NCERT Solutions Class 8 Maths Chapter 5 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 8 · M AT H S

NCERT Solutions

Chapter 5: Number Play

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

112 – 135 25 94 English

Solutions, notes, sample papers & more at 75 pages

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 5: Number Play
Chapter 5 of Ganita Prakash Part I is about divisibility and about the kind of argument that settles a question
for every number at once. It starts with Anshu's sums of consecutive numbers, moves to parity and multiples
of 4, and ends by proving why the divisibility shortcuts for 9, 3 and 11 actually work.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) 112 – 135

SECTIONS QUESTIONS

25 94

MEDIUM

English

In-text Questions — Page 112
5.1 Is This a Multiple Of? — Sum of Consecutive Numbers

MATH TALK

Q1 “Can I write every natural number as a sum of consecutive numbers?”

No. The powers of 2 — 1, 2, 4, 8, 16, 32, … — cannot be written as a sum of two or more
consecutive natural numbers. Every other natural number can.
Write a run of k consecutive numbers starting at a:

a + (a + 1) + … + (a + k – 1)

= ka + (1 + 2 + … + (k – 1))

= ka + k(k – 1)/2

So 2N = k(2a + k – 1)

Why it happens: look at the two factors k and (2a + k – 1). If k is even, then 2a + k – 1
is odd; if k is odd, then 2a + k – 1 is even. So one of them is always odd, and for k ≥
2 both are bigger than 1. Hence 2N — and therefore N — must have an odd factor
greater than 1. A power of 2 has no such factor, so it can never be written this way.

Check the smallest cases:

Page 1 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

N ODD FACTOR > 1? SUM OF CONSECUTIVE NUMBERS

6 3 1+2+3

7 7 3+4

8 none impossible

9 3, 9 4 + 5 and 2 + 3 + 4

16 none impossible

Q2 “Which numbers can I write as the sum of consecutive numbers in more than one
way?”

Numbers that have three or more odd factors. In fact the number of ways (using two or more
consecutive natural numbers) is exactly one less than the number of odd factors.

15 → odd factors 1, 3, 5, 15 (four of them) → 4 – 1 = 3 ways

15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5 ✓ (exactly what Anshu found)

NUMBER ODD FACTORS NUMBER OF WAYS

7 1, 7 1 — only 3 + 4

9 1, 3, 9 2 — 4 + 5, 2 + 3 + 4

10 1, 5 1—1+2+3+4

12 1, 3 1—3+4+5

45 1, 3, 5, 9, 15, 45 5

Why it happens: from Q1, every way of writing N corresponds to a factorisation 2N
= k × (2a + k – 1) in which one factor is odd. Each odd factor of N (other than the one
that gives k = 1) produces exactly one such run.

Page 2 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q3 “Ohh, I know all odd numbers can be written as a sum of two consecutive numbers.
Can we write all even numbers as a sum of consecutive numbers?”

Anshu's first claim is easy to justify, and the answer to his question is no — the even numbers 2,
4, 8, 16, 32, … (the powers of 2) cannot be written this way.

Any odd number = 2n + 1

2n + 1 = n + (n + 1) — two consecutive numbers

e.g. 35 = 17 + 18, 101 = 50 + 51

An even number can never be a sum of two consecutive numbers, because n + (n + 1) is always
odd. So it needs a run of three or more:

6 = 1 + 2 + 3 10 = 1 + 2 + 3 + 4
12 = 3 + 4 + 5 14 = 2 + 3 + 4 + 5

18 = 5 + 6 + 7 20 = 2 + 3 + 4 + 5 + 6

Check it yourself: try hard to write 8, 16 or 32 as a sum of consecutive numbers.
You will not succeed — they have no odd factor bigger than 1.

Q4 “Can I write 0 as a sum of consecutive numbers? Maybe I should use negative
numbers.”

Yes, once negative numbers are allowed. Take any run that is symmetric about 0 — the positive
and negative parts cancel.

0 = (–1) + 0 + 1

0 = (–2) + (–1) + 0 + 1 + 2

0 = (–3) + (–2) + (–1) + 0 + 1 + 2 + 3

Page 3 of 75

as e
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: in the run from –m to m, every negative number is paired with its

m l a se
opposite, and 0 sits in the middle. Each pair adds to 0, so the whole sum is 0. Notice
o
that such.a a g
c run always has an odd number of terms (2m + 1 of them).
se m
g l a
aTry This: with negatives allowed, every number becomes a sum of consecutive

co m
ag
numbers — even 8. For example 8 = (–7) + (–6) + … + 6 + 7 + 8, since everything from
m .
–7 to 7 cancels.
as e
a g l

. c om
Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and ‘–’ signs in
em all of them.
Q5

m a s
between the numbers. How many different possibilities exist? Write

. co a gl
as em
g l

a Eight possibilities. There are 3 gaps between the four numbers and each gap takes either sign,
so the count is 2 × 2 × 2 = 2³ = 8.
m a s
m .co agl
EXPRESSION
l a se VALUE

3+4+5+6 a g 18

co m
m .
3+4+5–6 6

m ase
.co
3+4–5+6 8
a g l
a s e3m
a gl
+4–5–6 –4

3–4+5+6 10
se m
com g l a
m . a
e
3–4+5–6 –2

a s
3–4–5+6
agl 0

3–4–5–6 – 12
co m
m .
m as e
cobefore g l
The tree in the book lists them in exactly this order: branch on the sign before 4, then before 5,
. a
s e m
then 6.
a
agl c
Did you notice? Every one of the eight values is an even number.
m .
m a s e
m . co agl
l a se
g 113
In-text Questions —aPage

co m
m .
m as e
.co


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Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

5.1 Is This a Multiple Of? — Sum of Consecutive Numbers

MATH TALK

Q1 Evaluate each expression and write the result next to it. Do you notice anything
interesting?

The eight values for 3, 4, 5, 6 are 18, 6, 8, – 4, 10, – 2, 0, – 12 — and the interesting thing is that
every one of them is even.

EXPRESSION VALUE EXPRESSION VALUE

3+4+5+6 18 3–4+5+6 10

3+4+5–6 6 3–4+5–6 –2

3+4–5+6 8 3–4–5+6 0

3+4–5–6 –4 3–4–5–6 – 12

A second thing worth noticing: the eight values are 12 apart at the extremes and fall in steps of
2 — no odd number ever appears, and no value is repeated.

Q2 Now, take four other consecutive numbers. Place the ‘+’ and ‘–’ signs as you have
done before. Find out the results of each expression. What do you observe?

Taking 5, 6, 7, 8:

EXPRESSION VALUE EXPRESSION VALUE

5+6+7+8 26 5–6+7+8 14

5+6+7–8 10 5–6+7–8 –2

5+6–7+8 12 5–6–7+8 0

5+6–7–8 –4 5–6–7–8 – 16

Observation: again all eight values are even. And three of them — 0, – 2 and – 4 — are exactly
the same as before.

Page 5 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q3 Repeat this for one more set of 4 consecutive numbers. Share your findings.

Taking 7, 8, 9, 10:

EXPRESSION VALUE EXPRESSION VALUE

7 + 8 + 9 + 10 34 7 – 8 + 9 + 10 18

7 + 8 + 9 – 10 14 7 – 8 + 9 – 10 –2

7 + 8 – 9 + 10 16 7 – 8 – 9 + 10 0

7 + 8 – 9 – 10 –4 7 – 8 – 9 – 10 – 20

The three constant answers appear once more. Writing the numbers as n, n + 1, n + 2, n + 3
shows why they must:

n – (n + 1) – (n + 2) + (n + 3) = 0

n – (n + 1) + (n + 2) – (n + 3) = – 2

n + (n + 1) – (n + 2) – (n + 3) = – 4

Why it happens: in each of these three, two of the n's are added and two are
subtracted, so all the n's cancel. What is left is a fixed number that does not depend
on which four consecutive numbers you started with.

Page 6 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q4 Do these patterns occur no matter which 4 consecutive numbers are chosen? Is
there a way to find out through reasoning? Hint: Use algebra and describe the 8
expressions in a general form.

+ 6 3+4+5+6
5 −
+
6 3+4+5−6
4
−
+ 6
+
5 −
6
3
+ 6
−
5 −
+
6
4
−
+ 6
5 −
6

The diagram on page 113: the eight ways of putting a ‘+’ or a ‘−’ sign before 4, before 5
and before 6, starting from 3. The first two branches are written out in full.

Yes — and algebra settles it in one step, without testing more sets.

The general expression is

n ± (n + 1) ± (n + 2) ± (n + 3)

= n(1 ± 1 ± 1 ± 1) + (0 ± 1 ± 2 ± 3)

Look at the two brackets separately.

1 ± 1 ± 1 ± 1 is one of 4, 2, 0, – 2 — always even. So the first part is an even multiple of n.
0 ± 1 ± 2 ± 3 starts from 1 + 2 + 3 = 6, which is even, and changing any sign alters it by twice
that number — again an even change. So the second part is always even too.

even + even = even

Page 7 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Why it happens: the value can only change by an even amount, so it can never cross
from even to odd. That is why the answer is even for every choice of four consecutive
numbers — the three fixed values 0, – 2, – 4 turn up every single time, and no one in
the class will ever report an odd result.

In-text Questions — Page 114
5.1 Is This a Multiple Of? — Sum of Consecutive Numbers

MATH TALK

Q1 Now take any 4 numbers, place ‘+’ and ‘–’ signs in the eight different ways, and
evaluate the resulting expression. What do you observe about their parities? Repeat
this with other sets of 4 numbers.

All eight expressions have the same parity — and it is the parity of a + b + c + d.

FOUR NUMBERS A+B+C+D A+B–C–D A–B–C–D PARITY OF ALL 8

2, 5, 7, 4 18 –4 – 14 even

1, 2, 3, 5 11 –5 –9 odd

10, 4, 6, 8 28 0 –8 even

3, 3, 3, 3 12 0 –6 even

With four consecutive numbers two of them are odd and two are even, so a + b + c + d is always
even — that is the special case we met on the last page. With any four numbers the shared
parity can be odd instead, but it is still shared by all eight.

Q2 Is there a way to explain why this happens? Hint: Think of the rules for parity of the
sum or difference of two numbers.

Yes. The key is that switching one sign changes the value by an even number, so it can never
change the parity.

Page 8 of 75

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Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
Take a + b – c – d and replace + b by – b:
m l as
.co
(a + b – c – d) – (a – b – c – d)
m a g
l a se
g
=a+b–c–d–a+b+c+d
a= 2b — an even number

. c om ag
s e m cannot have different parities — either both are
If two numbers differ by an even number, they
a
agl
even or both are odd.

Why it happens: starting from any one of the eight expressions you can reach the
co m
other seven by switching one, two or three signs. Every switch changes the value by
em.
m l as
.co g
2 × (one of the numbers), which is even. A chain of even changes is still an even
m a
se
change, so all eight expressions share one parity.

g l a
a The second explanation in the book reaches the same place through the parity rules:
om a s
e
. c
m odd ± even = odd agl
a s
odd ± odd = even even ± even = even

agl
In every rule the '+' case and the '–' case give the same answer. So a ± b has one parity

co m
.
whichever sign is used; then a ± b ± c has one parity; then a ± b ± c ± d does too.

e m
m l as
.co a g
a s em Replace any negative sign in the expression a + b – c – d with a positive sign and find
agl
Q3
the difference between the two numbers. What do you conclude from this
observation?
se m
com g l a
m. a
ase
agl

Change – c to + c:

co m
(a + b + c – d) – (a + b – c – d)
m .
m as e
.co
=a+b+c–d–a–b+c+d
a g l
se m = 2c — an even number
g l a
a c
m .
m a s e
agl
Change – d to + d instead and the difference is 2d, again even.
. co
e m
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Conclusion: it makes no difference which sign you switch or which way you switch it.
Turning + into – changes the value by – 2 × (that number); turning – into + changes it
by + 2 × (that number). Both are even. So the parity of the expression is fixed the
moment the four numbers are chosen — the signs cannot touch it.

In-text Questions — Page 115
Breaking Even

Q1 Is the phenomenon of all the expressions having the same parity limited to taking 4
numbers? What do you think?

No — it holds for any count of numbers. Nothing in the argument used the number four.

For n numbers a₁ ± a₂ ± … ± aₙ

there are 2ⁿ⁻¹ expressions,

and switching one sign changes the value by 2aᵢ — still even

So all 2n–1 expressions carry the parity of a₁ + a₂ + … + an.

NUMBERS HOW MANY EXPRESSIONS SHARED PARITY

3, 4, 5 2² = 4 even (3 + 4 + 5 = 12)

3, 4, 5, 6 2³ = 8 even (18)

3, 4, 5, 6, 7 2⁴ = 16 odd (25)

Check it yourself: 3 – 4 + 5 = 4, 3 + 4 – 5 = 2, 3 – 4 – 5 = – 6 — all even, matching 3 + 4
+ 5 = 12.

Page 10 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q2 We know how to identify even numbers. Without computing them, find out which
of the following arithmetic expressions are even. 43 + 37, 672 – 348, 4 × 347 × 3, 708 –
477, 809 + 214, 119 × 303, 543 – 479, 513³

Four of them are even: 43 + 37, 672 – 348, 4 × 347 × 3 and 543 – 479.

EXPRESSION PARITY REASONING EVEN?

43 + 37 odd + odd = even Yes

672 – 348 even – even = even Yes

4 × 347 × 3 a product with the factor 4 in it must be even Yes

708 – 477 even – odd = odd No

809 + 214 odd + even = odd No

119 × 303 odd × odd = odd No

543 – 479 odd – odd = even Yes

513³ odd × odd × odd = odd No

Check it yourself (afterwards!): 80, 324, 4164, 231, 1023, 36057, 64 and 135005697.
The parity rules got every one right without any of that work.

Q3 Using our understanding of how parity behaves under different operations, identify
which of the following algebraic expressions give an even number for any integer
values for the letter-numbers. 2a + 2b, 3g + 5h, 4m + 2n, 2u – 4v, 13k – 5k, 6m – 3n, x²
+ 2, b² + 1, 4k × 3j

Five of the nine are always even: 2a + 2b, 4m + 2n, 2u – 4v, 13k – 5k and 4k × 3j. Each of them
has 2 as a factor of the whole expression.

Page 11 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

EXPRESSION REWRITTEN ALWAYS EVEN?

2a + 2b 2(a + b) Yes

3g + 5h — No

4m + 2n 2(2m + n) Yes

2u – 4v 2(u – 2v) Yes

13k – 5k 8k = 2(4k) Yes

6m – 3n 3(2m – n) No

x² + 2 — No

b² + 1 — No

4k × 3j 12kj = 2(6kj) Yes

Why it happens: once an expression is written as 2 × (something whole), 2 is a
factor of it no matter what the letters stand for. The four that fail have no such factor
— a single counterexample kills each one: 3g + 5h = 3 when g = 1, h = 0; 6m – 3n = 3
when m = n = 1; x² + 2 = 11 when x = 3; b² + 1 = 5 when b = 2.

In-text Questions — Page 116
Breaking Even / Pairs to Make Fours

Q1 Similarly, determine and explain which of the other expressions always give even
numbers. Write a couple of examples and non-examples, as appropriate, for each
expression.

Handle each one the way the book handles 4m + 2q and x² + 2 — either pull out the factor 2, or
produce a non-example.

Page 12 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

EXPRESSION EXPLANATION EXAMPLES / NON-EXAMPLES

2a + 2b = 2(a + b), so 2 is a factor a = 3, b = 5 → 16; a = 4, b = – 7 → – 6.
Never odd.

3g + 5h Odd × odd stays odd; the sum of two odds Example g = 1, h = 1 → 8 (even). Non-
is even but odd + even is odd example g = 1, h = 0 → 3 (odd).

2u – 4v = 2(u – 2v) u = 5, v = 1 → 6; u = – 3, v = 2 → – 14.
Never odd.

13k – 5k = 8k = 2(4k) — the letters collect first k = 1 → 8; k = – 3 → – 24. Never odd.

6m – 3n = 3(2m – n); 3 is odd, so the parity follows Example m = 1, n = 2 → 0 (even). Non-
2m – n, which follows n example m = 1, n = 1 → 3 (odd).

b² + 1 b² is even when b is even and odd when b is Example b = 1 → 2 (even). Non-example
odd, so b² + 1 flips with b b = 2 → 5 (odd).

4k × 3j = 12kj = 2(6kj) k = 1, j = 1 → 12; k = 2, j = – 1 → – 24.
Never odd.

Why it happens: an expression is even for every integer value exactly when 2 can be
taken out as a factor of the whole expression. If it cannot, some choice of the letters
will make it odd — and one non-example is enough to settle the matter.

Q2 Write a few algebraic expressions which always give an even number.

Any expression with 2 as a factor of the whole thing will do:

2n 6p – 4q 10a + 8b

2(m + n) + 4 14x × 5y 8k² – 6k

Here is a subtler one that is not of the obvious form 2 × (something):

k² + k = k(k + 1)

Page 13 of 75

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Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: k and k + 1 are consecutive integers, so one of them is even. A

m as e
l
product with an even factor is even. Test it: k = 3 → 12, k = 4 → 20, k = – 5 → 20.

m .co a g
l a se
a g
Q3 Take a pair of even numbers. Add them. Is the sum divisible by 4? Try this with

co m
ag
different pairs of even numbers. When is the sum a multiple of 4, and when is it

m .
e
not? Is there a general rule or a pattern?

g l as
ANSWER a
Sometimes yes, sometimes no — and the pattern is exact.
co m
e m.
c o m EACH ÷ 4 LEAVES g l as
.
PAIR SUM MULTIPLE OF 4?

m a
l a se
g
4+8 0 and 0 12 Yes

a
0 and 0
s
12 + 16 28 Yes

m a
m .co agl
se
2+6 2 and 2 8 Yes

g l a
6 + 10 2 and 2
a 16 Yes

co m
2+4 2 and 0 6 No

m .
as e
com
0 and 2
l
8 + 10 18 No

. a g
m
ase
agl of 4 exactly when the two remainders are the same.
The rule: divide each even number by 4 — the remainder is either 0 or 2. The sum is a multiple

se m
com g l a
.
Why it happens: the two remainders add to 0, 2 or 4. Only 0 and 4 are multiples of
m a
ase
4. So both must be 0, or both must be 2 — never one of each.

agl

co m
In-text Questions — Page 117
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 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Pairs to Make Fours

Q1 When will two even numbers add up to give a multiple of 4? This problem is similar
to the question of identifying when adding two numbers will result in an even
number. Can you see this?

Exactly when both are multiples of 4, or neither is. There are three cases to examine, and
algebra settles all three:

CASE ALGEBRA MULTIPLE OF 4? EXAMPLE

both multiples of 4 4p + 4q = 4(p + q) Always 12 + 16 = 28 = 4 × 7

neither a multiple of 4 (4p + 2) + (4q + 2) = 4(p + q + 1) Always 6 + 10 = 16 = 4 × 4

one of each 4p + (4q + 2) = 4(p + q) + 2 Never 8 + 10 = 18 = 4 × 4 + 2

Yes, it is the same problem one level up. Compare the two tables:

even + even = even ↔ 4p + 4q is a multiple of 4

odd + odd = even ↔ (4p + 2) + (4q + 2) is a multiple of 4

even + odd = odd ↔ 4p + (4q + 2) is not a multiple of 4

Why it happens: in the parity question every whole number leaves remainder 0 or 1
on division by 2, and 'same remainder' wins. Here every even number leaves
remainder 0 or 2 on division by 4, and again 'same remainder' wins. The structure is
identical — only the divisor has changed.

Q2 What happens when we add a multiple of 4 to an even number that is not a multiple
of 4? Is it similar to the case of the parity of the sum of an even and an odd number?

The sum is never a multiple of 4 — it always leaves a remainder of 2.

Page 15 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

4p + (4q + 2)

= 4p + 4q + 2

= 4(p + q) + 2

Yes, it is exactly similar. Adding an even and an odd number can never give an even number;
adding a multiple of 4 and a non-multiple of 4 can never give a multiple of 4. In both cases the
two remainders are different (0 and 1 there, 0 and 2 here), so the leftover cannot be cleared.

Check it yourself: 20 + 6 = 26 = 4 × 6 + 2; 8 + 22 = 30 = 4 × 7 + 2; 100 + 2 = 102 = 4 ×
25 + 2.

Q3 Look at the following expressions and the visualisation. Write the corresponding
explanation and examples. 4p and (4q + 2) = 4p + (4q + 2) = 4p + 4q + 2 = 4 (p + q) + 2.

Explanation. A multiple of 4 can be written as 4p, and an even number that is not a multiple of
4 as 4q + 2. Adding them gives 4(p + q) + 2, so the sum is a multiple of 4 plus a leftover 2. It is
even, but it is never a multiple of 4.
The picture. Lay the first number out as p complete rows of 4 and the second as q complete
rows of 4 with 2 counters left over. Push them together: the full rows stack into (p + q) rows of 4,
and the 2 loose counters are still loose — there are not enough of them to make another row.

4p (here p = 3) 4q + 2 (q = 2) 4(p + q) + 2

+ =

2 left over

12 + 10 = 22 = 4 × 5 +
2

Full rows of 4 stack together; the loose 2 can never complete a row.

Page 16 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Examples.

12 + 6 = 18 = 4(3 + 1) + 2

8 + 2 = 10 = 4(2 + 0) + 2

16 + 22 = 38 = 4(4 + 5) + 2

20 + 10 = 30 = 4(5 + 2) + 2

In-text Questions — Page 118
Always, Sometimes, or Never

Q1 1. If 8 exactly divides two numbers separately, it must exactly divide their sum.
Statement 1 is always true. Determine if it is true with subtraction.

Always true for addition, and always true for subtraction as well.

The two numbers are multiples of 8: 8a and 8b

8a + 8b = 8(a + b) → a multiple of 8

8a – 8b = 8(a – b) → also a multiple of 8

TWO MULTIPLES OF 8 SUM DIFFERENCE

8 and 16 24 = 8 × 3 – 8 = 8 × (– 1)

16 and 56 72 = 8 × 9 – 40 = 8 × (– 5)

80 and 120 200 = 8 × 25 – 40 = 8 × (– 5)

Why it happens: multiples of 8 are what you get by adding 8 over and over. Adding
two such piles, or taking one away from the other, still leaves a whole number of 8s
— no partial group is ever created. In symbols, 8 comes out as a common factor
either way.

Page 17 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Tip: this is the general fact if a | M and a | N, then a | M + N and a | M – N.
Nothing about 8 was special.

In-text Questions — Page 119
Always, Sometimes, or Never

Q1 2. If a number is divisible by 8, then 8 also divides any two numbers (separately)
that add up to the number.

Sometimes true. The word that breaks it is any.

72 = 48 + 24 = 8 × 6 + 8 × 3 → both parts divisible by 8 ✓

72 = 50 + 22 → neither part divisible by 8 ✗
72 = 64 + 8 → both divisible by 8 ✓

72 = 70 + 2 → neither divisible by 8 ✗

Why it happens: a multiple of 8 can certainly be split as 8a + 8b, but it can also be
split as p + q where p and q are anything at all — you just cut the pile wherever you
like. Since some splits work and others do not, the statement is sometimes true.

Tip: a correct version does exist. If 8 divides the total and 8 divides one part, then 8
must divide the other part too — because the other part is (total – part), a difference
of two multiples of 8.

Q2 3. If a number is divisible by 7, then all multiples of that number will be divisible by
7.

Always true.

Page 18 of 75

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Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
e m.
The number is divisible by 7, so write it as 7j
m l as
.co
Its multiples are (7j) × m = 7(jm)
m a g
l a se
g
7 is a factor, so every multiple is divisible by 7
a
Take 14 = 7 × 2 (so j = 2). Its multiples:
com
e m . ag
g l as
a
28 = (7 × 2) × 2 = 7 × 4

70 = (7 × 2) × 5 = 7 × 10

co m
m.
154 = (7 × 2) × 11 = 7 × 22

o m l a se
g it m times gives mj
.c in the book makes the same point: 7j is j rows of 7, and taking
a
m
seof 7 — still whole rows, nothing left over.
The picture
l a
ag
rows

a s
com agl
Tip: in general, if A is divisible by k, then all multiples of A are divisible by k. This
.
em
is the first bullet of the chapter summary.
a s
agl

co m
.
In-text Questions — Page 120
e m
m l as
Always, Sometimes, or Never

.co a g
a s em 4. If a number is divisible by 12, then the number is also divisible by all the factors
agl Q1
of 12.

se m
com g l a
m . a
ase

agl
Always true.

The number is a multiple of 12, say 12m
co m
m .
as e
com
12m = 2 × 6 × m = 3 × 4 × m = 1 × 12 × m
.So a g l
se m 1, 2, 3, 4, 6 and 12 each appear as a factor

g l a
a c
m .
e
In general, if f is a factor of 12 then 12 = f × t for some whole number t, and so

m a s
12m = (f × t) × m = f × (tm) em
. c o agl
g l as
a
Take 24, which is divisible by 12. Its factors include every factor of 12: 1, 2, 3, 4, 6, 8, 12, 24.

co m
m .
m ase
.co


a g l Page 19 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Why it happens (the picture): a multiple of 12 is m rows of 12. A factor of 12 fills
one row of 12 exactly — so it fills every row exactly, and therefore fills the whole
array with nothing left over.

Tip: in general, if A is divisible by k, then A is divisible by all the factors of k.

Q2 5. If a number is divisible by 7, then it is also divisible by any multiple of 7.

Sometimes true. 42 is divisible by 7 and by 14, but not by 28.

42 = 7 × 6, and 14 = 7 × 2 → 42 ÷ 14 = 3 ✓
42 = 7 × 6, and 28 = 7 × 4 → 42 ÷ 28 = 1.5 ✗

The exact condition falls out of the algebra:

7k is divisible by 7m if and only if m is a factor of k

If k = ym, then 7k ÷ 7m = 7ym ÷ 7m = y

For 42, k = 6. The factors of 6 are 1, 2, 3, 6 — so 42 is divisible by 7 × 1 = 7, 7 × 2 = 14, 7 × 3 = 21
and 7 × 6 = 42, but by no other multiple of 7.

Tip: statement 3 and statement 5 look alike but point opposite ways. Going up to
multiples of the number always works; going across to other multiples of 7 does not.

In-text Questions — Page 121

Page 20 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Always, Sometimes, or Never / What Remains?

MATH TALK

Q1 Examine each of the following statements, and determine whether it is ‘Always
true’, ‘Sometimes true’, ‘Never true’. 6. If a number is divisible by both 9 and 4, it
must be divisible by 36.

Always true.

9 = 3², 4 = 2² — they share no prime

LCM (9, 4) = 3² × 2² = 36

If the number is divisible by 9, its prime factorisation contains 3 × 3. If it is also divisible by 4, the
factorisation contains 2 × 2. Since these are different primes, both blocks sit in the factorisation
together, so 3 × 3 × 2 × 2 = 36 divides the number.

Examples: 36, 72, 108, 180, 900 — each divisible by 9, by 4, and by 36

Tip: the general rule is if A is divisible by k and also by m, then A is divisible by
LCM (k, m).

Q2 7. If a number is divisible by both 6 and 4, it must be divisible by 24.

Sometimes true. The counterexample is small: 12.

12 ÷ 6 = 2 ✓ 12 ÷ 4 = 3 ✓ 12 ÷ 24 ✗

Why it happens: 6 = 2 × 3 and 4 = 2 × 2 share a factor 2. The rule promises only LCM
(6, 4) = 2² × 3 = 12, not 24. So every number divisible by 6 and 4 is a multiple of 12 —
and multiples of 12 are alternately 12, 24, 36, 48, …, only half of which are multiples
of 24.

Page 21 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

So the statement holds for 24, 48, 72 but fails for 12, 36, 60. Compare it with statement 6, where
9 and 4 had no common factor and the product was the LCM.

Q3 8. When you add an odd number to an even number we get a multiple of 6.

Never true. The sum of an odd and an even number is odd, while every multiple of 6 is even.

Suppose (2n) + (2m + 1) = 6j

2n + 2m = 6j – 1

2(n + m) = 6j – 1

The left side is even; the right side is 1 less than an even number, so it is odd. An even number
can never equal an odd number, so no values of n, m, j exist.

Did you know? Anshu's doubt in the margin — “Can I write an even and an odd
number as 2n and 2n+1 instead?” — is worth answering: no, not here. Writing 2n and
2n + 1 forces the two numbers to be consecutive. Using 2n and 2m + 1 keeps them
independent, which is what the statement needs.

Q4 Find a number that has a remainder of 3 when divided by 5. Write more such
numbers.

3, 8, 13, 18, 23, 28, 33, 38, … — each one is 3 more than a multiple of 5.

K 0 1 2 3 4 5

5K + 3 3 8 13 18 23 28

Check: 18 = 5 × 3 + 3 28 = 5 × 5 + 3 103 = 5 × 20 + 3

The gap between consecutive numbers in the list is always 5, because moving to the next
multiple of 5 keeps the leftover 3 unchanged.

Page 22 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q5 Which algebraic expression(s) capture all such numbers? (i) 3k + 5 (ii) 3k – 5 (iii) 3k/5
(iv) 5k + 3 (v) 5k – 2 (vi) 5k – 3

(iv) 5k + 3 and (v) 5k – 2. They look different but describe exactly the same set of numbers.

5k – 2 = 5k – 5 + 3 = 5(k – 1) + 3

OPTION FIRST FEW VALUES REMAINDER ON ÷ 5

(i) 3k + 5 5, 8, 11, 14, 17 varies — 0, 3, 1, 4, 2

(ii) 3k – 5 – 2, 1, 4, 7, 10 varies

(iii) 3k/5 0.6, 1.2, 1.8, … not a whole number

(iv) 5k + 3 3, 8, 13, 18, 23 3 — always ✓
(v) 5k – 2 3, 8, 13, 18, 23 (k ≥ 1) 3 — always ✓
(vi) 5k – 3 2, 7, 12, 17, 22 2 — always ✗

Why it happens: to leave remainder 3 on division by 5, a number must be a multiple
of 5 with 3 added. Only expressions of the form 5k + c, where c itself leaves
remainder 3 on division by 5, can do it. Here c = 3 in (iv) and c = – 2 in (v), and – 2 = – 5
+ 3.

In-text Questions — Page 122
What Remains?

Q1 Are there other expressions that generate numbers that are 3 more than a multiple
of 5?

Infinitely many. Any expression 5k + c works, provided c itself leaves a remainder of 3 on
division by 5.

Page 23 of 75

ase
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
em.
5k + 8 (8 = 5 + 3)
m l as
.co
5k + 13 (13 = 10 + 3)
m a g
l a se
g
5k – 7 (– 7 = – 10 + 3)
a5k – 12 (– 12 = – 15 + 3)

. c om ag
s e m18, 23, … — only the starting value of k shifts.
All of them run through the same list 3, 8, 13,

a g la
EXPRESSION k=0 k=1 k=2 k=3

5K + 3
. co18m
m
3 8 13

m as e
5K + 8
.co
8 13
a
18
g l 23

se m
g l a
a
5K – 7 –7 –2 3 8

om a s
. c agl
Tip: what does not work is changing the 5. An expression like 10k + 3 gives only 3, 13,

a s em 8, 18, 28. To capture all of them the
23, … — correct numbers, but it misses

agl
coefficient must be exactly 5.

co m
m .
Figure it m as e
. caoMultiple Of?
Out — Page 122
a g l
s em
5.1 Is This
a
agl
Q1 The sum of four consecutive numbers is 34. What are these numbers?
se m
com g l a
m . a
ase
agl

The numbers are 7, 8, 9 and 10.

com
Let the numbers be n, n + 1, n + 2, n + 3
m .
m as e
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n + (n + 1) + (n + 2) + (n + 3) = 34
a g l
se m 4n + 6 = 34
g l a
a c
4n = 28
m .
m a s e
. co agl
n=7
e m
g l as
a
Check: 7 + 8 + 9 + 10 = 34 ✓

co m
m .
m ase
.co


a g l Page 24 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Tip: 4n + 6 is always even, so four consecutive numbers can never add to an odd
total. Try 35 and you will get no answer at all.

Q2 Suppose p is the greatest of five consecutive numbers. Describe the other four
numbers in terms of p.

The other four are p – 1, p – 2, p – 3 and p – 4.

In increasing order the five numbers are

p – 4, p – 3, p – 2, p – 1, p

Their sum is (p – 4) + (p – 3) + (p – 2) + (p – 1) + p = 5p – 10 = 5(p – 2), so the sum of any five
consecutive numbers is a multiple of 5.

Check it yourself: take p = 20. The numbers are 16, 17, 18, 19, 20 and their sum is
90 = 5 × 18 = 5(20 – 2) ✓

Q3 For each statement below, determine whether it is always true, sometimes true, or
never true. Explain your answer. Mention examples and non-examples as
appropriate. Justify your claim using algebra. (i) The sum of two even numbers is a
multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The
sum of a multiple of 6 and a multiple of 9 is a multiple of 3. (v) The sum of a multiple
of 6 and a multiple of 3 is a multiple of 9.

(i) Sometimes true.

2m + 2n = 2(m + n) — a multiple of 3 only when 3 divides m + n

Example: 2 + 4 = 6 ✓ and 4 + 8 = 12 ✓. Non-example: 2 + 6 = 8 ✗ and 6 + 8 = 14 ✗.
(ii) Sometimes true.
Example: 20 is not divisible by 18, and it is not divisible by 9 either ✓. Non-example: 27 is not
divisible by 18, yet 27 = 9 × 3 is divisible by 9 ✗.

Page 25 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Why it happens: the multiples of 18 are every second multiple of 9 (9, 18, 27, 36, 45,
54 …). So the odd multiples of 9 — 9, 27, 45, 63 — all break this statement. What is
always true is the other direction: if a number is not divisible by 9, it cannot be
divisible by 18.

(iii) Sometimes true.
Example: 8 and 10 are not divisible by 6, and 8 + 10 = 18 is divisible by 6 — so the statement fails
here. Non-example (where it holds): 8 and 9 are not divisible by 6, and 8 + 9 = 17 is not divisible
by 6.

Write the two numbers as 6a + r and 6b + s, with r, s ≠ 0

Sum = 6(a + b) + (r + s)

Divisible by 6 exactly when r + s = 6 (or 12)

So 5 and 7 (remainders 5 and 1) give 12 ✓, while 5 and 8 (remainders 5 and 2) give 13 ✗.
(iv) Always true.

6x + 9y = 3(2x + 3y) — 3 is a factor of the whole sum

Examples: 12 + 9 = 21 = 3 × 7 ✓; 18 + 27 = 45 = 3 × 15 ✓; 6 + 90 = 96 = 3 × 32 ✓.

Why it happens: 3 divides 6 and 3 divides 9, so 3 divides both numbers — and a
divisor of two numbers always divides their sum.

(v) Sometimes true.

6x + 3y = 3(2x + y) — a multiple of 9 only when 3 divides 2x + y

Example: 18 + 9 = 27 = 9 × 3 ✓. Non-example: 12 + 9 = 21 ✗ and 6 + 6 = 12 ✗.

Q4 Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder
of 2 when divided by 4. Write an algebraic expression to describe all such numbers.

Such numbers are 2, 14, 26, 38, 50, 62, … and they are exactly the numbers of the form 12n + 2.

Page 26 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

N leaves remainder 2 on ÷ 3 → N – 2 is a multiple of 3

N leaves remainder 2 on ÷ 4 → N – 2 is a multiple of 4

So N – 2 is a common multiple of 3 and 4

LCM (3, 4) = 12

N – 2 = 12n → N = 12n + 2

N N÷3 N÷4

14 4 remainder 2 ✓ 3 remainder 2 ✓
26 8 remainder 2 ✓ 6 remainder 2 ✓
38 12 remainder 2 ✓ 9 remainder 2 ✓

Why it happens: the two conditions say the same thing about N – 2 — it must be a
multiple of 3 and a multiple of 4. By the rule proved on page 121, it must then be a
multiple of LCM (3, 4) = 12. The list steps by 12 each time.

Q5 “I hold some pebbles, not too many, When I group them in 3’s, one stays with me.
Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my
view. Group them by 5, yet one’s still around, But grouping by seven, perfection is
found. More than one hundred would be far too bold, Can you tell me the number of
pebbles I hold?”

There are 91 pebbles.
First turn each line into a condition:

LINE OF THE RIDDLE CONDITION

groups of 3, one stays remainder 1 on ÷ 3

pairing won't do, an odd pebble remains remainder 1 on ÷ 2

group by 5, one's still around remainder 1 on ÷ 5

grouping by seven, perfection remainder 0 on ÷ 7

more than one hundred is too bold less than 100

Page 27 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Remainder 1 on ÷ 2, ÷ 3 and ÷ 5

→ N – 1 is a common multiple of 2, 3 and 5

LCM (2, 3, 5) = 30
N = 30k + 1 → 1, 31, 61, 91, 121, …

Below 100 and divisible by 7: 91 = 7 × 13 ✓

Check all five lines against 91: 91 = 3 × 30 + 1 ✓; 91 is odd ✓; 91 = 5 × 18 + 1 ✓; 91 = 7 × 13
exactly ✓; 91 < 100 ✓.

Why the shortcut works: instead of testing every number below 100, notice that
the first three lines all say the same thing about N – 1. That collapses three
conditions into one — and only four candidates survive.

Q6 Tathagat has written several numbers that leave a remainder of 2 when divided by
6. He claims, “If you add any three such numbers, the sum will always be a multiple
of 6.” Is Tathagat’s claim true?

Yes, Tathagat's claim is true.

The three numbers are 6a + 2, 6b + 2 and 6c + 2
Sum = 6a + 6b + 6c + 6

= 6(a + b + c + 1) — a multiple of 6

Why it happens: each number carries a leftover of 2, and the three leftovers add to
6 — exactly one more complete group. So no remainder survives.

THREE NUMBERS SUM ÷6

2 + 8 + 14 24 4 ✓
8 + 20 + 32 60 10 ✓
2+2+2 6 1 ✓

Page 28 of 75

as e
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Try This: what about adding just two such numbers? (6a + 2) + (6b + 2) = 6(a + b) + 4

m as e
l
— never a multiple of 6. Three is exactly the right count.

m .co a g
l a se
a g
Q7 When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a

co m
ag
remainder of 5. Without calculating, can you say what remainders the following

m .
e
expressions will leave when divided by 7? Show the solution both algebraically and

g l
visually. (i) 4779 + 661 (ii) 4779 – 661 as
a

co m

(i) remainder 1 (ii) remainder 2.
em.
m l as
.co a g
Algebraically. Write each number as a multiple of 7 plus its remainder:

se m
g l a
a 4779 = 7p + 5 661 = 7q + 3

m a s
m .co agl
se
(i) 4779 + 661 = 7p + 7q + 8
g l a
= 7(p + q) + 7 + 1 a
m
= 7(p + q + 1) + 1 → remainder 1
. co
e m
com– 661 = 7p + 5 – 7q – 3
(ii).4779 g l as
m a
ase= 7(p – q) + 2 → remainder 2
agl
se m
com
Visually. Think of each number as counters arranged in rows of 7, with the loose ones at the
g l a
m . a
ase
bottom.

agl
4779 = full rows of 7, then 5 loose
Adding: the 5 and the.c3om loose ones
m
682 rows of 7
a s emfull row of 7,
gl
make 8 — one more
. co a
e m with 1 left over.
g l as
a → 1
c
661 = full rows of 7, then 3 loose
m .
m Subtracting: take the 3 loose ones a s e
94 rows of 7
em
. c o agl
as away from the 5 — 2 are left.
agl

co m
m .
m ase
.co


a g l Page 29 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Only the loose counters decide the remainder; the full rows cancel out.

Check it yourself: 4779 + 661 = 5440 = 7 × 777 + 1 ✓ and 4779 – 661 = 4118 = 7 × 588
+ 2 ✓ — but no division was needed to say so.

Q8 Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3
when divided by 4, and a remainder of 4 when divided by 5. What is the smallest
such number? Can you give a simple explanation of why it is the smallest?

The smallest such number is 59.
Look at the three conditions side by side: each remainder is exactly 1 less than its divisor.

DIVISOR REMAINDER MEANING

3 2 1 short of a multiple of 3

4 3 1 short of a multiple of 4

5 4 1 short of a multiple of 5

So N + 1 is a multiple of 3, of 4 and of 5

N + 1 is a multiple of LCM (3, 4, 5) = 3 × 4 × 5 = 60

Smallest such N + 1 is 60 → N = 59

Check: 59 = 3 × 19 + 2 ✓, 59 = 4 × 14 + 3 ✓, 59 = 5 × 11 + 4 ✓.

Why 59 is the smallest: N + 1 must be a common multiple of 3, 4 and 5, and 60 is
the smallest positive common multiple there is. Any smaller candidate would force N
+ 1 to be a smaller common multiple, which does not exist. The next numbers in the
family are 119, 179, 239 — each 60 more than the last.

In-text Questions — Page 123

Page 30 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

5.2 Checking Divisibility Quickly

Q1 Similarly, explain using algebra why the divisibility shortcuts for 5, 2, 4, and 8 work.

Write the number in general form as a sum of place values:

… + 10000e + 1000d + 100c + 10b + a

Now split it in the right place for each divisor.
Divisibility by 5 and by 2. Every place value above the units — 10, 100, 1000, … — is a multiple
of 10, so it is a multiple of both 5 and 2.

N = (a multiple of 10) + a

So N is divisible by 5 only when a is 0 or 5, and by 2 only when a is even. Only the units digit
matters.
Divisibility by 4. 100 = 4 × 25, and every higher place value is a multiple of 100. So

N = (a multiple of 100) + 10b + a

= (a multiple of 4) + (the last two digits)

Hence N is divisible by 4 exactly when the two-digit number formed by the last two digits is.
For 2856 → 56 = 4 × 14 ✓; for 1586 → 86 = 4 × 21 + 2 ✗.
Divisibility by 8. 1000 = 8 × 125, and every higher place value is a multiple of 1000. So

N = (a multiple of 1000) + 100c + 10b + a

= (a multiple of 8) + (the last three digits)

Hence N is divisible by 8 exactly when the last three digits form a multiple of 8. For 2856 → 856
= 8 × 107 ✓; for 6686 → 686 = 8 × 85 + 6 ✗.

Why the pattern: in each case we cut the number at the first place value that the
divisor already divides. Everything to the left is guaranteed to be a multiple of the
divisor, so only the small piece on the right needs checking. 10 works for 2 and 5; 100
is needed for 4; 1000 is needed for 8.

Page 31 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q2 Can you say, without actually calculating, which of these numbers are divisible by 9:
999, 909, 900, 90, 990?

All of them.

999 = 9 × 111 909 = 9 × 101

900 = 9 × 100 90 = 9 × 10 990 = 9 × 110

Why it happens: each of these numbers is built only from the digits 9 and 0.
Expanding any such number, every term is either 9 × (a place value) or 0 × (a place
value) — and each term is a multiple of 9. A sum of multiples of 9 is a multiple of 9.

In-text Questions — Page 124
A Shortcut for Divisibility by 9

Q1 Can we say that any number made up of only the digits ‘0’ and ‘9’, in any order, will
always be divisible by 9?

Yes. If every digit is 0 or 9, then in the expanded form every single term is a multiple of 9.

99009 = 9 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 9 × 1

= 9 × (10000 + 1000 + 1)

= 9 × 11001 = 99009 ✓

The same works for 90, 909, 9900, 90909, 999000 — take 9 out as a common factor and what
remains is a whole number.

Tip: this shortcut finds some multiples of 9 but not all of them — 18, 27, 36 and 405
are multiples of 9 with no 9 in sight. Unlike 2, 5 and 10, the units digit alone tells you
nothing about 9: both 99 and 109 end in 9, but only 99 is a multiple of 9.

Page 32 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q2 Is 10 divisible by 9? If not, what is the remainder? Check the divisibility of other
multiples of 10 (10, 20, 30, ...) by 9.

No — 10 leaves a remainder of 1, because 10 = 9 + 1.

MULTIPLE OF 10 10 20 30 40 50 60 70 80 90

REMAINDER ÷ 9 1 2 3 4 5 6 7 8 0

So for any multiple of 10 the remainder is the number of tens.

Why it happens: 10 = 9 + 1, so b tens is 10b = 9b + b. The 9b part is a whole number
of 9s and disappears; the b is what is left over. (At b = 9 the leftover 9 forms one
more complete group, which is why 90 leaves 0.)

Q3 Similarly, look at the remainder when the multiples of 100 (100, 200, 300, … ) are
divided by 9. What do you notice?

The remainder is the number of hundreds.

MULTIPLE OF 100 100 200 300 400 500 700 900

REMAINDER ÷ 9 1 2 3 4 5 7 0

100 = 99 + 1, and 99 is a multiple of 9

So 100c = 99c + c → remainder c

Check 700: 700 = 693 + 7, and 693 = 9 × 77 ✓ so the remainder is 7.

Did you notice? The tens behaved this way because 10 = 9 + 1; the hundreds behave
this way because 100 = 99 + 1. The same trick is about to work for every place value.

Page 33 of 75

as e
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Using this observation, find the remainder when 427 is divided by 9.
e
Q4

m l as
.co a g
a
s em
a l
Thegremainder is 4.

co m
. ag
427 has 4 hundreds → contributes remainder 4
em
427 has 2 tens → contributes remainder 2
g l as
427 has 7 units → contributes remainder 7
a

co m
m.
Total leftover = 4 + 2 + 7 = 13

m as e
.co
13 makes one more group of 9, leaving 4
a g l
a s em
a l 427 = 9 × 47 + 4 = 423 + 4 ✓
gCheck:
m a s
comultiple agl
Why it happens: 400 = 396 + 4 and 20 = 18 + 2, where 396 and 18 are multiples of 9.

m .
as e
So 427 = (396 + 18) + (4 + 2 + 7) — a big of 9 plus the digit sum. Only the

a g
digit sum can leave a remainder.l

co m
m .
as e
coformDivisibility by 9
In-text Questions — Page 125
. a g l
em
A Shortcut

a s
agl
m
Will this work with bigger numbers?
se
Q1

com g l a
m . a
ase

agl
Yes — for every place value, however large. That is the whole point of the pattern:

co m
1=0+1
m .
m as e
.co
10 = 9 + 1
a g l
se m 100 = 99 + 1
g l a
a c
.
1000 = 999 + 1

s e m
m a
co agl
10000 = 9999 + 1, and so on

m .
ase
a g l
Each place value is 1 more than a multiple of 9, so each digit contributes exactly itself to the
remainder. Take 7309:

co m
m .
m ase
.co


a g l Page 34 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

7 × 1000 + 3 × 100 + 0 × 10 + 9 × 1

= 7 × (999 + 1) + 3 × (99 + 1) + 0 × (9 + 1) + 9 × (0 + 1)

= (7 × 999 + 3 × 99 + 0 × 9 + 9 × 0) + (7 + 3 + 0 + 9)

= (a multiple of 9) + 19

Now shrink 19 the same way: 1 + 9 = 10, then 1 + 0 = 1. So 7309 is 1 more than a multiple of 9,
and 7309 ÷ 9 leaves remainder 1.
Check: 7309 = 9 × 812 + 1 = 7308 + 1 ✓

Tip: the rule to remember — a number is divisible by 9 if and only if the sum of
its digits is divisible by 9, and adding the digits repeatedly down to one digit gives
the remainder (with 9 standing for remainder 0).

Q2 Look at each of the following statements. Which are correct and why? (i) If a
number is divisible by 9, then the sum of its digits is divisible by 9. (ii) If the sum of
the digits of a number is divisible by 9, then the number is divisible by 9. (iii) If a
number is not divisible by 9, then the sum of its digits is not divisible by 9. (iv) If the
sum of the digits of a number is not divisible by 9, then the number is not divisible
by 9.

All four are correct.
Everything follows from one fact established on this page:

N = (a multiple of 9) + (sum of the digits of N)

So N and its digit sum leave the same remainder on division by 9

STATEMENT FORM CORRECT?

(i) 9 | N ⇒ 9 | digit sum the 'only if' half Yes

(ii) 9 | digit sum ⇒ 9 | N the 'if' half — the converse of (i) Yes

(iii) 9 ∤ N ⇒ 9 ∤ digit sum contrapositive of (ii) Yes

(iv) 9 ∤ digit sum ⇒ 9 ∤ N contrapositive of (i) Yes

Page 35 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Why all four survive: a statement and its contrapositive are always either both true
or both false, so (iii) rides on (ii) and (iv) rides on (i). What is not automatic is that (i)
and (ii) are both true — a statement and its converse usually part company. Here
they do not, because the remainders are equal, not merely related.

Check it yourself: 405 → 9 ✓ and 405 = 9 × 45 ✓ (both (i) and (ii) hold). 8888 → 32 ✗
and 8888 = 9 × 987 + 5 ✗ (both (iii) and (iv) hold).

Figure it Out — Page 126
A Shortcut for Divisibility by 9

Q1 Find, without dividing, whether the following numbers are divisible by 9. (i) 123 (ii)
405 (iii) 8888 (iv) 93547 (v) 358095

Only (ii) 405 is divisible by 9.

NUMBER SUM OF DIGITS REDUCED DIVISIBLE BY 9?

123 1+2+3=6 6 No (remainder 6)

405 4+0+5=9 9 Yes

8888 8 + 8 + 8 + 8 = 32 3+2=5 No (remainder 5)

93547 9 + 3 + 5 + 4 + 7 = 28 2 + 8 = 10 → 1 No (remainder 1)

358095 3 + 5 + 8 + 0 + 9 + 5 = 30 3+0=3 No (remainder 3)

Tip: while adding, throw away any 9s and any pairs that make 9. For 358095 the 9
goes, 3 + 5 + 8 + 5 = 21, then 2 + 1 = 3 — same answer, less work.

Q2 Find the smallest multiple of 9 with no odd digits.

The smallest one is 288.

Page 36 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Allowed digits: 0, 2, 4, 6, 8 — all even

A sum of even digits is even

So the digit sum must be an even multiple of 9 → 18, 36, …

Two digits? The largest possible digit sum is 8 + 8 = 16, which is less than 18. Impossible.
Three digits? Keep the hundreds digit as small as possible, then the tens:

Hundreds digit 2 → the other two even digits must add to 16

Only 8 + 8 = 16 works

→ 288

Check: 2 + 8 + 8 = 18, a multiple of 9, and 288 = 9 × 32 ✓. Every digit is even ✓. Nothing smaller
can work, because any number below 288 with all even digits has at most three digits and either
starts with 2 (forcing 8 and 8) or is a two-digit number.

Q3 Find the multiple of 9 that is closest to the number 6000.

The closest multiple of 9 is 6003.

6 + 0 + 0 + 0 = 6, so 6000 is 6 more than a multiple of 9

Multiple of 9 just below: 6000 – 6 = 5994 (= 9 × 666)
Multiple of 9 just above: 5994 + 9 = 6003 (= 9 × 667)

6000 – 5994 = 6 6003 – 6000 = 3

6003 is only 3 away, so it is the closer of the two.

Tip: the digit sum told us the remainder is 6 without any division. Whenever the
remainder is more than 4½, the multiple above is the nearer one.

Page 37 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q4 How many multiples of 9 are there between the numbers 4300 and 4400?

There are 11 of them.

4 + 3 + 0 + 0 = 7, so 4300 is 7 more than a multiple of 9

First multiple of 9 above 4300: 4300 + (9 – 7) = 4302 = 9 × 478

Last multiple of 9 below 4400: 4 + 4 + 0 + 0 = 8, so 4400 – 8 = 4392 = 9 × 488

Count = 488 – 478 + 1 = 11

The eleven numbers are 4302, 4311, 4320, 4329, 4338, 4347, 4356, 4365, 4374, 4383, 4392.

Why 11 and not 100 ÷ 9 ≈ 11.1: in a stretch of 100 consecutive numbers there are
either 11 or 12 multiples of 9, depending on where the stretch starts. Counting the
first and last one exactly is the safe way.

In-text Questions — Page 126
A Shortcut for Divisibility by 3

Q1 The shortcut to find the divisibility by 3 is similar to the method for 9. A number is
divisible by 3 if the sum of its digits is divisible by 3. Explore the remainders when
powers of 10 are divided by 3. Explain why this method works.

Every power of 10 leaves a remainder of 1 when divided by 3.

Page 38 of 75

as e
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
POWER OF 10 WRITTEN AS REMAINDER ÷ 3

m as e
1
.co
0+1 1
a g l
sem
g l a
a
10 9+1=3×3+1 1

100 99 + 1 = 3 × 33 + 1 1

co m
m . ag
se
1000 999 + 1 = 3 × 333 + 1 1

9999 + g l a
10000
a 1 = 3 × 3333 + 1 1

co m
m.
Why the method works: 9, 99, 999, 9999 are all multiples of 9 — and since 9 is itself

ase
com l
a multiple of 3, they are all multiples of 3 as well. So splitting a number the same way
. a g
m
ase
as before,

agl … + 1000d + 100c + 10b + a
m a s
.co agl
= (999d + 99c + 9b) + (d + c + b + a)

se m
g l a
a
the first bracket is a multiple of 3, so the number and its digit sum leave the same
remainder on division by 3.

co m
m .
m as e
.co
Examples: 15 → 1 + 5 = 6 ✓ and 15 = 3 × 5
a g l
s m → 8 + 7 = 15 ✓ and 87 = 3 × 29
e87
gl a
a
m
429714 → 4+2+9+7+1+4 = 27 ✓ and 429714 = 3 × 143238
a se
. com a g l
e m
g l as multiple of 9 is a multiple of 3, but not the other
Tip: this also explains why every
way round. 15, 33 and 87a have digit sums 6, 6 and 15 — divisible by 3, not by 9.

co m
m .
as e
comQuestions — Page 127
.
In-text
a g l
m
ase
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 39 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

A Shortcut for Divisibility by 11

MATH TALK

Q1 Using these observations, can you tell whether the number 462 is divisible by 11?

Yes, 462 is divisible by 11.
Use the alternating behaviour of the place values: 1 and 100 are 1 more than a multiple of 11,
while 10 is 1 less.

PART NEAREST MULTIPLE OF 11 EXCESS / SHORT

400 (4 hundreds) 396 = 11 × 36 4 more

60 (6 tens) 66 = 11 × 6 6 short

2 (2 units) 0 = 11 × 0 2 more

Total excess – total short = (4 + 2) – 6 = 0

Nothing is left over, so 462 is a multiple of 11. Check: 462 = 11 × 42 ✓

Q2 What could be a general method or shortcut to check divisibility by 11?

Here is the shortcut, in the two forms the chapter uses.
Form 1 — excess and short.

1. Add the digits sitting in the places 1, 100, 10000, … (these are 1 more than a multiple of 11).
Call this the excess.
2. Add the digits sitting in the places 10, 1000, 100000, … (these are 1 less than a multiple of 11).
Call this the short.
3. Compute excess – short. If it is 0 or a multiple of 11, the number is divisible by 11; otherwise
it tells you how far off you are.

Form 2 — alternating signs. Put '+' before the units digit, '–' before the tens digit, '+' before the
hundreds digit, and so on. Add up. Same answer, less writing.

Page 40 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

462 → + 2 – 6 + 4 = 0 → divisible ✓

320185 → + 5 – 8 + 1 – 0 + 2 – 3 = – 3 → 3 short of a multiple of 11

Why it works: 1 = 11 × 0 + 1, 10 = 11 × 1 – 1, 100 = 11 × 9 + 1, 1000 = 11 × 91 – 1, and
this alternation continues for every higher place. So each digit contributes either +
itself or – itself to the leftover, and the alternating sum is the leftover.

In-text Questions — Page 128
A Shortcut for Divisibility by 11

Q1 If this difference is 11 or a multiple of 11, what does that say about the remainder
obtained when the number is divisible by 11?

The remainder is 0 — the number is exactly divisible by 11.

In the book's example, 320185:

excess = 2 + 1 + 5 = 8, short = 3 + 0 + 8 = 11

8 – 11 = – 3 → 3 short of a multiple of 11

Now suppose the difference had come out as 11, or 22, or – 11. That leftover is itself a whole
number of 11s, so it joins the multiple of 11 already built up and nothing at all remains.

Why it happens: the number equals (a multiple of 11) + (the difference). If the
difference is also a multiple of 11, the total is a multiple of 11 plus a multiple of 11 —
still a multiple of 11, by the rule that a | M and a | N gives a | M + N.

Tip: if the difference is negative, add 11 (or a multiple of 11) to it to read off the
remainder. A difference of – 3 means remainder 11 – 3 = 8.

Page 41 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q2 Using this shortcut, find out whether the following numbers are divisible by 11.
Further, find the remainder if the number is not divisible by 11. (i) 158 (ii) 841 (iii) 481
(iv) 5529 (v) 90904 (vi) 857076

Attach alternating '+' and '–' signs starting from the units digit.

NUMBER ALTERNATING SUM VALUE CONCLUSION

(i) 158 8–5+1 4 Not divisible; remainder 4

(ii) 841 1–4+8 5 Not divisible; remainder 5

(iii) 481 1–8+4 –3 Not divisible; remainder 8 (= 11 – 3)

(iv) 5529 9–2+5–5 7 Not divisible; remainder 7

(v) 90904 4–0+9–0+9 22 Divisible by 11

(vi) 857076 6–7+0–7+5–8 – 11 Divisible by 11

Verification by actual division:

158 = 11 × 14 + 4 ✓ 841 = 11 × 76 + 5 ✓

481 = 11 × 43 + 8 ✓ 5529 = 11 × 502 + 7 ✓

90904 = 11 × 8264 ✓ 857076 = 11 × 77916 ✓

Tip: in (iii) the alternating sum was negative. A negative answer of – r means the
number is r short of a multiple of 11, so the remainder is 11 – r. In (v) the answer 22
is a multiple of 11, which counts as 0 left over.

In-text Questions — Page 129

Page 42 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

A Shortcut for Divisibility by 11 / More on Divisibility Shortcuts

MATH TALK

Q1 Look at the following procedure — 1. Place alternating ‘+’ and ‘–’ signs before every
digit starting from the unit’s digit. 2. Evaluate the expression. 3. The result denotes
the remainder obtained when the number is divided by 11. Is this method similar to
or different from the method we saw just before?

It is the same method, written more compactly.

EARLIER METHOD (320185) ALTERNATING-SIGN METHOD

excess = 2 + 1 + 5 = 8 (places 1, 100, 10000) these digits get '+'

short = 3 + 0 + 8 = 11 (places 10, 1000, 100000) these digits get '–'

8 – 11 = – 3 –3+2–8+1–0+5=–3

Why they agree: writing (excess) – (short) means adding some digits and
subtracting the others. Which digits get which sign is decided by the place value —
and place values alternate between 1 more and 1 less than a multiple of 11. So
attaching alternating signs from the units place is the excess-minus-short
calculation, just written in one line.

Tip: the compact version is faster and less error-prone, but the first version is the
one that explains why. Both give – 3 for 320185: 3 short of, or 8 more than, a multiple
of 11.

Page 43 of 75

as e
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Fill in the following table. Find a quick way to do this?
e
Q2

com g l as
m . a
ase
NUMBER DIVISIBLE BY

agl 2 3 4 5 6 8 9 10 11

m
.co
128 Yes No Yes No No Yes No No No

ag
a sem
agl
990

1586

co m
m.
275

m as e
.co
6686
a g l
a s em639210
a gl
s
429714

m a
m .co agl
se
2856

g l a
3060
a
406839
co m
m .
m as e
.co a g l
se m
a

a g l
m
NUMBER 2 3 4 5 6 8 9 10 11

a se
128 Yes No
.
Yes
com No No Yes No No No
a g l
e m
Yeslas No
ag
990 Yes Yes Yes No Yes Yes Yes

1586 Yes No No No No No No No No

co m
m .
ase
275 No No No Yes No No No No Yes

. com agNol
e m
as
6686 Yes No No No No No No No

a g l
c
639210 Yes Yes No Yes Yes No No Yes Yes

m .
m a s e
.co gl
429714 Yes Yes No No Yes No Yes No No

m Yes a
2856 Yes
l a
Yes
se No Yes Yes No No No

3060 Yes a g Yes Yes Yes Yes No Yes Yes No

co m
m .
m as e
.co


a g l Page 44 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

406839 No Yes No No No No No No No

The quick way — do only four checks per row and read off the rest:

2 — units digit even. 5 — units digit 0 or 5. 10 — units digit 0 (so 10 = 2 and 5 together).
3 and 9 — one digit sum serves both. If the digit sum is a multiple of 9 it is automatically a
multiple of 3.
6 — no new work: it is just 2 and 3.
4 — last two digits; 8 — last three digits.
11 — alternating sum of the digits from the units place.

639210 → digit sum 21 → multiple of 3 but not 9

→ alternating sum 0 – 1 + 2 – 9 + 3 – 6 = – 11 → multiple of 11 ✓

→ last two digits 10 → not a multiple of 4

Note on the book's sample row: the printed table on page 129 marks 128 as not
divisible by 4. That is a misprint — 128 = 4 × 32, so the entry should be Yes. (Indeed
128 = 2⁷, so it is divisible by 2, 4, 8, 16, 32, 64 and 128.)

Q3 How can we find out if a number is divisible by 6?

Check divisibility by 2 and by 3 — both must hold.

6 = 2 × 3, and 2 and 3 have no common factor

LCM (2, 3) = 6

Why it works: if a number is divisible by 2 and also by 3, then by the rule from page
121 it is divisible by LCM (2, 3) = 6. Conversely, if it is divisible by 6 it is divisible by
every factor of 6, so by 2 and 3. The two tests are equivalent.

Page 45 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q4 Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3
on these numbers and divide each number by 6 to verify — 38, 225, 186, 64.

Yes, it works every time.

NUMBER ÷ 2? DIGIT SUM → ÷ 3? PREDICTION ACTUAL ÷ 6

38 Yes 11 → No Not divisible 38 = 6 × 6 + 2 ✓
225 No 9 → Yes Not divisible 225 = 6 × 37 + 3 ✓
186 Yes 15 → Yes Divisible 186 = 6 × 31 ✓
64 Yes 10 → No Not divisible 64 = 6 × 10 + 4 ✓

Notice that 38 and 64 pass the test for 2 but fail for 3, and 225 does the opposite. Both tests
must be passed.

In-text Questions — Page 130
Divisibility Shortcuts for Other Numbers / Digital Roots

MATH TALK

Q1 How about checking divisibility by 24? Will checking the divisibility by its factors, 4
and 6, work? Why or why not?

No, it does not work. The smallest counterexample is 12.

12 ÷ 4 = 3 ✓ 12 ÷ 6 = 2 ✓ but 12 ÷ 24 ✗

Why not: 4 = 2 × 2 and 6 = 2 × 3 share the factor 2. Being divisible by both only
guarantees divisibility by LCM (4, 6) = 12, not by 24. Other numbers that pass 4 and 6
but fail 24: 36, 60, 84.

The chapter's replacement test is to check divisibility by 3 and by 8 instead.

Page 46 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q2 Explain using prime factorisation why checking divisibility by 3 and 8 works for
checking divisibility by 24, but checking divisibility by 4 and 6 is not sufficient for
checking divisibility by 24.

Write everything in prime factors:

24 = 2 × 2 × 2 × 3 = 2³ × 3

Using 3 and 8. Divisibility by 8 puts 2³ into the number's prime factorisation; divisibility by 3
puts a 3 there. These use different primes, so both blocks sit side by side:

the number contains 2³ × 3 = 24 ✓

LCM (3, 8) = 3 × 8 = 24

Using 4 and 6. Divisibility by 4 puts 2² in; divisibility by 6 puts 2 × 3 in. But the 2 inside 6 can be
the same 2 that is already inside 4 — the two demands overlap.

4 = 2², 6 = 2 × 3

LCM (4, 6) = 2² × 3 = 12, one factor of 2 short of 24

The general point: two tests k and m together prove divisibility by LCM (k, m), never
by more. So a pair of tests works for a number N only when LCM (k, m) = N — which
happens when k and m are coprime and multiply to N.

Q3 What property do you think this digital root will have? Recall that we did this while
finding the divisibility shortcut for 9.

The digital root of a number is its remainder on division by 9 — except that a multiple of 9 has
digital root 9 rather than 0.

Page 47 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

NUMBER DIGITAL ROOT REMAINDER ÷ 9

489710 29 → 11 → 2 2

427 13 → 4 4

7309 19 → 10 → 1 1

405 9 0

Why it happens: replacing a number by its digit sum never changes the remainder
on division by 9, because the number equals (a multiple of 9) + (its digit sum).
Repeating the step keeps the remainder fixed all the way down to a single digit —
and the only single digits are 1 to 9, so remainder 0 shows up as 9.

Q4 Between the numbers 600 and 700, which numbers have the digital root: (i) 5, (ii) 7,
(iii) 3?

Every number here is 6ab, so the digit sum is 6 + a + b. Work out what a + b must be in each
case.
(i) Digital root 5 → 6 + a + b must be 5, 14 or 23 → a + b = 8 or 17

608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698

(ii) Digital root 7 → 6 + a + b must be 7 or 16 → a + b = 1 or 10

601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691

(iii) Digital root 3 → 6 + a + b must be 12 or 21 → a + b = 6 or 15

606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696

Did you notice? Each list has 11 numbers and they step by 9. That is no accident —
adding 9 leaves the remainder on division by 9 unchanged, so it keeps the digital
root the same.

Page 48 of 75

as e
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Write the digital roots of any 12 consecutive numbers. What do you observe?
e
Q5

m l as
.co a g
a
s em
agl25 to 36:
Take

m
.co ag
NUMBER 25 26 27 28 29 30 31 32 33 34 35 36

a s1em 2
agl
DIGITAL ROOT 7 8 9 3 4 5 6 7 8 9

Observation: the digital roots go up by 1 each time, and after 9 they start again at 1. The

co m
m.
pattern repeats every 9 numbers — that is why the 12 roots show the first three of them twice.

m as e
.co
Why it happens: adding 1 to the number adds 1 to its remainder on division by 9,
a g l
a s emthe remainder returns to 0 — which the digital root records as 9.
gl
until
a
a s
. com of 9 is always 9. Now, find the digital roots agl
em
We saw that the digital root of multiples
s
Q6

g l a
of some consecutive multiples of (i) 3, (ii) 4, and (iii) 6.
a
m

. co
se m21
com
MULTIPLES OF 3
l a
3 6 9 12 15 18 24 27

. ROOT g
a9
e m
as
agl
DIGITAL 3 6 9 3 6 3 6 9

(i) The roots cycle 3, 6, 9 — only three values, repeating every 3 multiples.
se m
com g l a
m . a
ase
MULTIPLES OF 4 4 8 12 16 20 24 28 32 36

DIGITAL ROOT agl4 8 3 7 2 6 1 5 9

co m
(ii) The roots cycle 4, 8, 3, 7, 2, 6, 1, 5, 9 — all nine values appear before repeating.
m .
m l a s24e
.co ag
em
MULTIPLES OF 6 6 12 18 30 36

a s
agl DIGITAL ROOT 6 3 9 6 3 9

.c
s e m
m a
co agl
(iii) The roots cycle 6, 3, 9 — again only three values.

m .
as e
a g l

co m
m .
m as e
.co


a g l Page 49 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Why the cycle lengths differ: each step adds the multiplier to the root, working
modulo 9. For 3 and 6 the step shares the factor 3 with 9, so only the multiples of 3
among the roots are ever reached — three of them. For 4, which shares no factor
with 9, the steps eventually land on every root, so the cycle has length 9.

Q7 What are the digital roots of numbers that are 1 more than a multiple of 6? What do
you notice? Try to explain the patterns noticed.

The roots cycle 7, 4, 1.

6K + 1 7 13 19 25 31 37 43 49 55

DIGITAL ROOT 7 4 1 7 4 1 7 4 1

What you notice: only three of the nine possible roots ever appear, and they are exactly the
roots that leave remainder 1 on division by 3.

Explanation: moving from one such number to the next adds 6, so the digital root
moves on by 6 and then drops back by 9 when it overshoots: 7 → 13 gives 4, 4 → 10
gives 1, 1 → 7. Since 6 and 9 share the factor 3, the steps can only reach roots that
differ from 7 by a multiple of 3 — namely 7, 4 and 1. The cycle length is 9 ÷ 3 = 3,
exactly as it was for the multiples of 3 and of 6.

Try This: repeat with numbers that are 2 more than a multiple of 6 — 8, 14, 20, 26, …
The roots cycle 8, 5, 2. Same length, shifted by one.

Q8 I’m made of digits, each tiniest and odd, No shared ground with root #1 — how odd!
My digits count, their sum, my root — All point to one bold number’s pursuit — The
largest odd single-digit I proudly claim. What’s my number? What’s my name?

The number is 11,11,11,111 — nine 1s in a row.
Its name in the Indian system: eleven crore eleven lakh eleven thousand one hundred
eleven.
Read the clues one by one:

Page 50 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

CLUE WHAT IT FIXES

“made of digits, each tiniest and odd” every digit is 1 — the smallest odd digit

“the largest odd single-digit I proudly claim” the target value is 9

“my digits count … all point to” 9 there are 9 digits

“their sum … all point to” 9 1 + 1 + … + 1 (nine times) = 9 ✓
“my root … all point to” 9 digit sum 9, so the digital root is 9 ✓

11,11,11,111 = 111111111

digits: 9 digit sum: 9 digital root: 9

and 111111111 = 9 × 12345679 ✓

Did you know? That last division is a small curiosity in itself — 111111111 ÷ 9 =
12345679, the digits 1 to 9 with the 8 missing.

Figure it Out — Page 131
Digital Roots

MATH TALK

Q1 The digital root of an 8-digit number is 5. What will be the digital root of 10 more
than that number?

The new digital root is 6.

Digital root 5 means the number is 5 more than a multiple of 9

10 = 9 + 1, so adding 10 adds 1 to the leftover

New leftover = 5 + 1 = 6

Test it with an actual 8-digit number:

Page 51 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

50000000 → digit sum 5 → root 5

50000000 + 10 = 50000010 → digit sum 6 → root 6 ✓

14000000 → digit sum 5 → root 5

14000010 → digit sum 6 → root 6 ✓

Tip: if the starting root had been 8 or 9, the answer would wrap round: 8 + 1 = 9, and
9 + 1 = 10 → 1. Here 5 + 1 = 6 is safely under 9, so no wrap is needed.

Q2 Write any number. Generate a sequence of numbers by repeatedly adding 11. What
would be the digital roots of this sequence of numbers? Share your observations.

Start at 10 and keep adding 11:

NUMBER 10 21 32 43 54 65 76 87 98 109 120

DIGITAL ROOT 1 3 5 7 9 2 4 6 8 1 3

Observations.

The digital root goes up by 2 at every step, wrapping round after 9.
All nine roots appear, and the whole pattern repeats after 9 terms.
Reading the roots in order gives 1, 3, 5, 7, 9, 2, 4, 6, 8 — the odd roots first, then the even
ones.

Why 2: 11 = 9 + 2, so adding 11 adds 2 to the remainder on division by 9. Since 2 and
9 share no factor, repeatedly stepping by 2 lands on every one of the nine roots
before returning to the start.

Check it yourself: begin at 7 instead — 7, 18, 29, 40, 51, … The roots are 7, 9, 2, 4, 6,
… Same step of 2, different starting point.

Page 52 of 75

Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q3 What will be the digital root of the number 9a + 36b + 13?

The digital root is 4, whatever whole numbers a and b are.

9a + 36b + 13

= 9a + 36b + 9 + 4

= 9(a + 4b + 1) + 4

The bracket is multiplied by 9, so that part is a multiple of 9 and contributes nothing to the
leftover. What remains is 4.

Check with a = 2, b = 1: 18 + 36 + 13 = 67 → 6 + 7 = 13 → 1 + 3 = 4 ✓

Check with a = 0, b = 0: 13 → 1 + 3 = 4 ✓

Check with a = 5, b = 3: 45 + 108 + 13 = 166 → 13 → 4 ✓

Q4 Make conjectures by examining if there are any patterns or relations between (i)
the parity of a number and its digital root. (ii) the digital root of a number and the
remainder obtained when the number is divided by 3 or 9.

(i) Parity and digital root — there is no relation.

NUMBER PARITY DIGITAL ROOT PARITY OF THE ROOT

12 even 3 odd

24 even 6 even

15 odd 6 even

27 odd 9 odd

All four combinations occur, so knowing the parity tells you nothing about the digital root and
vice versa.

Page 53 of 75

ase
Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

co m
m.
Why: parity is about the remainder on division by 2, the digital root about the

m completely independent. as e
l
remainder on division by 9. Since 2 and 9 share no factor, the two pieces of
. c o a g
m
information are
as e
g l
a Digital root and remainders — there is a clean relation both times.
(ii)

m
Division by 9: the remainder equals the digital root, except that a digital root of 9 means

. co ag
m
remainder 0.

as e
g l
Division by 3: the remainder is what the digital root itself leaves on division by 3.
a
DIGITAL ROOT REMAINDER ÷ 9 REMAINDER ÷ 3

co m
e m.
las
1, 4, 7 1, 4, 7 1

2, 5, 8 .co
m 2 ag
s em
2, 5, 8

a
agl3, 6 3, 6 0

m a s
.co agl
9 0 0

a s em
l leave the same remainder on division by 9, and —
Why: a number and its digitgsum
a
because 9 is a multiple of 3 — the same remainder on division by 3 as well.
Repeating the digit sum keeps both remainders fixed. So the single digit at the end
co m
m .
carries both answers.
m as e
.co a g l
sem
g l a
a In-text Questions — Page 131
se m
5.3 Digits in Disguise
com g l a
m . a
ase
MATH TALK

agl
Solve the cryptarithms given below. (i) A1 + 1B = B0 (ii) AB + 37 = 6A (iii) ON + ON +
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ON = PO (iv) QR + QR + QR = PRR
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em(i) A = 7, B = 9 → 71 + 19 = 90

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Units: 1 + B must end in 0 → B = 9, carry 1

Tens: A + 1 + 1 = B = 9 → A =m 7.
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Check: 71 + 19 = 90 ✓ag

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Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

(ii) A = 2, B = 5 → 25 + 37 = 62

Tens: A + 3 + (carry) = 6

If the carry is 0: A = 3, and then units 3 = B + 7 gives B = – 4 ✗
So the carry is 1: A + 4 = 6 → A = 2

Units: B + 7 = A + 10 = 12 → B = 5

Check: 25 + 37 = 62 ✓

(iii) Three solutions, one of which is O = 3, N = 1, P = 9. The puzzle is 3 × ON = PO with a two-
digit product.

Units: 3N must end in O

The product must stay below 100, so ON ≤ 33 → O = 1, 2 or 3

O N (SO THAT 3N ENDS IN O) CHECK

1 7 (3 × 7 = 21) 17 × 3 = 51 = PO ✓ →P=5
2 4 (3 × 4 = 12) 24 × 3 = 72 = PO ✓ →P=7
3 1 (3 × 1 = 3) 31 × 3 = 93 = PO ✓ →P=9

All three keep the letters on different digits, so all three are valid: 17 + 17 + 17 = 51, 24 + 24 + 24
= 72 and 31 + 31 + 31 = 93. (The book's answer key lists only the last one.)
(iv) Q = 8, R = 5, P = 2 → 85 + 85 + 85 = 255

3 × QR = PRR, a three-digit number, so QR ≥ 34

Units: 3R must end in R → 2R ends in 0 → R = 0 or 5

R = 0 forces 30Q = 100P, impossible for digits

So R = 5: 3(10Q + 5) = 100P + 55

30Q + 15 = 100P + 55 → 3Q = 10P + 4
P = 2 gives 3Q = 24 → Q = 8

Check: 85 × 3 = 255 ✓

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Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

Q2 (v) PQ × 8 = RS. Guna says, “Oh, this means a 2-digit number multiplied by 8 should
give another 2-digit number. I know that 10 × 8 = 80. But the units digits of 10 and 80
are the same, which we don’t want. For the same reason PQ cannot be 11 as P and Q
correspond to different digits. 12 × 8 = 96 fits all the conditions”. Can PQ be 13?
Think.

No, PQ cannot be 13 — and 12 is the only value that works.

13 × 8 = 104, a 3-digit number ✗

For the product to stay at two digits we need PQ × 8 ≤ 99, that is PQ ≤ 12. So only 10, 11 and 12
are candidates:

PQ PQ × 8 ALLOWED?

10 80 No — Q and S would both be 0

11 88 No — P and Q would both be 1

12 96 Yes — P = 1, Q = 2, R = 9, S = 6

Why nothing bigger works: multiplication by 8 is increasing, so once 13 × 8 crosses
100, every larger two-digit number does too. One check at the boundary rules out all
87 remaining possibilities at once — that is the economy Guna is using.

In-text Questions — Page 132
5.3 Digits in Disguise

Q1 (vi) Try this now: GH × H = 9K. This means a 2-digit number multiplied by a 1-digit
number gives another 2-digit number in the 90s. Observe the letters corresponding
to the units digits in this cryptarithm. Pick the solution to this question from the
options given below: 11 × 9 = 99, 12 × 8 = 96, 46 × 2 = 92, 24 × 4 = 96, 47 × 2 = 94, 31 × 3
= 93, 16 × 6 = 96.

The solution is 24 × 4 = 96, giving G = 2, H = 4, K = 6.
The letter to watch is H: it is both the units digit of the two-digit number and the multiplier. So
the multiplier must equal the units digit.

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Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

OPTION UNITS DIGIT VS MULTIPLIER FITS GH × H = 9K?

11 × 9 = 99 1 ≠ 9 No

12 × 8 = 96 2 ≠ 8 No

46 × 2 = 92 6 ≠ 2 No

24 × 4 = 96 4=4✓ Yes

47 × 2 = 94 7 ≠ 2 No

31 × 3 = 93 1 ≠ 3 No

16 × 6 = 96 6 = 6 ✓, but then H = 6 and K = 6 No — two letters cannot share a digit

Tip: 16 × 6 = 96 is the near miss worth studying. It passes the units-digit test but
breaks the basic rule of a cryptarithm — different letters must stand for different
digits.

Q2 (vii) Here is one more: BYE × 6 = RAY. Anshu says, “Since the product is a 3-digit
number, B can’t be 2 or more. If B = 2, i.e., 2 hundreds, the product will be more than
1200. So, B = 1.” What can you say about ‘Y’? What digits are possible/not possible?

Y must be even and less than 7, so the only possibilities are Y = 0, 2, 4 or 6. Testing them
leaves exactly one solution.
Two facts pin Y down straight away:

Y is even. The product RAY ends in Y, and 6 × (anything) is even. So Y cannot be 1, 3, 5, 7 or 9.
Y is less than 7. If Y = 7, then BYE is at least 170, and 170 × 6 = 1020 — a 4-digit product. So
7, 8 and 9 are out.

Now try each surviving value, remembering B = 1 and that 6 × E must end in Y:

Y E FORCED BY “6 × E ENDS IN Y” OUTCOME

0 E = 5 (6 × 5 = 30) 105 × 6 = 630 → R = 6, A = 3 ✓ all letters different
2 E = 7 (6 × 7 = 42) 127 × 6 = 762 → R = 7 clashes with E = 7 ✗

4 E = 9 (6 × 9 = 54) 149 × 6 = 894 → A = 9 clashes with E = 9 ✗

6 E = 1 or 6 both clash — E = 1 is B, E = 6 is Y ✗

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Class 8 Maths Chapter 5 Number Play AglaSem · NCERT Solutions

So B = 1, Y = 0, E = 5, R = 6, A = 3 and 105 × 6 = 630 ✓

Tip: Y = 0 is allowed because Y is never the leading digit of either number. Only B and
R are leading digits, and neither of them may be 0.

Q3 Solve the following: (i) UT × 3 = PUT (ii) AB × 5 = BC (iii) L2N × 2 = 2NP (iv) XY × 4 = ZX
(v) PP × QQ = PRP (vi) JK × 6 = KKK

(i) U = 5, T = 0, P = 1 → 50 × 3 = 150

3(10U + T) = 100P + 10U + T

30U + 3T = 100P + 10U + T

20U + 2T = 100P → 10U + T = 50P

So UT = 50P; P = 1 gives UT = 50 (P = 2 would need UT = 100)

(ii) A = 1, B = 9, C = 5 → 19 × 5 = 95

A 2-digit × 5 stays 2-digit only if AB ≤ 19, so A = 1

(10 + B) × 5 = 10B + C
50 + 5B = 10B + C → C = 50 – 5B

C must be a digit → 50 – 5B ≤ 9 → B = 9, C = 5

(iii) Two solutions: L = 1, N = 5, P = 0 and L = 1, N = 4, P = 8.

2(100L + 20 + N) = 200 + 10N + P

200L + 40 + 2N = 200 + 10N + P

200L + 40 = 200 + 8N + P

L = 1 → 8N + P = 40 (L = 2 would need 8N + P = 240)

N = 5 gives P = 0 → 125 × 2 = 250 ✓ N = 4 gives P = 8 → 124 × 2 = 248 ✓. Both keep every letter
on its own digit and away from the printed 2. (The book's answer key lists only 125 × 2 = 250.)
(iv) X = 2, Y = 3, Z = 9 → 23 × 4 = 92

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Class 8 Maths Chapter 5 Number Play
a g l AglaSem · NCERT Solutions

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4(10X + Y) = 10Z + X → 39X + 4Y = 10Z
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2-digit product → XY ≤ 24, so X = 1 or 2
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X = 1: 39 + 4Y = 10Z needs 4Y to end in 1 — impossible, 4Y is even
aX = 2: 78 + 4Y = 10Z needs 4Y to end in 2 → Y = 3 → Z = 9

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(v) Three solutions, one of which is P = 2, m
a s e Q = 1, R = 4.

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PP = 11P and QQ = 11Q, so PP × QQ = 121 × P × Q

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The product is 3-digit → P × Q ≤ 8

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121 × 2 = 242, 121 × 3 = 363, 121 × 4 = 484 — each of the form PRP
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a PRODUCT READING IT AS PRP CRYPTARITHM

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242 P = 2, R = 4, so Q = 1 22 × 11 = 242 ✓
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363 P = 3, R = 6, so Q = 1 33 × 11 = 363

484 P = 4, R = 8, so Q = 1 44 × 11 = 484 ✓
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121 × 1 = 121 fails because it forces P = Q = 1, and 121 × 5 = 605 onwards m
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ends with
comthe same digit. a gla
a s emJ = 7, K = 4 → 74 × 6 = 444
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(vi)

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KKK = 111K, so 6(10J + K) = 111K
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60J + 6K = 111K

60J = 105K → 4J = 7K
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K must be a multiple of 4 → K = 4, J = 7 (K = 8 would give J = 14)
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a s em it Out — Page 132
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