

Row 1: (m + 3) + (m – 4) + (m + 1) = 3m

Row 2: (m – 2) + m + (m + 2) = 3m

Row 3: (m – 1) + (m + 4) + (m – 3) = 3m

Column 1: (m + 3) + (m – 2) + (m – 1) = 3m

Diagonal: (m + 3) + m + (m – 3) = 3m ✓

Why it happens: the nine numbers m – 4, m – 3, …, m + 4 are just nine consecutive
numbers with m in the middle. In every line the two numbers on either side of m
cancel each other — one is as much above m as the other is below.

Q2 Once the generalised form is obtained, share your observations with the class.

Some observations worth sharing:

The magic sum is always 3m — three times the centre number.
The centre m is the middle number of the nine consecutive numbers used.
The four corners are m + 3, m + 1, m – 1, m – 3 — the numbers that are odd distances from
m.
The four edge-middles are m – 4, m + 2, m – 2, m + 4 — the numbers that are even
distances from m.
Any two cells opposite each other through the centre add to 2m, for example (m + 3) + (m –
3) = 2m.

Page 35 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

To make a magic square with any centre, just choose m — the pattern of + and – never
changes.

Try This: put m = 100 and you get 103 96 101 / 98 100 102 / 99 104 97, with
magic sum 300.

Figure it Out — Page 137
Generalising a 3 × 3 Magic Square

TRY THIS

Q1 Using this generalised form, find a magic square if the centre number is 25.

Put m = 25 in the generalised form.

GENERALISED M+3 M–4 M+1

Value 28 21 26

GENERALISED M–2 M M+2

Value 23 25 27

GENERALISED M–1 M+4 M–3

Value 24 29 22

The magic square is:

28 21 26

23 25 27

24 29 22

Magic sum = 3m = 3 × 25 = 75

Rows: 28 + 21 + 26 = 75, 23 + 25 + 27 = 75, 24 + 29 + 22 = 75 ✓

Columns: 28 + 23 + 24 = 75, 21 + 25 + 29 = 75, 26 + 27 + 22 = 75 ✓

Diagonals: 28 + 25 + 22 = 75, 26 + 25 + 24 = 75 ✓

Page 36 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Check it yourself: the nine entries are 21 to 29 — nine consecutive numbers with 25
in the middle, exactly as expected.

Q2 What is the expression obtained by adding the 3 terms of any row, column or
diagonal?

The expression is 3m.

Row 1: (m + 3) + (m – 4) + (m + 1) = 3m + 3 – 4 + 1 = 3m

Row 2: (m – 2) + m + (m + 2) = 3m – 2 + 2 = 3m
Column 2: (m – 4) + m + (m + 4) = 3m – 4 + 4 = 3m

Diagonal: (m + 1) + m + (m – 1) = 3m

Why it happens: each line has three m's, and the added and subtracted numbers
always cancel out. That is exactly why the square is magic — and it also shows the
magic sum is always three times the centre number.

Q3 Write the result obtained by— (a) adding 1 to every term in the generalised form.
(b) doubling every term in the generalised form

(a) Adding 1 to every term

m+4 m–3 m+2

m–1 m+1 m+3

m m+5 m–2

Line sum = (m + 4) + (m – 3) + (m + 2) = 3m + 3

So the magic sum increases by 3, and the new centre is m + 1.

(b) Doubling every term

Page 37 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

2m + 6 2m – 8 2m + 2

2m – 4 2m 2m + 4

2m – 2 2m + 8 2m – 6

Line sum = (2m + 6) + (2m – 8) + (2m + 2) = 6m

So the magic sum is doubled, and the new centre is 2m.

Why it happens: both grids are still magic because every cell was changed in the
same way. Notice that after doubling the entries are no longer consecutive — they
jump in steps of 2.

Q4 Create a magic square whose magic sum is 60.

Magic sum = 3m, so choose m = 20.

3m = 60

m = 60 ÷ 3 = 20

23 16 21

18 20 22

19 24 17

Rows: 23 + 16 + 21 = 60, 18 + 20 + 22 = 60, 19 + 24 + 17 = 60 ✓

Columns: 23 + 18 + 19 = 60, 16 + 20 + 24 = 60, 21 + 22 + 17 = 60 ✓

Diagonals: 23 + 20 + 17 = 60, 21 + 20 + 19 = 60 ✓

Tip: to make a magic square with any magic sum S that is a multiple of 3, simply take
m = S ÷ 3 and fill in the generalised form.

Page 38 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Is it possible to get a magic square by filling nine non-consecutive numbers?
e
Q5

m l as
.co a g
a
s em
aglit is possible. Multiply every entry of a known magic square by the same number.
Yes,
Multiplying 8 1 6 / 3 5 7 / 4 9 2 by 3 gives nine multiples of 3 — which are certainly not

co m
. ag
consecutive:

e m
g l as
a
24 3 18

m
9 15 21

co
e m.
as
12 27 6

m l
.co a g
a s em 24 + 3 + 18 = 45, 9 + 15 + 21 = 45, 12 + 27 + 6 = 45 ✓
a gl Rows:

Columns: 24 + 9 + 12 = 45, 3 + 15 + 27 = 45, 18 + 21 + 6 = 45 ✓
m a s
Diagonals: 24 + 15 + 6 = 45, 18 + 15 + 12 = 45 ✓
m .co agl
l a se
g
Magic sum = 15 × 3 = 45
a

. com
Why it happens: what a magic square really needs is the pattern of differences from
the centre, not consecutive numbers. Here every difference has e
s m been tripled:
m a
simply
3 ×.c5o= 15 is the centre, and the entries are 15 ± 3, 15 ± 6, a gl± 9, 15 ± 12.
sem
15

a
agl
m
Try This: any nine numbers of the form m – 4d, m – 3d, …, m + 4d work for any step

a se
com l
d. Take m = 20 and d = 5 to get 35 0 25 / 10 20 30 / 15 40 5, with magic sum 60.
. a g
m
ase
agl
In-text Questions — Page 137
co m
The First-ever 4 × 4 Magic Square
m .
o m l a se
g Chautīsā Yantra?
.c Chauṭīs means 34. Why do you think they called it the
a
se m
a
Q1

agl c
m .
e

om a s
. c agl
Because every row, every column and every diagonal of that 4 × 4 square adds up to 34 —

s e mthe yantra.
so the number 34 is the identity of
a
agl

co m
m .
m ase
.co


a g l Page 39 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

7 12 1 14

2 13 8 11

16 3 10 5

9 6 15 4

Rows: 7 + 12 + 1 + 14 = 34, 2 + 13 + 8 + 11 = 34, 16 + 3 + 10 + 5 = 34, 9 + 6 + 15 + 4 =

34

Columns: 7 + 2 + 16 + 9 = 34, 12 + 13 + 3 + 6 = 34, 1 + 8 + 10 + 15 = 34, 14 + 11 + 5 + 4

= 34

Diagonals: 7 + 13 + 10 + 4 = 34, 14 + 8 + 3 + 9 = 34

Why it happens: the square uses the numbers 1 to 16, whose total is 1 + 2 + … + 16
= 136. Spread equally over 4 rows, each row must get 136 ÷ 4 = 34. So 34 is forced.

Did you know? this inscription is from the 10th century Pārśhvanath Jain temple at
Khajuraho — the earliest recorded 4 × 4 magic square in the world.

Q2 Every row, column and diagonal in this magic square adds up to 34. Can you find
other patterns of four numbers in the square that add up to 34?

Yes — this square is extraordinarily rich. Here are patterns that all add to 34:

Page 40 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

PATTERN NUMBERS SUM

The four corners 7 + 14 + 9 + 4 34

The centre 2 × 2 block 13 + 8 + 3 + 10 34

Top-left 2 × 2 block 7 + 12 + 2 + 13 34

Top-right 2 × 2 block 1 + 14 + 8 + 11 34

Bottom-left 2 × 2 block 16 + 3 + 9 + 6 34

Bottom-right 2 × 2 block 10 + 5 + 15 + 4 34

Middle two of the top and bottom rows 12 + 1 + 6 + 15 34

Middle two of the left and right columns 2 + 16 + 11 + 5 34

A broken diagonal 12 + 8 + 5 + 9 34

Another broken diagonal 1 + 11 + 16 + 6 34

A “kite” shape 7 + 8 + 15 + 4 34

Its mirror 14 + 13 + 3 + 4 34

Why it happens: the Chautīsā Yantra is what mathematicians call a most-perfect
magic square. Every 2 × 2 block, and every set of four cells that “wraps around” the
edges like a broken diagonal, also totals 34. That is far more than an ordinary magic
square promises.

Try This: pick any 2 × 2 block anywhere in the grid, letting it wrap around the sides,
and add the four numbers. You will get 34 every single time.

In-text Questions — Pages 140–141

Page 41 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Section 6.4 Nature’s Favourite Sequence: The Virahāṅka–Fibonacci Numbers!

Q1 How many rhythms are there with 8 beats consisting of short syllables (1 beat) and
long syllables (2 beats)? Here are some possibilities: long long long long / short
short short short short short short short / short long long short long / long long
short short long. Can you find others?

Yes, many more. A short syllable takes 1 beat and a long syllable takes 2 beats, so every rhythm
is a way of making 8 out of 1s and 2s.

RHYTHM AS A SUM

short short long long long 1+1+2+2+2

long short long short long 2+1+2+1+2

long long long short short 2+2+2+1+1

short long short long long 1+2+1+2+2

long short short short long long 2+1+1+1+2+2

short short short short long long 1+1+1+1+2+2

Altogether there are 34 such 8-beat rhythms.

Tip: counting them one by one is slow and easy to get wrong. The systematic
method in the book gives the answer without listing.

Q2 Phrased more mathematically: In how many different ways can one write a number,
say 8, as a sum of 1’s and 2’s? For example, we have: 8 = 2 + 2 + 2 + 2, 8 = 1 + 1 + 1 + 1 +
1 + 1 + 1 + 1, 8 = 1 + 2 + 2 + 1 + 2, 8 = 2 + 2 + 1 + 1 + 2, etc. Do you see other ways?

There are 34 ways. Build them up step by step instead of listing them all.

N 1 2 3 4 5 6 7 8

Number of ways 1 2 3 5 8 13 21 34

Page 42 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

ways(3) = ways(2) + ways(1) = 2 + 1 = 3

ways(4) = ways(3) + ways(2) = 3 + 2 = 5

ways(5) = ways(4) + ways(3) = 5 + 3 = 8

ways(6) = ways(5) + ways(4) = 8 + 5 = 13

ways(7) = ways(6) + ways(5) = 13 + 8 = 21

ways(8) = ways(7) + ways(6) = 21 + 13 = 34

Some more ways for 8: 1 + 1 + 2 + 2 + 2, 2 + 1 + 2 + 1 + 2, 2 + 2 + 1 + 2 + 1, 1 + 2 + 1 + 1 + 1 + 2,
2 + 1 + 1 + 1 + 1 + 2.

Why it happens: every sum for 8 must begin with either a 1 or a 2. If it begins with
1, the rest makes 7; if it begins with 2, the rest makes 6. So the count for 8 is the
count for 7 plus the count for 6.

Q3 Try writing the number 5 as a sum of 1s and 2s in all possible ways in your
notebook! How many ways did you find? (You should find 8 different ways!) Can you
figure out the answer without listing down all the possibilities? Can you try it for n
= 8?

There are exactly 8 ways for n = 5:

1+1+1+1+1 1+1+1+2 1+1+2+1 1+2+1+1

1+2+2 2+1+1+1 2+1+2 2+2+1

Without listing: put a “1 +” in front of every 4-beat rhythm (5 of them) and a “2 +” in front of
every 3-beat rhythm (3 of them).

ways(5) = ways(4) + ways(3) = 5 + 3 = 8

ways(8) = ways(7) + ways(6) = 21 + 13 = 34

Why it happens: the first syllable is either short (1 beat) or long (2 beats). Those two
cases never overlap and together they cover everything, so the counts simply add.
This gives the Virahāṅka sequence 1, 2, 3, 5, 8, 13, 21, 34, …

Page 43 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Use the systematic method to write down all 6-beat rhythms, i.e., write 6 as the
e
Q4

m l as
.co
sum of 1’s and 2’s in all possible ways. Did you get 13 ways?

a g
se m
g l a
a

Yes — exactly 13 ways. Write “1 +” before each of the eight 5-beat rhythms and “2 +” before each

m
of the five 4-beat rhythms.
. co ag
e m
STARTING WITH 1 + (8 WAYS)
g l as STARTING WITH 2 + (5 WAYS)

1+1+1+1+1+1
a 2+1+1+1+1

co m
m.
1+1+1+1+2 2+1+1+2

m as e
.co a g l
m
1+1+1+2+1 2+1+2+1

l a se
a g 1+1+2+1+1 2+2+1+1

m a s
1+1+2+2 2+2+2

m .co — agl
se
1+2+1+1+1

g l a
1+2+1+2 a —

co m
.
1+2+2+1 —

e m
om g l as
8m+.c5 = 13 ✓ a
ase
agl
se m
com a
Check it yourself: add up the beats in any row — every single sum comes to 6.

. a g l
m
gl ase
a
In-text Questions — Page 142

co m
.
Section 6.4 The Virahāṅka–Fibonacci Numbers

em
c o mWrite the next number in the sequence, after 55. glas
m . a
e
Q1

a s
agl c
m .
m a s e
co agl
The next number is 89.

m .
ase
a g l

co m
m .
m ase
.co


a g l Page 44 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Rule: each number = sum of the two numbers before it.

1 + 2 = 3 • 2 + 3 = 5 • 3 + 5 = 8 • 5 + 8 = 13

8 + 13 = 21 • 13 + 21 = 34 • 21 + 34 = 55

34 + 55 = 89

Check it yourself: the rule holds for every number already printed, which is why we
can trust it for the next one.

Q2 Write the next 3 numbers in the sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ____, ____, ____,
… If you have to write one more number in the sequence above, can you tell
whether it will be an odd number or an even number (without adding the two
previous numbers)?

The next three numbers are 144, 233, 377.

55 + 89 = 144

89 + 144 = 233

144 + 233 = 377

The number after 377 will be even, and you can tell without adding.

Parities so far: 1(O), 2(E), 3(O), 5(O), 8(E), 13(O), 21(O), 34(E), 55(O), 89(O), 144(E),

233(O), 377(O)

Last two are odd and odd → odd + odd = even

Why it happens: the parities repeat in the block odd, odd, even over and over. After
two odds the next term must be even. (The actual value is 233 + 377 = 610, which is
indeed even.)

Page 45 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q3 What is the parity of each number in the sequence? Do you notice any pattern in
the sequence of parities?

POSITION 1 2 3 4 5 6 7 8 9 10 11 12

Number 1 2 3 5 8 13 21 34 55 89 144 233

Parity O E O O E O O E O O E O

The pattern is O, E, O, O, E, O, O, E, … — after the first term the block odd, odd, even repeats
forever.

O+E=O

E+O=O

O + O = E … and the cycle starts again

So a term is even exactly when its position is 2nd, 5th, 8th, 11th, 14th, … — that is, when the
position leaves remainder 2 on division by 3.

Why it happens: the parity of a term depends only on the parities of the two before
it. Since there are only a few parity pairs, the pattern must repeat — and here it
repeats every three terms.

Q4 How many petals do you see on each of these flowers?

Counting the petals in the three daisy pictures gives 13, 21 and 34.

FLOWER PETALS COUNTED IS IT A VIRAHĀṄKA NUMBER?

First daisy 13 Yes — 6th term

Second daisy 21 Yes — 7th term

Third daisy 34 Yes — 8th term

And 13, 21, 34 are three numbers standing next to each other in the sequence 1, 2, 3, 5, 8, 13,
21, 34, 55, …

Page 46 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Why it happens: petals, sunflower spirals and pine-cone spirals grow in a way that
packs seeds and petals most efficiently, and this packing throws up Virahāṅka–
Fibonacci numbers again and again in nature.

Did you know? next time you see a marigold, a sunflower or a pineapple, count the
spirals — you will very often land on 8, 13, 21 or 34.

In-text Questions — Page 143
Section 6.5 Digits in Disguise

Q1 What could U and T be? Can T be 2? Can it be 3?

The puzzle is T + T + T = UT, that is, 3 × T must be a 2-digit number whose units digit is again T.

T 3×T UNITS DIGIT SAME AS T?

2 6 6 No (and 6 is not 2-digit)

3 9 9 No

4 12 2 No

5 15 5 Yes

6 18 8 No

7 21 1 No

8 24 4 No

9 27 7 No

So T = 5 and U = 1, giving UT = 15.

5 + 5 + 5 = 15 ✓
T cannot be 2 (3 × 2 = 6 is not even a 2-digit number) and T cannot be 3 (3 × 3 = 9, again a single
digit, and the units digit is 9, not 3).

Page 47 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Why it happens: 3T ending in T means 3T – T = 2T ends in 0, so 2T must be a
multiple of 10. For a single digit T that forces T = 0 or T = 5, and T = 0 would not give
a 2-digit sum.

Q2 Here K2 means that the number is a 2-digit number having the digit ‘2’ in the units
place and ‘K’ in the tens place. K2 is added to itself to give a 3-digit sum HMM. What
digit should the letter M correspond to?

M = 4, and the full solution is K = 7, H = 1, so 72 + 72 = 144.

Units column: 2 + 2 = 4, so M = 4 with no carry.

Tens column must also show M = 4:

K + K = 4 or 14 → 2K = 14 → K = 7 (2K = 4 gives K = 2, but then 22 + 22 = 44 is not a

3-digit number)

Carry 1 into the hundreds column → H = 1

So 72 + 72 = 144 ✓

Why it happens: the units digit is decided first and it fixes M straight away. The tens
digit of the answer must then match that same M, which pins down K.

Q3 What about H? Can it be 2? Can it be 3?

H must be 1. It cannot be 2 or 3.

K2 is at most 92, so K2 + K2 is at most 92 + 92 = 184.

A 3-digit sum below 200 must begin with 1.

So H = 1, and H = 2 or H = 3 is impossible.

Page 48 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: doubling a 2-digit number can never reach 200. The hundreds digit

m ase
l
of the answer can only come from a carry of 1, so H = 1 always.

m .co a g
l a se
a g
Q4 Find out what each letter stands for: YY + Z = ZOO; B5 + 3D = ED5; KP + KP = PRR; C1 +

co m
ag
C = 1FF

m .
as e
a g l
Puzzle 1: YY + Z = ZOO

co m
se m.
m a
YY is at most 99 and Z at most 9, so the sum is at most 108.
o l
g ZOO = 100.
.c sum at most 108 must be 100 – 108, so Z = 1 and O = 0, giving
a
e m
las
A 3-digit

ag YY + 1 = 100 → YY = 99 → Y = 9
Check: 99 + 1 = 100 ✓
m a s
m .co agl
l a se
g
Puzzle 2: B5 + 3D = ED5
a
om
Units: 5 + D must end in 5 → D = 0, no carry.
. c
s e
Tens digit of the answer is D = 0, so B + 3 must end in 0 → B + 3 = 10 →m B = 7, carry 1.
. com E = 1 a gla
em
Hundreds:
a s
agl Check: 75 + 30 = 105 ✓
se m
com g l a
. a
Puzzle 3: KP + KP = PRR
m
ase
agl
Doubling a 2-digit number is at most 198, so P = 1.

m
2 × (10K + 1) = 100 + 11R
. co
em
as
20K + 2 = 100 + 11R → 20K – 98 = 11R

. co=m6 gives 120 – 98 = 22 = 11 × 2 → R = 2 a g l
em Check: 61 + 61 = 122 ✓
K

a s
agl
.c
s e m
m a
co agl
Puzzle 4: C1 + C = 1FF

m .
ase
a g l

com
m .
m ase
.co


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Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

(10C + 1) + C = 11C + 1, and 1FF = 100 + 11F

11C + 1 = 100 + 11F → 11(C – F) = 99 → C – F = 9

So C = 9 and F = 0
Check: 91 + 9 = 100 ✓

CRYPTARITHM LETTERS SUM IN FIGURES

YY + Z = ZOO Y = 9, Z = 1, O = 0 99 + 1 = 100

B5 + 3D = ED5 B = 7, D = 0, E = 1 75 + 30 = 105

KP + KP = PRR K = 6, P = 1, R = 2 61 + 61 = 122

C1 + C = 1FF C = 9, F = 0 91 + 9 = 100

Why it happens: always start where the information is tightest — usually the units
column (it has no incoming carry) or the leading digit of a longer answer (it can only
be a carry, so it is almost always 1).

Figure it Out — Pages 143–144
Section 6.5 Digits in Disguise

Q1 A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off?
Why?

The bulb will be OFF.

Each toggle changes the state: ON → OFF → ON → OFF …

After an even number of toggles the bulb is back to ON.

After an odd number of toggles the bulb is OFF.

77 is odd → the bulb is OFF.

Page 50 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

TOGGLES 0 1 2 3 4 … 77

State ON OFF ON OFF ON … OFF

Why it happens: the two toggles “ON → OFF” and “OFF → ON” cancel each other,
exactly like a pair of dots. 77 toggles make 38 such pairs with one toggle left over,
and that last toggle switches the bulb off.

Q2 Liswini has a large old encyclopaedia. When she opened it, several loose pages fell
out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of
the page numbers of the loose sheets be 6000? Why or why not?

Parity says yes, but a closer check says no — the sum can never be exactly 6000.
Step 1 — the parity test. Each sheet carries two page numbers, one odd and one even.

One sheet: odd + even = odd

50 sheets: 50 odd numbers added together
50 is an even count of odd numbers → the total is even

6000 is even, so parity does not rule it out.

Step 2 — look closer. If a sheet is the kth leaf, its two pages are 2k – 1 and 2k.

Sum for one sheet = (2k – 1) + 2k = 4k – 1

Sum for 50 sheets = 4 × (k1 + k2 + … + k50) – 50

If this equals 6000: 4 × (sum of k's) = 6050

sum of k's = 6050 ÷ 4 = 1512.5 — not a whole number!

So no set of 50 sheets can total 6000. The total always leaves remainder 2 when divided by 4,
while 6000 leaves remainder 0.

Why it happens: the parity test is a good first filter, but it only checks divisibility by 2.
Checking divisibility by 4 is a sharper test, and it settles this question.

Page 51 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Check it yourself: the closest reachable totals near 6000 are 5998 and 6002 — both
leave remainder 2 on division by 4.

Q3 Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the
circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even
numbers (‘e’) to satisfy the parity of the row and column sums.

o

e

e e o

The 2 × 3 grid printed with this question on page 144.

The circles read: row 1 → o, row 2 → e, column 1 → e, column 2 → e, column 3 → o.
One arrangement that works:

Page 52 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

C1 C2 C3 ROW SUM REQUIRED

R1 1 2 4 7 o ✓

R2 3 6 5 14 e ✓

Column sum 4 8 9 — —

Required e ✓ e ✓ o ✓ — —

Odd numbers used: 1, 3, 5 (three of them). Even numbers used: 2, 4, 6 (three of them). ✓
In terms of parity the pattern is:

Row 1: o e e → o + e + e = odd ✓

Row 2: o e o → o + e + o = even ✓

Column 1: o + o = even ✓ Column 2: e + e = even ✓ Column 3: e + o = odd ✓

Try This: another pattern that works is e o e in row 1 and e o o in row 2 — for
example 2, 1, 4 over 6, 3, 5.

Q4 Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero.
Use negative numbers, as needed.

Magic sum = 3m, so take m = 0 in the generalised form.

3 –4 1

–2 0 2

–1 4 –3

Rows: 3 – 4 + 1 = 0, –2 + 0 + 2 = 0, –1 + 4 – 3 = 0 ✓

Columns: 3 – 2 – 1 = 0, –4 + 0 + 4 = 0, 1 + 2 – 3 = 0 ✓

Diagonals: 3 + 0 – 3 = 0, 1 + 0 – 1 = 0 ✓

Page 53 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the nine entries are – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4 — nine consecutive

m l a se
numbers with 0 in the middle. Every number is matched by its opposite, so all lines
o
c
cancel to .zero. a g
se m
g l a
aTip: a quick way to get it — take 8 1 6 / 3 5 7 / 4 9 2 and subtract 5 from every entry.

com
m . ag
l a se
Q5 g ‘odd’ or ‘even’: (a) Sum of an odd number of even
Fill in the following blanksawith
numbers is ______ (b) Sum of an even number of odd numbers is ______ (c) Sum of an

co m
m.
even number of even numbers is ______ (d) Sum of an odd number of odd numbers is
______
m as e
.co a g l
se m
l a

a g
SUM OF … ANSWER EXAMPLE

m a s
m.co agl
se
(a) an odd number of even numbers even 2 + 4 + 6 = 12

l a
(b)
ag
an even number of odd numbers even 1 + 3 + 5 + 7 = 16

om
(c) an even number of even numbers even 2+4=6

. c
em
com as
an odd number of odd numbers 1+3+5=9
l
(d) odd

. a g
e m
as Why it happens: even numbers never leave a stray dot, so however many you add,
agl the sum stays even — that settles (a) and (c). Odd numbers each leave one stray dot;
se m
com
those strays cancel two at a time, so an even count gives even (b) and an odd count
g l a
. a
em
leaves one dot over, giving odd (d).
a s
agl

co m
.
Q6 What is the parity of the sum of the numbers from 1 to 100?

e m
com g l as
. a
semThe sum is even.

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

co m
m .
m as e
.co


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Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

From 1 to 100 there are 50 odd numbers (1, 3, 5, …, 99) and 50 even numbers.

The 50 even numbers add to an even total.

50 odd numbers = an even count of odd numbers → even total.

even + even = even

Checking with the actual value:

1 + 2 + 3 + … + 100 = (1 + 100) × 100 ÷ 2 = 101 × 50 = 5050 — even ✓

Tip: the parity was found without computing 5050 at all — just by counting how
many odd numbers there are.

Q7 Two consecutive numbers in the Virahāṅka sequence are 987 and 1597. What are
the next 2 numbers in the sequence? What are the previous 2 numbers in the
sequence?

Next two: 2584 and 4181. Previous two: 610 and 377.

Going forward (add the two before):

987 + 1597 = 2584

1597 + 2584 = 4181

Going backward (subtract):

1597 – 987 = 610

987 – 610 = 377

So the stretch of the sequence is:

…, 377, 610, 987, 1597, 2584, 4181, …

Page 55 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Why it happens: the rule a + b = c can be read backwards as c – b = a, so the same
rule runs in both directions.

Check it yourself: 377 + 610 = 987 ✓ and 610 + 987 = 1597 ✓.

Q8 Angaan wants to climb an 8-step staircase. His playful rule is that he can take either
1 step or 2 steps at a time. For example, one of his paths is 1, 2, 2, 1, 2. In how many
different ways can he reach the top?

Angaan can reach the top in 34 different ways.

STAIRCASE OF N STEPS 1 2 3 4 5 6 7 8

Number of ways 1 2 3 5 8 13 21 34

ways(3) = ways(2) + ways(1) = 2 + 1 = 3

ways(4) = 3 + 2 = 5

ways(5) = 5 + 3 = 8

ways(6) = 8 + 5 = 13

ways(7) = 13 + 8 = 21

ways(8) = 21 + 13 = 34

Why it happens: Angaan's last move is either a 1-step (so he was on step 7) or a 2-
step (so he was on step 6). These two cases never overlap, so the counts add —
exactly the Virahāṅka rule. It is the very same problem as writing 8 as a sum of 1s
and 2s.

Did you know? this is why 8 beats of short and long syllables also give 34 rhythms
— the poem and the staircase are the same puzzle in different clothes.

Page 56 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

Q9 What is the parity of the 20th term of the Virahāṅka sequence?

The 20th term is even.

Parity pattern: O, E, O, O, E, O, O, E, O, O, E, …

After the first term the block odd, odd, even repeats.

Even terms sit at positions 2, 5, 8, 11, 14, 17, 20, …

Each of these is 3 more than the previous, and 20 = 2 + 3 × 6 → the 20th term is even.

Confirming with the actual sequence:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946

The 20th term is 10946, which is even ✓

Why it happens: parity depends only on the two previous parities, so it must fall
into a repeating cycle. Here the cycle has length 3, so you only need to know the
position number's remainder on division by 3.

Q10 Identify the statements that are true. (a) The expression 4m – 1 always gives odd
numbers. (b) All even numbers can be expressed as 6j – 4. (c) Both expressions 2p +
1 and 2q – 1 describe all odd numbers. (d) The expression 2f + 3 gives both even and
odd numbers.

Only statement (a) is true.

Page 57 of 60

Class 7 Maths Chapter 6 Number Play AglaSem · NCERT Solutions

STATEMENT TRUE / REASON
FALSE

(a) 4m – 1 always gives odd True 4m is always even, and even – 1 = odd. Values: 3,
numbers 7, 11, 15, …

(b) All even numbers can be False 6j – 4 gives 2, 8, 14, 20, … It skips 4, 6, 10, 12, 16,
written as 6j – 4 …

(c) Both 2p + 1 and 2q – 1 False With letter-numbers 1, 2, 3, …, 2q – 1 gives 1, 3,
describe all odd numbers 5, … but 2p + 1 gives 3, 5, 7, … and misses 1.

(d) 2f + 3 gives both even and False 2f is even and 3 is odd, so 2f + 3 is always odd: 5,
odd numbers 7, 9, 11, …

Why it happens: for (b), 6j – 4 = 2(3j – 2), so it only produces even numbers whose
half is of the form 3j – 2 — it jumps in steps of 6 and misses most even numbers. For
(d) the parity is locked: an even part plus an odd part is always odd.

Tip: in (c), if the letter-number were allowed to be 0 as well, then 2p + 1 with p = 0, 1,
2, … would also give 1, 3, 5, … and the statement would become true. Here we follow
the book, where a letter-number stands for a counting number 1, 2, 3, …

Q11 Solve this cryptarithm: UT + TA = TAT

U = 9, T = 1, A = 0, that is, 91 + 10 = 101.

Page 58 of 60

as e
Class 7 Maths Chapter 6 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
Write the numbers using place value:
m l as
.co
UT = 10U + T • TA = 10T + A • TAT = 100T + 10A + T = 101T + 10A
m a g
l a se
g
a(10U + T) + (10T + A) = 101T + 10A

co m
. ag
10U + 11T + A = 101T + 10A
e m
10U = 90T + 9A
g l as
10U = 9 × (10T + A) a

co m
em.
m l as
.co
The right side is a multiple of 9, so 10U must be a multiple of 9 → U = 0 or U = 9.
a g
a s em
U = 0 is impossible (UT would not be a 2-digit number), so U = 9.

a gl 90 = 9 × (10T + A) → 10T + A = 10 → T = 1, A = 0
m a s
Check the addition:
m .co agl
l a se
91 a g
+10
co m
m .
as e
om l
——

1m0.1c ✓ a g
a se
agl
se m
co1.mOnce T = 1, the units column gives T + A a
Why it happens: TAT is a 3-digit answer from adding two 2-digit numbers, so its
leading digit T can only be a carry.of
a g l
a s emin 1, so A = 0 — and then U falls out at once.
agl
ending in T, that is 1 + A ending

co m
m .
as e
. com at a glance a g l
sem
Chapter
a
agl Numbers can carry information about an arrangement without telling you the actual values
c
— each child calls out how many children ahead of them are taller.
m .
m a s e
co agl
Parity means being even or odd. An even number can be arranged in pairs; an odd number

m .
e
always leaves one out.

g l as
Parity rules: even + even = even, odd + odd = even, even + odd = odd. A sum of an odd count
a
of odd numbers is odd; of an even count of odd numbers is even.

co m
m .
m ase
.co


a g l Page 59 of 60