NCERT Solutions Class 6 Maths Chapter 5 Prime Time

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

C L A S S 6 · M AT H S

NCERT Solutions

Chapter 5: Prime Time

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

107 – 128 14 79 English

Solutions, notes, sample papers & more at 70 pages

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

CLASS 6 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 5: Prime Time
Complete, step-by-step NCERT Solutions for Class 6 Maths Chapter 5 Prime Time from the NCERT textbook
Ganita Prakash. Every Figure it Out question from pages 108 to 126 is solved, along with all the in-text
questions — the Idli-Vada game, Jump Jackpot, prime and composite numbers, the Sieve of Eratosthenes, co-
prime numbers, prime factorisation and the divisibility tests.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 6) 107 – 128

SECTIONS QUESTIONS

14 79

MEDIUM

English

In-text Question — Page 107
Section 5.1 Common Multiples and Common Factors — Idli-Vada Game

Q1 Which is the first number for which the players should say, ‘idli-vada’? It is 15, which
is a multiple of 3, and also a multiple of 5. Find out other such numbers that are
multiples of both 3 and 5. These numbers are called _____________________________.

The players say ‘idli-vada’ exactly at the numbers that are in the 3 table and in the 5 table.

Multiples of 3 → 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, …

Multiples of 5 → 5, 10, 15, 20, 25, 30, 35, 40, 45, …

The numbers that appear in both lists are

15, 30, 45, 60, 75, 90, 105, 120, …

These are exactly the numbers in the 15 table. So the blank is filled as:
These numbers are called common multiples of 3 and 5.

Page 1 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Why they are the multiples of 15: a number that is a multiple of 3 and of 5 must
contain both a 3 and a 5 as factors, so it must be a multiple of 3 × 5 = 15. That is why
every 15th number in the game is an ‘idli-vada’ number.

Tip: to find the common multiples of any two numbers quickly, write the table of the
bigger one and tick the numbers that the smaller one also divides.

Figure it Out — Page 108
Section 5.1 Common Multiples and Common Factors

Q1 At what number is ‘idli-vada’ said for the 10th time?

‘Idli-vada’ is said at the common multiples of 3 and 5, that is, at the multiples of 15.

1st time → 15 2nd → 30 3rd → 45 4th → 60 5th → 75

6th → 90 7th → 105 8th → 120 9th → 135 10th → 150

You do not have to count them one by one:

10th ‘idli-vada’ = 15 × 10 = 150

Answer: at the number 150.

Q2 If the game is played for the numbers 1 to 90, find out: a. How many times would
the children say ‘idli’ (including the times they say ‘idli-vada’)? b. How many times
would the children say ‘vada’ (including the times they say ‘idli-vada’)? c. How many
times would the children say ‘idli-vada’?

Just count how many multiples of 3, of 5 and of 15 there are up to 90.

Page 2 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

a. ‘idli’ → multiples of 3 up to 90 = 90 ÷ 3 = 30 times

b. ‘vada’ → multiples of 5 up to 90 = 90 ÷ 5 = 18 times

c. ‘idli-vada’ → multiples of 15 up to 90 = 90 ÷ 15 = 6 times

The 6 ‘idli-vada’ numbers are 15, 30, 45, 60, 75 and 90.

Why the divisions work: the multiples of 3 up to 90 are 3 × 1, 3 × 2, …, 3 × 30 —
there are as many of them as there are counting numbers up to 30. So the count is
simply 90 ÷ 3.

Careful: the question says including the ‘idli-vada’ turns. If you wanted the number
of turns where only ‘idli’ is said, it would be 30 − 6 = 24, and only ‘vada’ would be 18 −
6 = 12.

Q3 What if the game was played till 900? How would your answers change?

900 is exactly 10 times 90, so every count becomes 10 times bigger.

WHAT IS SAID NUMBERS UP TO 90 UP TO 900

‘idli’ multiples of 3 30 900 ÷ 3 = 300

‘vada’ multiples of 5 18 900 ÷ 5 = 180

‘idli-vada’ multiples of 15 6 900 ÷ 15 = 60

The 10th ‘idli-vada’ is still at 150, but now the game goes on till the 60th one, at 900.

Check it yourself: only-‘idli’ turns = 300 − 60 = 240, and only-‘vada’ turns = 180 − 60 =
120.

Page 3 of 70

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Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
Is this figure somehow related to the ‘idli-vada’ game? Hint: Imagine playing the
e
Q4

m l as
.co
game till 30. Draw the figure if the game is played till 60.

a g
se m
g l a
a
m
.co
Multiples of 3 Multiples of 5
ag
asem
agl
18
21 10 5
co m
m.
30

as e
com
3 24
. a g l
m
ase
25

agl
9 20
15
12
s
27
m a
m .co agl
l a se
a g Common multiples
of 3 and 5

co m
m .
e
Fig. 5.1 (page 108) — the numbers up to 30 sorted into “multiples of 3”, “multiples of 5”

m l as
.co
and the overlap.

a g
se m
g l a
a
se m

com g l a
.
Yes — Fig. 5.1 is a picture of the game played till 30.
m a
ase
agl
The left circle holds the multiples of 3 — the numbers for which the players say ‘idli’ (3, 9,
12, 18, 21, 24, 27).
The right circle holds the multiples of 5 — the numbers for which they say ‘vada’ (5, 10, 20,

c o m
.
25).

s e m where the players
The overlapping middle part holds 15 and 30 — the common multiples,

. om say ‘idli-vada’.
cmust a gla
asem
l
Playing till 60 gives the same picture with more numbers in it:
a g
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 4 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Multiples of 3 (idli) Multiples of 5 (vada)

3 6 9 12 5 10 20
15 30
18 21 24 27
25 35 40
33 36 39 42
45 60
50 55
48 51 54 57

common multiples

The ‘idli-vada’ game played up to 60. Left only → ‘idli’, right only → ‘vada’, overlap → ‘idli-vada’.

‘idli’ only (16 numbers): 3, 6, 9, 12, 18, 21, 24, 27, 33, 36, 39, 42, 48, 51, 54, 57

‘vada’ only (8 numbers): 5, 10, 20, 25, 35, 40, 50, 55

‘idli-vada’ (4 numbers): 15, 30, 45, 60

Check the counting: multiples of 3 up to 60 = 20, multiples of 5 = 12, common
multiples = 4. So 20 − 4 = 16 numbers are ‘idli’ only and 12 − 4 = 8 are ‘vada’ only. ✔

In-text Questions — Pages 108 & 109

Page 5 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Playing with other number pairs

MATH TALK

Q1 Let us now play the ‘idli-vada’ game with different pairs of numbers: a. 2 and 5, b. 3
and 7, c. 4 and 6. We will say ‘idli’ for multiples of the smaller number, ‘vada’ for
multiples of the larger number and ‘idli-vada’ for common multiples. Draw a figure
similar to Fig. 5.1 if the game is played up to 60.

Multiples of 3 Multiples of 5

18
21 10 5
30
3 24

25
9 20
15
12
27

Common multiples
of 3 and 5

Fig. 5.1 (page 108) — the numbers up to 30 sorted into “multiples of 3”, “multiples of 5”
and the overlap.

For each pair, first find the first common multiple. Then every next common multiple is obtained
by adding that number again.

Page 6 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

PAIR ‘IDLI’ (SMALLER ‘VADA’ (LARGER ‘IDLI-VADA’ UP TO
NUMBER) NUMBER) 60

a. 2 and multiples of 2 multiples of 5 10, 20, 30, 40, 50, 60
5

b. 3 and multiples of 3 multiples of 7 21, 42
7

c. 4 and 6 multiples of 4 multiples of 6 12, 24, 36, 48, 60

Now the three figures. In each one, write the ‘idli’-only numbers in the left circle, the ‘vada’-only
numbers in the right circle and the common multiples in the overlap.

a. 2 and 5 — first common multiple 10

‘idli’ only: 2, 4, 6, 8, 12, 14, 16, 18, 22, 24, 26, 28, 32, 34, 36, 38, 42, 44, 46, 48, 52, 54, 56,

58 (24 numbers)

‘vada’ only: 5, 15, 25, 35, 45, 55 (6 numbers)

‘idli-vada’: 10, 20, 30, 40, 50, 60 (6 numbers)

b. 3 and 7 — first common multiple 21

‘idli’ only: 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60 (18 numbers)

‘vada’ only: 7, 14, 28, 35, 49, 56 (6 numbers)

‘idli-vada’: 21, 42 (2 numbers)

c. 4 and 6 — first common multiple 12

‘idli’ only: 4, 8, 16, 20, 28, 32, 40, 44, 52, 56 (10 numbers)

‘vada’ only: 6, 18, 30, 42, 54 (5 numbers)

‘idli-vada’: 12, 24, 36, 48, 60 (5 numbers)

Something worth noticing: for 2 and 5 the first ‘idli-vada’ is 10 = 2 × 5, and for 3 and
7 it is 21 = 3 × 7 — the product itself. But for 4 and 6 it is only 12, which is less than 4
× 6 = 24. The reason is that 4 and 6 share the common factor 2. You will meet this
idea again on page 116.

Page 7 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q2 Yesterday, we played this game with two numbers. We ended up saying just ‘idli’ or
‘idli-vada’ and nobody said just ‘vada’! One of the numbers was 4. Which of the
following could be the other number: 2, 3, 5, 8, 10?

Answer: 8.
Nobody said just ‘vada’. That means every multiple of the other number was also a multiple of 4
— so it always turned into an ‘idli-vada’. This happens only when the other number is itself a
multiple of 4.

OTHER NUMBER A ‘VADA’ NUMBER THAT IS NOT A MULTIPLE OF 4 WORKS?

2 2, 6, 10, … are multiples of 2 but not of 4 No

3 3, 6, 9, … are not multiples of 4 No

5 5, 10, 15, … are not multiples of 4 No

8 8, 16, 24, 32, … every one is a multiple of 4 Yes ✔
10 10, 30, 50, … are not multiples of 4 No

With 4 and 8: 4 → idli, 8 → idli-vada, 12 → idli, 16 → idli-vada, 20 → idli, 24 → idli-

vada, …

Why: ‘vada’ is said at multiples of 8, and every multiple of 8 (8, 16, 24, …) is also a
multiple of 4. So a plain ‘vada’ never gets a chance — it is always upgraded to ‘idli-
vada’.

In-text Questions — Page 110

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Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

Jump Jackpot — common factors
co m
e m.
m as
MATH TALK

.co a g l
a s em jump size can reach both 15 and 30? There are multiple jump sizes possible.
gl
What
a
Q1
Try to find them all.

com
m . ag
l a se
g To reach both treasures, the jump size must be a
Jumpy starts at 0 and lands only on multiples of his jump size. So a jump size works for a
number only if it is a factor of that a
number.
common factor of 15 and 30.

co m
em.
m l as
.co g
Factors of 15 → 1, 3, 5, 15
m a
l a se
Factors of 30 → 1, 2, 3, 5, 6, 10, 15, 30

a g
s
The numbers in both lists are
m a
m .co agl
l a se
g
1, 3, 5 and 15
a
m
Answer: jump sizes 1, 3, 5 and 15 — four in all.
. co
e m
m l as
.co a g
Jump 3 → 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 ✔

s m 5 → 5, 10, 15, 20, 25, 30 ✔
eJump
gl a
a
m
Jump 15 → 15, 30 ✔

a se
. com a g l
m
ase
agl
Why every factor of 15 works here: 15 is itself a factor of 30, so anything that
divides 15 also divides 30. That is why all four factors of 15 make it.

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

co m
m .
m ase
.co


a g l Page 9 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q2 Look at the table below. What do you notice? In the table, 1. Is there anything
common among the shaded numbers?

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70

The table on page 110 — some numbers are shaded (green) and some are circled.

The shaded numbers in the table (31 to 70) are

33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69

Yes — every one of them is in the 3 table.

33 = 3 × 11 36 = 3 × 12 39 = 3 × 13 42 = 3 × 14 45 = 3 × 15

48 = 3 × 16 51 = 3 × 17 54 = 3 × 18 57 = 3 × 19 60 = 3 × 20

63 = 3 × 21 66 = 3 × 22 69 = 3 × 23

All the shaded numbers are multiples of 3. They also step up by 3 each time — 33, 36, 39, … —
so they sit in a slanting pattern in the table.

Tip: a number is a multiple of 3 exactly when the sum of its digits is a multiple of 3.
Check 57 → 5 + 7 = 12, and 12 is in the 3 table. ✔

Q3 2. Is there anything common among the circled numbers?

The circled numbers are

Page 10 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

32, 36, 40, 44, 48, 52, 56, 60, 64, 68

Each one goes up by 4 from the previous one, and each is in the 4 table.

32 = 4 × 8 36 = 4 × 9 40 = 4 × 10 44 = 4 × 11 48 = 4 × 12

52 = 4 × 13 56 = 4 × 14 60 = 4 × 15 64 = 4 × 16 68 = 4 × 17

All the circled numbers are multiples of 4.

Tip: a number is a multiple of 4 exactly when the number made by its last two digits
is a multiple of 4. You will prove this yourself on page 124.

Q4 3. Which numbers are both shaded and circled? What are these numbers called?

Only three numbers wear both marks:

36, 48 and 60

They are common multiples of 3 and 4.

36 = 3 × 12 = 4 × 9 48 = 3 × 16 = 4 × 12 60 = 3 × 20 = 4 × 15

Why they are the multiples of 12: a number that is a multiple of 3 and also of 4
must be a multiple of 3 × 4 = 12 (because 3 and 4 have no common factor other than
1). The multiples of 12 lying between 31 and 70 are 36, 48 and 60 — exactly the three
we found. ✔

Try This: if the table went on to 100, the next number that is both shaded and circled
would be 72, then 84, then 96.

Figure it Out — Pages 110 & 111

Page 11 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Section 5.1 Common Multiples and Common Factors

MATH TALK TRY THIS

Q1 Find all multiples of 40 that lie between 310 and 410.

Write the 40 table and pick the numbers that fall in between.

40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440, …

Between 310 and 410 we get

320, 360 and 400

A quicker way: 310 ÷ 40 = 7 remainder 30, so the first multiple after 310 is 40 × 8 =
320. Then keep adding 40: 320, 360, 400. The next one, 440, is already past 410.

Q2 Who am I? a. I am a number less than 40. One of my factors is 7. The sum of my
digits is 8. b. I am a number less than 100. Two of my factors are 3 and 5. One of my
digits is 1 more than the other.

a. The number is 35.
“One of my factors is 7” means the number is a multiple of 7. The multiples of 7 below 40 are:

MULTIPLE OF 7 7 14 21 28 35

Sum of digits 7 1+4=5 2+1=3 2 + 8 = 10 3+5=8 ✔

b. The number is 45.
Having both 3 and 5 as factors means the number is a multiple of 15. Below 100 these are:

MULTIPLE OF 15 15 30 45 60 75 90

Digits 1, 5 3, 0 4, 5 ✔ 6, 0 7, 5 9, 0

Differ by 1? No No Yes No No No

Page 12 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

4 and 5 → 5 = 4 + 1 ✔ and 45 = 3 × 15 = 5 × 9 ✔

Q3 A number for which the sum of all its factors is equal to twice the number is called a
perfect number. The number 28 is a perfect number. Its factors are 1, 2, 4, 7, 14 and
28. Their sum is 56 which is twice 28. Find a perfect number between 1 and 10.

The perfect number is 6.

Factors of 6 → 1, 2, 3, 6

1 + 2 + 3 + 6 = 12 = 2 × 6 ✔

Let us check that no other number from 1 to 10 works:

NUMBER FACTORS SUM OF FACTORS TWICE THE NUMBER PERFECT?

4 1, 2, 4 7 8 No

5 1, 5 6 10 No

6 1, 2, 3, 6 12 12 Yes ✔
8 1, 2, 4, 8 15 16 No

9 1, 3, 9 13 18 No

10 1, 2, 5, 10 18 20 No

Did you know? Perfect numbers are very rare. The first four are 6, 28, 496 and 8128.
No one has ever found an odd perfect number, and no one has been able to prove
that none exists!

Q4 Find the common factors of: a. 20 and 28 b. 35 and 50 c. 4, 8 and 12 d. 5, 15 and 25

List all the factors of each number and pick out the ones that appear everywhere.

Page 13 of 70

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Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
PART FACTORS COMMON FACTORS

m as e
a. 20 and 28
.co
20 → 1, 2, 4, 5, 10, 20 1, 2, 4
a g l
m
ase
28 → 1, 2, 4, 7, 14, 28

ab.g35l and 50 35 → 1, 5, 7, 35 1, 5
50 → 1, 2, 5, 10, 25, 50

co m
m . ag
se
c. 4, 8 and 12 4 → 1, 2, 4 1, 2, 4

12 → 1, 2,g l
8 → 1, 2, 4, 8
a
a 3, 4, 6, 12
d. 5, 15 and 25 5 → 1, 5 1, 5
co m
e m.
15 → 1, 3, 5, 15

m l as
.co
25 → 1, 5, 25

a g
a s em
a gl Always remember: 1 divides every number, so 1 is a common factor of any set of
numbers. That is why 1 appears in all four answers.

m a s
em
.co agl
a s
l are multiples of 25 but not multiples of 50.
Q5 agthat
Find any three numbers

co m
m .
m as e
.co
Write the 25 table and cross out the numbers that also appear in the 50 table.
a g l
a s em
agl 25 table → 25, 50, 75, 100, 125, 150, 175, 200, …
se m
com g l a
. a
Three such numbers: 25, 75 and 125. (175, 225, 275 … also work.)
m
ase
agl
Why the alternate ones survive: 50 = 25 × 2, so a multiple of 25 is also a multiple of
50 only when it is an even number of 25s. So take the odd multiples of 25 — 25 × 1,

co m
.
25 × 3, 25 × 5, … — and none of them will be a multiple of 50.

em
m l as
.co a g
a s em
agl Q6 Anshu and his friends play the ‘idli-vada’ game with two numbers, which are both
smaller than 10. The first time anybody says ‘idli-vada’ is after the number 50. What
.c
s e m
m a
could the two numbers be which are assigned ‘idli’ and ‘vada’?

e m . co agl
g l as
a
The first ‘idli-vada’ happens at the first common multiple of the two numbers. So we need two

m
numbers below 10 whose first common multiple is more than 50.

. co
em
m l as
.co a g
Page 14 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

PAIR FIRST COMMON MULTIPLE MORE THAN 50?

6 and 7 42 No

5 and 9 45 No

6 and 9 18 No

7 and 8 56 Yes ✔
7 and 9 63 Yes ✔
8 and 9 72 Yes ✔

Answer: the two numbers could be 7 and 8, or 7 and 9, or 8 and 9.

Why only these three: both numbers are below 10, so their product is at most 9 × 8
= 72. To push the first common multiple past 50, the two numbers must be large and
must not share a factor — otherwise the first common multiple drops well below the
product. 7, 8 and 9 are pairwise co-prime, so their first common multiples are the
products 56, 63 and 72.

Note: the answers printed at the back of the book give only 7, 8 and 8, 9 — the pair
7 and 9 works equally well (63 > 50).

Q7 In the treasure hunting game, Grumpy has kept treasures on 28 and 70. What jump
sizes will land on both the numbers?

Jumpy needs a common factor of 28 and 70.

Factors of 28 → 1, 2, 4, 7, 14, 28

Factors of 70 → 1, 2, 5, 7, 10, 14, 35, 70

The numbers in both lists are

1, 2, 7 and 14

Page 15 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Jump 7 → 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 ✔

Jump 14 → 14, 28, 42, 56, 70 ✔

Answer: jump sizes 1, 2, 7 and 14.

Shortcut: 28 = 2 × 2 × 7 and 70 = 2 × 5 × 7. The primes shared by both are 2 and 7, so
the common factors are 1, 2, 7 and 2 × 7 = 14.

Q8 In the diagram below, Guna has erased all the numbers except the common
multiples. Find out what those numbers could be and fill in the missing numbers in
the empty regions.

Multiples of ____ Multiples of ____

72

48

24

Common multiples

The diagram from page 111 — only the common multiples are left; every other region is
empty.

The overlap shows 24, 48 and 72. These go up by 24 each time, so the first common multiple of
the two numbers is 24.
So we need a pair whose first common multiple is 24. A neat choice is 8 and 12:

Page 16 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Multiples of 8 Multiples of 12

8 16 32 40 12 36 60
24 48

72 96
84
56 64 80 88

common multiples

Multiples of 8 on the left, multiples of 12 on the right; the common multiples 24, 48, 72, 96 sit in the
middle.

Multiples of 8 only → 8, 16, 32, 40, 56, 64, 80, 88 …

Multiples of 12 only → 12, 36, 60, 84 …

Common multiples → 24, 48, 72, 96 …

Other pairs also work, because they too have 24 as their first common multiple:

3 and 8 — left: 3, 6, 9, 12, 15, 18, 21, 27 …; right: 8, 16, 32, 40 …
6 and 8 — left: 6, 12, 18, 30, 36, 42 …; right: 8, 16, 32, 40 …
4 and 24, 2 and 24, 12 and 24 — here every multiple of 24 sits in the overlap.

How to be sure: the numbers left in the middle are 24, 48, 72 — the multiples of 24.
So whichever pair Guna chose, their first common multiple has to be exactly 24, no
smaller and no bigger.

Page 17 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q9 Find the smallest number that is a multiple of all the numbers from 1 to 10, except
for 7.

We need the smallest number divisible by 1, 2, 3, 4, 5, 6, 8, 9 and 10. Take the biggest power of
each prime that is needed.

To cover 8 we need 2 × 2 × 2

To cover 9 we need 3 × 3

To cover 5 (and 10) we need 5

Smallest number = 2 × 2 × 2 × 3 × 3 × 5 = 8 × 9 × 5 = 360

Check every number:

DIVIDE 360 BY 1 2 3 4 5 6 8 9 10

Answer 360 180 120 90 72 60 45 40 36

Every division is exact, so 360 is the answer.

Q10 Find the smallest number that is a multiple of all the numbers from 1 to 10.

This is the previous answer with the missing 7 put back.

Smallest number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 360 × 7 = 2520

Check:

DIVIDE 2520 BY 1 2 3 4 5 6 7 8 9 10

Answer 2520 1260 840 630 504 420 360 315 280 252

Page 18 of 70

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Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
Why we multiply only by 7: 360 already contains 2 three times, 3 twice and 5 once

m l a se
— enough for 1, 2, 3, 4, 5, 6, 8, 9 and 10. Only the prime 7 was missing, so
o g the answer
.c by 7 once is enough. Nothing smaller can work, because
a
m
multiplying

l a secontain 8 = 2 × 2 × 2, 9 = 3 × 3, 5 and 7.
ag
must

o m
c ag
Did you know? 2520 is the smallest number divisible by every number from 1 to 10.
.
mnumber from 1 to 20 is 232792560!
s e
The smallest number divisible by every
a
agl

co m
m.
In-text Questions — Pages 112 to 114
m as e
.co l
Section 5.2 Prime Numbers

a g
a s em
a glQ1 Guna wants to put 12 figs in each box and Anshu wants to put 7 figs in each box.
How many arrangements are possible? Think and find out the different ways how —

m a s
agl
1. Guna can arrange 12 figs in a rectangular manner. 2. Anshu can arrange 7 figs in a
rectangular manner.
m.co
l a se
ANSWER a g
m
A rectangular arrangement of rows × columns is possible only when rows × columns gives the

. co
m
number of figs. So we simply need the pairs of factors.

m as e
.co
1. Guna's 12 figs — 12 = 1 × 12 = 2 × 6 = 3 × 4, and each can be turned around:
a g l
se m
g l a
a 12 figs — six rectangular arrangements

se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co
12 × 1 6×2 4×3 3×4 2×6
a g l 1 × 12

se m Six rectangular arrangements of 12 figs. If you count a rectangle and its turned-around twin as the
g l a
a c
.
same, there are three different shapes.

s e m
com a
12 × 1, 6 × 2, 4 × 3, 3 × 4,em
. agl
a s 2 × 6, 1 × 12 → 6 arrangements
agl
2. Anshu's 7 figs — 7 can only be split as 1 × 7:

co m
m .
m ase
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Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

7 figs — only one rectangle
(turned two ways)

7×1 1×7

7 figs can be laid out in only one way — a single line, standing up or lying down.

7 × 1 or 1 × 7 → only one shape

Why the difference: 12 has six factors (1, 2, 3, 4, 6, 12) so it gives six arrangements,
but 7 has only two factors (1 and 7) so it gives just one line. This single fact is what
makes 7 a prime and 12 a composite number.

Q2 Observe the number of rows and columns in each of the arrangements. How are
they related to 12?

In every arrangement, rows × columns = 12. So the number of rows and the number of columns
are always factors of 12.

Page 20 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

ROWS 12 6 4 3 2 1

Columns 1 2 3 4 6 12

Rows × columns 12 12 12 12 12 12

Reading the top row we get 12, 6, 4, 3, 2, 1 — exactly the list of factors of 12.

Tip: this gives a lovely way to find all factors of a number — count how many
different rectangles you can build with that many counters or tamarind seeds.

Q3 How many prime numbers are there from 21 to 30? How many composite numbers
are there from 21 to 30?

Test each of the ten numbers from 21 to 30.

NUMBER 21 22 23 24 25 26 27 28 29 30

A factor other than 1 and itself 3 2 — 2 5 2 3 2 — 2

Prime / composite C C P C C C C C P C

Primes from 21 to 30 → 23 and 29 — that is 2 primes

Composites from 21 to 30 → 21, 22, 24, 25, 26, 27, 28, 30 → 8 composite numbers

2 + 8 = 10, and there are exactly 10 numbers from 21 to 30. ✔

Q4 Can we list all the prime numbers from 1 to 100? Here is an interesting way to find
prime numbers. Just follow the steps given and see what happens.

Yes. Write 1 to 100 in a grid and follow the five steps — cross out 1, then circle 2 and cross out
its later multiples, circle 3 and cross out its later multiples, then 5, then 7. After that everything
left is already circled.

Page 21 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

The Sieve of Eratosthenes — 25 primes below 100

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70

71 72 73 74 75 76 77 78 79 80

81 82 83 84 85 86 87 88 89 90

91 92 93 94 95 96 97 98 99 100

Circled = prime · crossed out = composite (1 is neither)

The Sieve of Eratosthenes. Only the circled numbers survive — those are the primes.

The 25 prime numbers below 100 are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29,

31, 37, 41, 43, 47, 53, 59, 61, 67, 71,

73, 79, 83, 89, 97

Tip: you can stop crossing out after 7. Any composite number below 100 must have
a factor of 10 or less (because 11 × 11 = 121 is already more than 100), so it is caught
by 2, 3, 5 or 7.

Page 22 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q5 Guna and Anshu started wondering how this simple method is able to find prime
numbers! Think how this method works. Read the steps given above again and see
what happens after each step is carried out.

The sieve works because of one simple idea: a composite number always has a smaller prime as
a factor, so it is bound to get crossed out at some step.
Follow what happens to a number, say 91:

91 is odd → survives step 2 (multiples of 2)

9 + 1 = 10, not a multiple of 3 → survives step 3

does not end in 0 or 5 → survives step 4

but 91 = 7 × 13 → crossed out when the multiples of 7 go

And a prime such as 37:

37 is not a multiple of 2, 3 or 5, and 6 × 6 = 36 < 37 < 49 = 7 × 7

so no number from 2 to 6 divides it → 37 is never crossed out → it stays circled

The reason in one line: each step removes the numbers that have that particular
prime as a factor. Since every composite number has at least one prime factor, every
composite gets removed. The numbers that stay behind have no prime factor
smaller than themselves — and that is exactly what being prime means.

Did you know? Eratosthenes lived about 2200 years ago in Greece. He also
measured the size of the Earth using shadows — and his answer was surprisingly
close to the correct value.

Figure it Out — Pages 114 & 115

Page 23 of 70

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Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

Section 5.2 Prime Numbers
co m
e m.
m l as
Q1
.co a g
We see that 2 is a prime and also an even number. Is there any other even prime?
m
l a se
ag

m
No. 2 is the only even prime number.

. co ag
e m
g l as
a
Take any even number bigger than 2, say 34.

34 = 2 × 17 → its factors are 1, 2, 17, 34 — that is four factors

co m
em.
m l as
.co g
Why it must be so: every even number has 2 as a factor. If the number is bigger

em a
s
than 2, then it has at least three factors — 1, 2 and the number itself. A prime is
a
gl allowed only two factors, so no even number above 2 can be prime. The number 2
a escapes because for 2 the “extra” factor 2 is the number itself.

m a s
m .co agl
a se
Fun name: mathematicians call 2 “the oddest prime” — because it is the only even
l
one! a g

co m
m .
as e
Q2
om
.csuccessive g l
Look at the list of primes till 100. What is the smallest difference between two
a
se m primes? What is the largest difference?

g l a
a
m

a se
com g l
Write the primes and the gaps between them:

m . a
ase
agl
2 1 3 2 5 2 7 4 11 2 13 4 17 2 19 4 23 6 29 2 31 6 37 4 41 2 43 4 47 6 53

53 6 59 2 61 6 67 4 71 2 73 6 79 4 83 6 89 8 97
co m
m .
as e
com difference = 8, between 89 and 97. g l
Smallest difference = 1, between 2 and 3.
. a
sem
Largest
a
agl c
m .
e
3−2=1 97 − 89 = 8
m a s
e m . co agl
g l as
a

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m .
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.co


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Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Why 1 happens only once: two numbers that differ by 1 are one odd and one even.
The only even prime is 2, so the only such pair can be 2 and 3. After that, all primes
are odd, so every gap is an even number — 2, 4, 6 or 8.

Check it yourself: the numbers 90 to 96 in between 89 and 97 are all composite —
that is the long stretch that creates the biggest gap.

Page 25 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q3 Are there an equal number of primes occurring in every row in the table on the
previous page? Which decades have the least number of primes? Which have the
most number of primes?

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

41 42 43 44 45 46 47 48 49 50

51 52 53 54 55 56 57 58 59 60

61 62 63 64 65 66 67 68 69 70

71 72 73 74 75 76 77 78 79 80

81 82 83 84 85 86 87 88 89 90

91 92 93 94 95 96 97 98 99 100

The table of 1 to 100 on page 113 — the primes are circled, everything else is crossed
out.

No, the primes are not spread out equally. Counting them row by row:

Page 26 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

DECADE PRIMES IN IT HOW MANY

1 – 10 2, 3, 5, 7 4

11 – 20 11, 13, 17, 19 4

21 – 30 23, 29 2

31 – 40 31, 37 2

41 – 50 41, 43, 47 3

51 – 60 53, 59 2

61 – 70 61, 67 2

71 – 80 71, 73, 79 3

81 – 90 83, 89 2

91 – 100 97 1

4 + 4 + 2 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 25 primes ✔

Least: the decade 91 – 100, with only one prime (97).
Most: the decades 1 – 10 and 11 – 20, with four primes each.

Why primes thin out: as numbers grow bigger there are more and more smaller
primes that could divide them, so it gets harder for a number to escape being
composite. Primes never stop, but they do spread further apart.

Q4 Which of the following numbers are prime: 23, 51, 37, 26?

For each number, look for a factor other than 1 and itself.

Page 27 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

NUMBER CHECK PRIME OR
COMPOSITE?

23 not even; 2 + 3 = 5 (not a multiple of 3); does not end in 0 or 5; 4 × 4 = Prime ✔
16 < 23 < 25

51 5 + 1 = 6, a multiple of 3 → 51 = 3 × 17 Composite

37 not even; 3 + 7 = 10 (not a multiple of 3); does not end in 0 or 5; not a Prime ✔
multiple of 7 either

26 even → 26 = 2 × 13 Composite

Answer: 23 and 37 are prime. 51 and 26 are composite.

Careful: 51 looks like a prime at first glance, but the digit-sum test catches it at once:
5 + 1 = 6, so 51 is in the 3 table.

Q5 Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.

The primes below 20 are 2, 3, 5, 7, 11, 13, 17, 19. Pick pairs whose sum ends in 0 or 5.

2+3=5=5×1✔

3 + 7 = 10 = 5 × 2 ✔

2 + 13 = 15 = 5 × 3 ✔

Three such pairs: (2, 3), (3, 7) and (2, 13).
There are more — try these too:

3 + 17 = 20 ✔ 7 + 13 = 20 ✔ 11 + 19 = 30 ✔ 13 + 17 = 30 ✔

Try This: notice that every pair above except (2, 3) and (2, 13) is a pair of two odd
primes, and their sum is always even. So an odd sum like 5 or 15 can only come from
a pair that includes the even prime 2.

Page 28 of 70

as e
Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1
e
Q6

m l as
.co
and 3. Find such pairs of prime numbers up to 100.

a g
se m
g l a
a

Take each two-digit prime, reverse its digits, and check whether the reversed number is also

m
prime.
. co ag
e m
PRIME REVERSED
g l as IS THE REVERSE PRIME?

13 31
a Yes ✔

co m
m.
17 71 Yes ✔
m as e
.co ✔
a g l
m
37 73 Yes

l a se
a g 79 97 Yes ✔

m a s
19 91 = 7 × 13 No

m .co No agl
se
23 32 (even)

g l a
29 92 (even) a No

co m
.
59 95 = 5 × 19 No

em
m l as
.co g
83 38 (even) No

em the pairs are (13, 31), (17, 71), (37, 73) and (79, 97). a
a s
gl
Answer:
a
se m
comso it is a palindromic prime rather than an a
Did you know? Such primes are called emirps — “prime” written backwards! The
. a g l
em
number 11 reads the same both ways,

a s
agl
emirp.

co m
m .
as e
Q7 Find seven consecutive composite numbers between 1 and 100.
m l
.co a g
a s emANSWER
agl Look for the biggest gap between two primes below 100. It comes between 89 and 97, and all
.c
the seven numbers in between are composite.
s e m
m a
em
. c o agl
NUMBER 90
l a s
91 92 93 94 95 96

A factor 2 × 45 ag 7 × 13 2 × 46 3 × 31 2 × 47 5 × 19 2 × 48

co m
m .
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.co


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Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Answer: 90, 91, 92, 93, 94, 95, 96.

Why 91 is the sneaky one: 90, 92, 94, 96 are even and 93, 95 are obviously in the 3
and 5 tables. Only 91 hides its factors — but 91 = 7 × 13, so it is composite too. That
is what makes this a run of seven composites in a row, the longest one below 100.

Q8 Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are
twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.

Run down the list of primes and note every pair that differs by 2.

(3, 5) (5, 7) (11, 13) (17, 19) (29, 31) (41, 43) (59, 61) (71, 73)

The book already gives (3, 5) and (17, 19), so the other twin primes are

(5, 7), (11, 13), (29, 31), (41, 43), (59, 61) and (71, 73)

There are 8 twin-prime pairs below 100 in all.

Did you know? Nobody knows whether twin primes go on for ever. This is one of the
most famous unsolved problems in mathematics, called the Twin Prime
Conjecture.

Page 30 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q9 Identify whether each statement is true or false. Explain. a. There is no prime
number whose units digit is 4. b. A product of primes can also be prime. c. Prime
numbers do not have any factors. d. All even numbers are composite numbers. e. 2
is a prime and so is the next number, 3. For every other prime, the next number is
composite.

STATEMENT TRUE / REASON
FALSE

a. There is no prime number True A number ending in 4 is even, so 2 is a factor of it. It is
whose units digit is 4. also bigger than 2, so it has at least three factors — 1, 2
and itself. Example: 14 = 2 × 7, 24 = 2 × 12.

b. A product of primes can False Multiply two or more primes and the answer picks up
also be prime. those primes as extra factors. 3 × 5 = 15, whose factors
are 1, 3, 5, 15 — four factors, so 15 is composite.

c. Prime numbers do not False Every prime has exactly two factors — 1 and itself. Factors
have any factors. of 13 are 1 and 13.

d. All even numbers are False 2 is even but prime. Every other even number is indeed
composite numbers. composite.

e. For every prime other True Every prime except 2 is odd, so the number just after it is
than 2, the next number is even and bigger than 2 — hence composite. 3 → 4, 5 → 6,
composite. 7 → 8, 11 → 12, 97 → 98.

The common thread: statements (a), (d) and (e) all rest on the single fact that 2 is
the only even prime. Once you hold on to that, all three become easy.

Q10 Which of the following numbers is the product of exactly three distinct prime
numbers: 45, 60, 91, 105, 330?

Find the prime factorisation of each and count the different primes in it.

Page 31 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

NUMBER PRIME HOW MANY PRIMES ARE EXACTLY THREE
FACTORISATION MULTIPLIED DISTINCT PRIMES?

45 3×3×5 3 primes, but only 2 different No
ones

60 2×2×3×5 4 primes, 3 different No

91 7 × 13 2 primes No

105 3×5×7 3 primes, all different Yes ✔
330 2 × 3 × 5 × 11 4 primes, all different No

3 × 5 × 7 = 15 × 7 = 105 ✔

Answer: 105.

Read the question carefully: 45 is a product of three primes but two of them are
the same, and 330 has three different primes plus one more. Only 105 is a product of
exactly three distinct primes.

Q11 How many three-digit prime numbers can you make using each of 2, 4 and 5 once?

Answer: none — zero such primes.
The digits 2, 4 and 5 can be arranged in six ways:

NUMBER 245 254 425 452 524 542

Last digit 5 4 5 2 4 2

Divisible by 5 2 5 2 2 2

245 = 5 × 49 254 = 2 × 127 425 = 5 × 85

452 = 2 × 226 524 = 2 × 262 542 = 2 × 271

Page 32 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Why not even one: whichever way you arrange them, the number must end in 2, 4
or 5. Ending in 2 or 4 makes it even (so 2 is a factor); ending in 5 makes it a multiple
of 5. In every case the number has a factor other than 1 and itself, so it is composite.

Q12 Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other
primes for which doubling and adding 1 gives another prime? Find at least five
such examples.

Yes, there are many. Double the prime and add 1, then check the answer.

PRIME P 2×P+1 IS IT PRIME?

2 2×2+1=5 Yes ✔
3 2×3+1=7 Yes ✔
5 2 × 5 + 1 = 11 Yes ✔
11 2 × 11 + 1 = 23 Yes ✔
23 2 × 23 + 1 = 47 Yes ✔
29 2 × 29 + 1 = 59 Yes ✔
41 2 × 41 + 1 = 83 Yes ✔
7 2 × 7 + 1 = 15 = 3 × 5 No

13 2 × 13 + 1 = 27 = 3 × 9 No

17 2 × 17 + 1 = 35 = 5 × 7 No

Five examples: 2 → 5, 5 → 11, 11 → 23, 23 → 47 and 29 → 59. (41 → 83 is a sixth.)

Did you know? Primes like 2, 3, 5, 11, 23, 29 and 41 — where doubling and adding 1
gives another prime — are called Sophie Germain primes, after the French
mathematician Sophie Germain, who used them in her work on Fermat's Last
Theorem.

In-text Questions — Pages 115 to 117

Page 33 of 70

as e
Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

Section 5.3 Co-prime numbers for safekeeping treasures
co m
em.
m as
MATH TALK

.co a g l
a s em should Grumpy place the treasures so that Jumpy cannot reach both the
gl
Where
a
Q1
treasures? Check if these pairs are safe: a. 15 and 39 b. 4 and 15 c. 18 and 29 d. 20
and 55

co m
m . ag
l a se
Remember the new rule — a jumpa g of 1 is not allowed. So a pair is safe only if the two
size
numbers have no common factor other than 1.

co m
e m.
com as
PAIR FACTORS COMMON FACTORS SAFE?

. a g l
m
ase
a. 15 and 39 15 → 1, 3, 5, 15 1, 3 Not safe — jump 3 reaches 15 and 39

agl
39 → 1, 3, 13, 39

✔
s
b. 4 and 15
a
4 → 1, 2, 4 only 1 Safe
m
.co agl
15 → 1, 3, 5, 15

s m1
eonly
c. 18 and 29
gl a ✔
a
18 → 1, 2, 3, 6, 9, 18 Safe
29 → 1, 29

d. 20 and 55
co m
.
20 → 1, 2, 4, 5, 10, 20 1, 5 Not safe — jump 5 reaches 20 and 55

em
as
55 → 1, 5, 11, 55

. com(4, 15) and (18, 29). a g l
s em
Safe pairs:

gl a
a
m
Why 29 makes a pair safe so easily: 29 is a prime, so its only factors are 1 and 29.

a se
com l
Since 29 does not divide 18, the only common factor left is 1.
. a g
m
ase
agl

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

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m .
m ase
.co


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Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q2 Which of the following pairs of numbers are co-prime? a. 18 and 35 b. 15 and 37 c. 30
and 415 d. 17 and 69 e. 81 and 18

PAIR PRIME FACTORS COMMON PRIME FACTOR? CO-PRIME?

a. 18 and 35 18 = 2 × 3 × 3 none Yes ✔
35 = 5 × 7

b. 15 and 37 15 = 3 × 5 none Yes ✔
37 is prime

c. 30 and 415 30 = 2 × 3 × 5 5 No
415 = 5 × 83

d. 17 and 69 17 is prime none Yes ✔
69 = 3 × 23

e. 81 and 18 81 = 3 × 3 × 3 × 3 3 No
18 = 2 × 3 × 3

Co-prime pairs: a (18 and 35), b (15 and 37) and d (17 and 69).

Tip: both numbers in a co-prime pair need not be prime. 18 and 35 are both
composite, yet they are co-prime, because they are built from completely different
primes.

Page 35 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q3 While playing the ‘idli-vada’ game with different number pairs, Anshu observed
something interesting! 1. Sometimes the first common multiple was the same as
the product of the two numbers. 2. At other times the first common multiple was
less than the product of the two numbers. Find examples for each of the above.
How is it related to the number pair being co-prime?

NUMBER FIRST COMMON PRODUCT SAME OR CO-
PAIR MULTIPLE LESS? PRIME?

3 and 5 15 15 Same Yes

3 and 7 21 21 Same Yes

4 and 9 36 36 Same Yes

4 and 6 12 24 Less No (share 2)

3 and 6 6 18 Less No (share 3)

6 and 15 30 90 Less No (share 3)

The rule Anshu spotted:

If the two numbers are co-prime, the first common multiple is exactly their product.
If they share a common factor, the first common multiple is less than the product — in
fact it is the product divided by the biggest common factor.

4 and 6 share 2 → first common multiple = 24 ÷ 2 = 12

6 and 15 share 3 → first common multiple = 90 ÷ 3 = 30

3 and 6 share 3 → first common multiple = 18 ÷ 3 = 6

Why: the product always is a common multiple. But when the two numbers share a
factor, that factor gets counted twice in the product, so a smaller common multiple
already exists. Co-prime numbers have nothing to share, so nothing smaller than
the product can work.

Page 36 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q4 Co-prime art. Observe the following thread art. The first diagram has 12 pegs and
the thread is tied to every fourth peg (we say that the thread-gap is 4). The second
diagram has 13 pegs and the thread-gap is 3. What about the other diagrams?
Observe these pictures, share and discuss your findings in class.

12 13 16
11 1 12 1 15 1
14 2
11 2
10 2
13 3

10 3
9 3 12 4

9 4 11 5
8 4
8 5 10 6
7 5 9 7
6 7 6 8

23 24 1
22 2
21 3
20 4
19 5

18 6

17 7
16 8
15 9
14 10
13 12 11

The four thread-art pictures on page 116. In each one the pegs are numbered round the
circle and the thread runs from peg to peg by a fixed thread-gap.

Read each picture by counting the pegs on the circle and the pegs the thread actually touches.

The four thread pictures of the book
12 13 16 23 24 1
11 1 12 1 15 1 22 2
14 2 21 3
10 2 11 2 20 4
13 3
19 5
10 3
9 3 12 4 18 6
17 7
9 4 11 5
8 4 16 8
8 5 10 6 15 9
7 5 9 7 14 10
6 7 6 8 13 12 11

12 pegs, thread-gap 4 13 pegs, thread-gap 3 16 pegs, thread-gap 6 24 pegs, thread-gap 6
thread touches 3 pegs thread touches 13 pegs thread touches 8 pegs thread touches 4 pegs

The four thread pictures of the book, redrawn. Notice how many pegs the thread reaches in each
one.

Page 37 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

DIAGRAM PEGS THREAD- BIGGEST PEGS THE SHAPE
GAP COMMON THREAD YOU SEE
FACTOR TOUCHES

1st 12 4 4 12 ÷ 4 = 3 a triangle

2nd 13 3 1 13 ÷ 1 = 13 (all) a 13-pointed
star

3rd 16 6 2 16 ÷ 2 = 8 an 8-pointed
star

4th 24 6 6 24 ÷ 6 = 4 a square

1st: 12 → 4 → 8 → back to 12. Only pegs 4, 8, 12 are used.

4th: 24 → 6 → 12 → 18 → back to 24. Only pegs 6, 12, 18, 24 are used.

The rule: number of pegs touched = (number of pegs) ÷ (biggest common factor of
the two numbers). When that common factor is 1 — that is, when the numbers are
co-prime — every single peg is touched.

Q5 In some diagrams, the thread is tied to every peg. In some, it is not. Is it related to
the two numbers (the number of pegs and the thread-gap) being co-prime?

Yes, exactly. The thread reaches every peg if and only if the number of pegs and the thread-gap
are co-prime.

PEGS AND GAP CO-PRIME? DOES THE THREAD REACH EVERY PEG?

12 and 4 No (share 4) No — only 3 of the 12 pegs

13 and 3 Yes Yes — all 13 pegs

16 and 6 No (share 2) No — only 8 of the 16 pegs

24 and 6 No (share 6) No — only 4 of the 24 pegs

Page 38 of 70

as e
Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the thread lands on peg numbers that are multiples of the gap,

m l a se
counted round and round the circle. It comes back to the starting peg for the first
o
m
time at the .c first common multiple of the two numbers. If they aareg co-prime, that
a e
scommon
g l
first multiple is the full product — so the thread has to go all the way round
athe circle (gap) times, visiting every peg. If they share a factor, it returns early and
misses the rest.
co m
e m . ag
g l as
a
Try This: take 13 pegs with a gap of 5, or 11 pegs with a gap of 4 — both are co-
prime pairs, so both will give a star that touches every peg.

co m
se m.
o m l a
gof 10 b. 10 pegs, thread-
m .c such pictures for the following: a. 15 pegs, thread-gap
Make
a
se gap of 7 c. 14 pegs, thread-gap of 6 d. 8 pegs, thread-gap of 3
Q6

l a
ag
s
m a
First decide, for each pair, how many pegs the thread will touch. Then draw.
m .co agl
l a se
14
15
1 a g 10
13
14
1
8
9 1
13 2 7 1

m
12 2

. co
12 3 8 2

m
11 3

ase
6 2

m
11 4

co agl
10 4
7 3

m .
10 5

e
9 5

as
5 3
9 6 6 4

l
8 6
8 7

g
5 7 4

a 15 pegs, thread-gap 10 10 pegs, thread-gap 7 14 pegs, thread-gap 6 8 pegs, thread-gap 3

m
thread touches 3 pegs thread touches 10 pegs thread touches 7 pegs thread touches 8 pegs

a se
cosomevery peg is used. g l
The four pictures asked for. Only (a) and (c) miss some pegs — in (b) and (d) the numbers are co-

m . a
ase
prime,

a gl
PART PEGS GAP CO-PRIME? PEGS TOUCHED PICTURE

co m
a. 15 10
m .
sae10-pointed star
No (share 5) 15 ÷ 5 = 3 a triangle on pegs 10, 5, 15

m l a
.co ag
em
b. 10 7 Yes all 10

a s
agl c. 14 6 No (share 2) 14 ÷ 2 = 7 a 7-pointed star on the even pegs

.c
s e m
m a
d. 8 3 Yes all 8 an 8-pointed star

e m . co agl
g l as
a

co m
m .
m as e
.co


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Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

a. 15 → 10 → 5 → back to 15 (only three pegs)

b. 10 → 7 → 4 → 1 → 8 → 5 → 2 → 9 → 6 → 3 → back to 10 (all ten pegs)

c. 14 → 6 → 12 → 4 → 10 → 2 → 8 → back to 14 (seven pegs, all even)
d. 8 → 3 → 6 → 1 → 4 → 7 → 2 → 5 → back to 8 (all eight pegs)

Tip for drawing: mark the pegs evenly on a circle with a compass, number them,
then join peg to peg counting the gap each time. Stop when you come back to
where you started.

In-text Questions — Pages 117 to 120
Section 5.4 Prime Factorisation

Q1 Teacher: Are 56 and 63 co-prime? Anshu: I can write 56 = 14 × 4 and 63 = 21 × 3 …
there are no common factors. The numbers are co-prime. Guna: Hold on. I can also
write 56 = 7 × 8 and 63 = 9 × 7 … so they are not co-prime. Clearly Guna is right, as 7
is a common factor. But where did Anshu go wrong?

Anshu's mistake was to think that one way of splitting a number shows all its factors. It does
not.

56 = 14 × 4 → tells us 14 and 4 are factors of 56

But the full list is: 1, 2, 4, 7, 8, 14, 28, 56

63 = 21 × 3 → tells us 21 and 3 are factors of 63

But the full list is: 1, 3, 7, 9, 21, 63

The number 7 is a factor of both, but Anshu's two splittings simply did not show it. Guna's
splittings (56 = 7 × 8 and 63 = 9 × 7) happened to show it.

Page 40 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

The lesson: to be sure about common factors we must break both numbers all the
way down to primes:

56 = 2 × 2 × 2 × 7 and 63 = 3 × 3 × 7

Now 7 cannot hide anywhere. 56 and 63 are not co-prime.

Q2 Try another example: 80 and 63. If we take 80 = 16 × 5 and 63 = 9 × 7, then there are
no common factors. Can we conclude that 80 and 63 are co-prime?

Not from that alone. Anshu's mistake warns us — one lucky splitting proves nothing, because
there may be other ways to factorise that reveal a shared factor.

80 = 40 × 2 = 20 × 4 = 10 × 8 = 16 × 5 = ???

63 = 9 × 7 = 3 × 21 = ???

The safe way is to go down to the primes:

80 = 2 × 2 × 2 × 2 × 5

63 = 3 × 3 × 7

The primes of 80 are 2 and 5; the primes of 63 are 3 and 7. Nothing is shared, and now there is
nowhere left to hide.
So yes — 80 and 63 really are co-prime, but the prime factorisation is what proves it.

What if they had a composite common factor? Say some composite number c
divided both. Then every prime inside c would also divide both — so that prime
would show up in both prime factorisations. Since no prime is shared, no common
factor bigger than 1 can exist at all.

Page 41 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q3 Does the order matter? Using this diagram, can you explain why 30 = 2 × 3 × 5, no
matter which way you multiply 2, 3, and 5?

The block of unit cubes shown on page 119 — 3 cubes across the front, 2 cubes high and
5 cubes deep.

The picture is a box built out of small cubes, 2 cubes tall, 3 cubes wide and 5 cubes deep.
Count the cubes in whichever order you like — the total never changes.

30 = 2 × 3 × 5 — five slabs, each of 2 rows and 3 columns

+ + + +

6 6 6 6 6

6 + 6 + 6 + 6 + 6 = 30

Page 42 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

The same 30 cubes, seen as five slabs of 2 rows and 3 columns each.

By slabs → (2 × 3) × 5 = 6 × 5 = 30

By columns → 2 × (3 × 5) = 2 × 15 = 30

Another way → (2 × 5) × 3 = 10 × 3 = 30

And yet another → (3 × 5) × 2 = 15 × 2 = 30

The box does not change when you turn it round, so the count cannot change either.

The two ideas hidden here: we may swap the numbers being multiplied (2 × 3 = 3 ×
2) and we may group them any way we like (2 × 3) × 5 = 2 × (3 × 5). In a later class you
will meet these as the commutativity and associativity of multiplication. Because
of them, the order of the prime factors does not matter — which is why we usually
write them in increasing order: 30 = 2 × 3 × 5, 225 = 3 × 3 × 5 × 5.

Q4 When we find the prime factorisation of a number, we first write it as a product of
two factors. For example, 72 = 12 × 6. Then, we find the prime factorisation of each
of the factors: 12 = 2 × 2 × 3 and 6 = 2 × 3. Now, can you say what the prime
factorisation of 72 is?

Put the two little factorisations side by side.

72 = 12 × 6

72 = (2 × 2 × 3) × (2 × 3)

72 = 2 × 2 × 3 × 2 × 3

Writing the primes in increasing order:

72 = 2 × 2 × 2 × 3 × 3

Multiply back to check:

2×2×2=8 3×3=9 8 × 9 = 72 ✔

Page 43 of 70

as e
Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
Tip: it does not matter which first split you choose. Start with 72 = 8 × 9 instead: 8 = 2

m ase
l
× 2 × 2 and 9 = 3 × 3, giving 72 = 2 × 2 × 2 × 3 × 3 — the very same answer.

m .co a g
l a se
ag
Q5 Observe how many times each prime factor occurs in the factorisation of 72.

com
ag
Compare it with how many times it occurs in the factorisations of 12 and 6 put

m .
e
together.

g l as
ANSWER a

co m
m.
PRIME TIMES IN 12 = 2 × 2 TIMES IN 6 = 2 × TOTAL TIMES IN 72 = 2 × 2 × 2 × 3

m as e
.co l
×3 3 ×3

a g
se m ✔
l a
2 2 1 2+1= 3

a g 3

✔
a s
com
3 1 1

agl
1+1= 2

.
2
m
gl ase
a
The counts match exactly. So when two numbers are multiplied, the prime factors of the
product are simply the prime factors of the two numbers put together.

c o m
Why this is so useful: it lets you find the prime factorisation of a m .
s e big product

. comever multiplying it out. For example
without
a gla
m
ase
agl 56 × 25 = (2 × 2 × 2 × 7) × (5 × 5) = 2 × 2 × 2 × 5 × 5 × 7

se m
com 4 of the next exercise. g l a
. a
em
You will use exactly this trick in question

a s
agl
Figure it Out — Page 120
co m
m .
m
Section 5.4 Prime Factorisation
as e
.co a g l
a s emQ1
gl
Find the prime factorisations of the following numbers: 64, 104, 105, 243, 320, 141,
a c
.
1728, 729, 1024, 1331, 1000.

s e m
m a

e m . co agl
g l as
Keep dividing by the smallest prime that works, until only primes are left.
a

co m
m .
m ase
.co


a g l Page 44 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

NUMBER HOW IT BREAKS UP PRIME FACTORISATION

64 64 = 2 × 32 = 2 × 2 × 16 = … 2×2×2×2×2×2

104 104 = 8 × 13 2 × 2 × 2 × 13

105 105 = 5 × 21 = 5 × 3 × 7 3×5×7

243 243 = 3 × 81 = 3 × 3 × 27 = … 3×3×3×3×3

320 320 = 64 × 5 2×2×2×2×2×2×5

141 1 + 4 + 1 = 6, so 3 divides it: 141 = 3 × 47 3 × 47

1728 1728 = 64 × 27 2×2×2×2×2×2×3×3×3

729 729 = 27 × 27 3×3×3×3×3×3

1024 1024 = 2 × 512 = 2 × 2 × 256 = … 2×2×2×2×2×2×2×2×2×2

1331 1331 = 11 × 121 = 11 × 11 × 11 11 × 11 × 11

1000 1000 = 10 × 10 × 10 = (2 × 5) × (2 × 5) × (2 × 5) 2×2×2×5×5×5

Checks: 26 = 64 ✔ 8 × 13 = 104 ✔ 3 × 5 × 7 = 105 ✔ 35 = 243 ✔

64 × 5 = 320 ✔ 3 × 47 = 141 ✔ 64 × 27 = 1728 ✔ 36 = 729 ✔

210 = 1024 ✔ 11 × 11 × 11 = 1331 ✔ 8 × 125 = 1000 ✔

Tip for 141: a number is divisible by 3 when its digits add up to a multiple of 3. Here
1 + 4 + 1 = 6, so 141 = 3 × 47, and 47 is prime.

Q2 The prime factorisation of a number has one 2, two 3s, and one 11. What is the
number?

Just multiply the primes as many times as they occur.

Page 45 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Number = 2 × 3 × 3 × 11

= 2 × 9 × 11

= 18 × 11
= 198

Answer: 198.

Check: 198 = 2 × 99 = 2 × 9 × 11 = 2 × 3 × 3 × 11 ✔ — one 2, two 3s and one 11,
exactly as asked.

Q3 Find three prime numbers, all less than 30, whose product is 1955.

Break 1955 down step by step. It ends in 5, so start with 5.

1955 ÷ 5 = 391

Now split 391. It is odd, 3 + 9 + 1 = 13 (not a multiple of 3), does not end in 0 or 5.

Try 7 → no. 11 → no. 13 → no. 17 → 391 ÷ 17 = 23 ✔

And 23 is prime.

1955 = 5 × 17 × 23

All three — 5, 17 and 23 — are prime and all are less than 30. ✔

Check: 17 × 23 = 391, and 391 × 5 = 1955 ✔

Q4 Find the prime factorisation of these numbers without multiplying first. a. 56 × 25 b.
108 × 75 c. 1000 × 81

Factorise each part separately and then write the two lists side by side — no big multiplication
needed.

Page 46 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

a. 56 × 25

56 = 2 × 2 × 2 × 7 25 = 5 × 5

56 × 25 = 2 × 2 × 2 × 5 × 5 × 7

b. 108 × 75

108 = 2 × 2 × 3 × 3 × 3 75 = 3 × 5 × 5

108 × 75 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

c. 1000 × 81

1000 = 2 × 2 × 2 × 5 × 5 × 5 81 = 3 × 3 × 3 × 3

1000 × 81 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 5

Check if you like: 56 × 25 = 1400, and 8 × 25 × 7 = 1400 ✔; 108 × 75 = 8100, and 4 ×
81 × 25 = 8100 ✔; 1000 × 81 = 81000, and 8 × 81 × 125 = 81000 ✔

Q5 What is the smallest number whose prime factorisation has: a. three different
prime numbers? b. four different prime numbers?

To keep the answer as small as possible, use the smallest primes and use each of them only
once.

a. smallest three primes → 2, 3, 5

2 × 3 × 5 = 30

b. smallest four primes → 2, 3, 5, 7

2 × 3 × 5 × 7 = 30 × 7 = 210

Page 47 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Why nothing smaller can work: the number must contain three (or four) different
primes. Swapping any of 2, 3, 5, 7 for a bigger prime only makes the product larger,
and repeating a prime also makes it larger without adding a new prime. So 30 and
210 are the smallest.

Try This: continue the pattern — the smallest number with five different primes is 2
× 3 × 5 × 7 × 11 = 2310.

Figure it Out — Page 122
Using prime factorisation

Q1 Are the following pairs of numbers co-prime? Guess first and then use prime
factorisation to verify your answer. a. 30 and 45 b. 57 and 85 c. 121 and 1331 d. 343
and 216

PAIR PRIME FACTORISATION COMMON PRIME FACTOR CO-PRIME?

a. 30 and 45 30 = 2 × 3 × 5 3 and 5 No
45 = 3 × 3 × 5

b. 57 and 85 57 = 3 × 19 none Yes ✔
85 = 5 × 17

c. 121 and 1331 121 = 11 × 11 11 No
1331 = 11 × 11 × 11

d. 343 and 216 343 = 7 × 7 × 7 none Yes ✔
216 = 2 × 2 × 2 × 3 × 3 × 3

Checks: 3 × 19 = 57 ✔ 5 × 17 = 85 ✔ 7 × 7 × 7 = 343 ✔ 6 × 6 × 6 = 216 ✔

A good guess first: 30 and 45 are both in the 5 table, so they cannot be co-prime.
121 and 1331 are both powers of 11. And 343 = 73 is odd while 216 = 63 is even — a
promising sign that they share nothing.

Page 48 of 70

as e
Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
Is the first number divisible by the second? Use prime factorisation. a. 225 and 27 b.
e
Q2

m l as
.co
96 and 24 c. 343 and 17 d. 999 and 99

a g
se m
g l a
a

The first number is divisible by the second only if the whole prime factorisation of the second

m
sits inside that of the first.
. co ag
m
se IS THE SECOND ONE
PART PRIME
FACTORISATIONS ag l a INCLUDED?
DIVISIBLE?

a. 225 ÷ 225 = 3 × 3 × 5 × 5 27 needs three 3s, 225 has only two
co
No
m
e m.
27 27 = 3 × 3 × 3

m l as
m
b. 96 ÷ 24
.co 96 = 2 × 2 × 2 × 2 × 2 × 3 a g
yes — 96 = (2 × 2 × 2 × 3) × 2 × 2 Yes ✔ (96 ÷ 24 =
l a se 24 = 2 × 2 × 2 × 3 4)

g
a c. 343 ÷ 17 343 = 7 × 7 × 7 17 does not appear in 343 at all No

m a s
agl
17 = 17

.c11odoes not appear in 999
d. 999 ÷ 999 = 3 × 3 × 3 × 37
a s em No
99 99 = 3 × 3 × 11
ag l

co m
Long-division check: 225 ÷ 27 = 8 remainder 9 ✘ 96 ÷ 24 = 4 remainder 0 ✔
m .
m as e
.co
343 ÷ 17 = 20 remainder 3 ✘ 999 ÷ 99 = 10 remainder 9 ✘
a g l
se m
g l a
a Watch part (a) closely: both 225 and 27 are made only of 3s and 5s, so it looks

se m
comfails. a
promising. But how many times a prime occurs matters. 27 wants three 3s and 225
. a g l
em
can supply only two — so the division

a s
agl
The first number has prime factorisation 2 × 3 × 7 and the second number has prime
co m
.
Q3
factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
e m
m l as
.co
mANSWER a g
s e
agla
.c
First number = 2 × 3 × 7 = 42
s e m
m a
Second number = 3 × 7 × 11 = 231
em . co agl
g l as
a
Are they co-prime? No. They share the primes 3 and 7, so 3, 7 and 21 are all common factors.

co m
m .
m as e
.co


a g l Page 49 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Does one divide the other? No, neither divides the other.

Does 42 divide 231? 42 needs a 2, but 231 = 3 × 7 × 11 has no 2 → No

Does 231 divide 42? 231 needs an 11, but 42 = 2 × 3 × 7 has no 11 → No

(Also 231 is bigger than 42, so it could never divide it.)

Check: 231 ÷ 42 = 5 remainder 21 ✘

Remember the two different tests: for co-prime we ask “is any prime shared?”; for
divisibility we ask “is the whole factorisation contained?” Two numbers can share
some primes and still not divide each other — exactly as here.

Q4 Guna says, “Any two prime numbers are co-prime?”. Is he right?

Yes, Guna is right — as long as the two primes are different.

Factors of a prime p → 1 and p only

Factors of a different prime q → 1 and q only

Since p ≠ q, the only factor they share is 1 → co-prime ✔

Examples:

2 and 3 → common factor only 1 ✔

3 and 11 → common factor only 1 ✔

17 and 89 → common factor only 1 ✔

The one exception to keep in mind: if the two primes are the same number, say 5
and 5, then 5 is a common factor and they are not co-prime. So the correct
statement is: any two different prime numbers are co-prime.

Page 50 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

In-text Questions — Pages 123 to 125
Section 5.5 Divisibility Tests

MATH TALK

Q1 The first few multiples of 10 are: 10, 20, 30, 40, … Is 125 a multiple of 10? Will this
number appear in the previous sequence? Why or why not? Can you now answer if
8560 is divisible by 10?

125 is not a multiple of 10, and it will never appear in that sequence.

10, 20, 30, …, 110, 120, 130, 140, … — the list jumps straight from 120 to 130

125 ÷ 10 = 12 remainder 5

Why: every multiple of 10 ends in 0, because 10 × any number puts a zero in the units place. But
125 ends in 5, so it can never be in the list.
Is 8560 divisible by 10? Yes.

8560 ends in 0 → 8560 = 856 × 10 → 8560 ÷ 10 = 856, remainder 0 ✔

Q2 Consider this statement: Numbers that are divisible by 10 are those that end with
‘0’. Do you agree?

Yes, we fully agree. This works both ways round.

Every multiple of 10 ends in 0 — because multiplying by 10 shifts every digit one place to
the left and drops a 0 into the units place. 7 × 10 = 70, 43 × 10 = 430.
Every number ending in 0 is a multiple of 10 — because you can rub off the last 0 and read
the rest as the quotient. 2560 = 256 × 10.

Numbers ending in 0 → 10, 40, 90, 250, 8560 → all divisible by 10 ✔

Numbers not ending in 0 → 125, 682, 8536 → none divisible by 10 ✘

Page 51 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

The reason behind it: any number can be split as (all digits except the last) × 10 +
(units digit). The first part is already a multiple of 10, so the whole number is a
multiple of 10 exactly when the units digit is 0.

Q3 Explore by listing down the multiples: 5, 10, 15, 20, 25, ... What do you observe about
these numbers? Do you see a pattern in the last digit? What is the largest number
less than 399 that is divisible by 5? Is 8560 divisible by 5?

List the multiples of 5 and look at the last digit:

5, 10, 15, 20, 25, 30, 35, 40, 45, 50, …

The last digit is always 0 or 5, turn and turn about.
Largest number less than 399 divisible by 5 = 395.

399 ÷ 5 = 79 remainder 4 → 5 × 79 = 395

(400 is also a multiple of 5, but it is bigger than 399.)

Is 8560 divisible by 5? Yes.

8560 ends in 0 → 8560 ÷ 5 = 1712, remainder 0 ✔

Q4 Consider this statement: Numbers that are divisible by 5 are those that end with
either a ‘0’ or a ‘5’. Do you agree?

Yes, we agree.

Ends in 0 or 5 → 25, 60, 195, 395, 8560 → all divisible by 5 ✔

Ends in any other digit → 78, 99, 173, 572, 8536 → none divisible by 5 ✘

Page 52 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Why it is true: split the number as (all digits except the last) × 10 + (units digit). Since
10 is a multiple of 5, the first part is always a multiple of 5. So the whole number is a
multiple of 5 exactly when the units digit is a multiple of 5 — and among the digits 0
to 9 only 0 and 5 are multiples of 5.

Tip: notice how the same trick explains both the 10 test and the 5 test. Only the last
digit matters for both, because 10 is divisible by both 10 and 5.

Q5 The first few multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... What do you
observe? Do you see a pattern in the last digit? Is 682 divisible by 2? Can we answer
this without doing the long division? Is 8560 divisible by 2? Why or why not?

Look at the last digits of the multiples of 2:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, …

The last digit is always one of 0, 2, 4, 6, 8, repeating in a cycle of five.
Is 682 divisible by 2? Yes — and we do not need long division.

682 ends in 2, which is even → 682 ÷ 2 = 341, remainder 0 ✔

Is 8560 divisible by 2? Yes, because it ends in 0.

8560 ÷ 2 = 4280, remainder 0 ✔

Q6 Consider this statement: Numbers that are divisible by 2 are those that end with ‘0’,
‘2’, ‘4’, ‘6’ or ‘8’. Do you agree? What are all the multiples of 2 between 399 and 411?

Yes, we agree — these are exactly the even numbers.

Page 53 of 70

as e
Class 6 Maths Chapter 5 Prime Time
a g l AglaSem · NCERT Solutions

co m
m.
Why: once again split the number as (all digits except the last) × 10 + (units digit). 10

as e
comis even — that is, 0, 2, 4, 6 or 8. l
is even, so the first part is always even. The number is therefore even exactly when
. a g
em
its units digit
a s
agl of 2 between 399 and 411:
Multiples

co m
400, 402, 404, 406, 408 and 410
e m . ag
g l as
a
That is six numbers — every alternate number in that stretch.

co m
em.
Check: 400 ÷ 2 = 200, 402 ÷ 2 = 201, 404 ÷ 2 = 202, 406 ÷ 2 = 203, 408 ÷ 2 = 204, 410 ÷
m l as
.co g
2 = 205. ✔
m a
l a se
a g
s
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers
om a
Q7

. c agl
between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you

a s
observe? Is 8536 divisible by 4?em
agl

co m
m .
se
STRETCH OF NUMBERS DIVISIBLE BY 4 LAST TWO DIGITS

com330 and 340 l a
m . ag32, 36
ase
Between 332, 336

agl Between 1730 and 1740 1732, 1736 32, 36

se m
Between 2030 and 2040
com
2032, 2036 32, 36
g l a
m . a
e
as the same two endings turn up — 32 and 36. The digits
g l
What we observe: in all three stretches
in front make no differenceaat all. So only the last two digits decide divisibility by 4.

m
Is 8536 divisible by 4? Yes.

. co
em
m l as
.co g
Last two digits = 36, and 36 ÷ 4 = 9 ✔
m So 8536 ÷ 4 = 2134, remainder 0 ✔ a
l a se
ag
.c
s e m
m a
co agl
Careful: the last single digit is of no use for 4. Both 12 and 22 end in 2, but 12 is a
multiple of 4 and 22 is not. m.
l a se
ag

com
m .
m ase
.co


a g l Page 54 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q8 Consider these statements: 1. Only the last two digits matter when deciding if a
given number is divisible by 4. 2. If the number formed by the last two digits is
divisible by 4, then the original number is divisible by 4. 3. If the original number is
divisible by 4, then the number formed by the last two digits is divisible by 4. Do you
agree? Why or why not?

Yes — all three statements are true.

Why: any number can be split as

Number = (the part in front) × 100 + (the last two digits)

and 100 = 4 × 25, so the first piece is always a multiple of 4, whatever the front digits
may be. So the whole number is a multiple of 4 exactly when the last two digits form
a multiple of 4 — which is precisely what all three statements say.

8536 = 85 × 100 + 36 = 85 × (4 × 25) + (4 × 9) = 4 × (2125 + 9) = 4 × 2134 ✔

4028 = 40 × 100 + 28 → 28 ÷ 4 = 7 ✔ 364 = 3 × 100 + 64 → 64 ÷ 4 = 16 ✔

But 8542 → last two digits 42, and 42 ÷ 4 leaves 2 → 8542 is not divisible by 4 ✘

Try This: pick any four-digit number, keep the last two digits and change the first
two as you like — 1132, 5532, 9932. All of them stay divisible by 4, because 32 is.

Page 55 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q9 Find numbers between 120 and 140 that are divisible by 8. Also find numbers
between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you
observe? Change the last two digits of 8560 so that the resulting number is a
multiple of 8.

STRETCH OF NUMBERS DIVISIBLE BY 8 LAST THREE DIGITS

Between 120 and 140 128, 136 128, 136

Between 1120 and 1140 1128, 1136 128, 136

Between 3120 and 3140 3128, 3136 128, 136

What we observe: the same two endings — 128 and 136 — appear every time. The thousands
digit changes nothing. So only the last three digits decide divisibility by 8.
Changing the last two digits of 8560:

8560 → last three digits 560, and 560 ÷ 8 = 70

So 8560 is already a multiple of 8 (8560 = 8 × 1070).

If we must change the last two digits, plenty of choices work — the last three digits just have to
stay a multiple of 8:

8504, 8512, 8520, 8528, 8536, 8544, 8552, 8568, 8576, 8584, 8592

For example 8552 = 8 × 1069 ✔ and 8536 = 8 × 1067 ✔

Note: the answer key at the back of the book gives 8552, which is correct — but it is
worth noticing that 8560 itself was already divisible by 8.

Page 56 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Q10 Consider these statements: 1. Only the last three digits matter when deciding if a
given number is divisible by 8. 2. If the number formed by the last three digits is
divisible by 8, then the original number is divisible by 8. 3. If the original number is
divisible by 8, then the number formed by the last three digits is divisible by 8. Do
you agree? Why or why not?

Yes — all three statements are true.

Why: split the number as

Number = (the part in front) × 1000 + (the last three digits)

and 1000 = 8 × 125, so the first piece is always a multiple of 8. Whatever happens in
front simply cannot change the answer — only the last three digits can.

8576 = 8 × 1000 + 576 → 576 ÷ 8 = 72 ✔

7648 = 7 × 1000 + 648 → 648 ÷ 8 = 81 ✔

5024 = 5 × 1000 + 024 → 24 ÷ 8 = 3 ✔

But 8570 → 570 ÷ 8 leaves remainder 2 → 8570 is not divisible by 8 ✘

See the pattern: 10 = 10 × 1 → one digit for 10 and 5; 100 = 4 × 25 → two digits for
4; 1000 = 8 × 125 → three digits for 8. Each time we double the divisor we need one
more digit.

Figure it Out — Pages 125 & 126

Page 57 of 70

Class 6 Maths Chapter 5 Prime Time AglaSem · NCERT Solutions

Section 5.5 Divisibility Tests

Q1 2024 is a leap year (as February has 29 days). Leap years occur in the years that are
multiples of 4, except for those years that are evenly divisible by 100 but not 400. a.
From the year you were born till now, which years were leap years? b. From the year
2024 till 2099, how many leap years are there?

a. Write down your birth year and keep adding until you reach a multiple of 4; after that every
4th year is a leap year (there is no “century” exception anywhere between 2001 and 2099).

Example: if you were born in 2013, the leap years since then are

2016, 2020, 2024 — and the next ones will be 2028, 2032, 2036, …

If you were born in 2014 → 2016, 2020, 2024

If you were born in 2012 → 2012, 2016, 2020, 2024

Test any year quickly with the rule for 4 — look at the last two digits: 16 ÷ 4 = 4 ✔, 20 ÷ 4 = 5 ✔,
24 ÷ 4 = 6 ✔, but 22 ÷ 4 leaves 2 ✘.
b. From 2024 till 2099 there are 19 leap years.

They are 2024, 2028, 2032, …, 2096

Count = (2096 − 2024) ÷ 4 + 1 = 72 ÷ 4 + 1 = 18 + 1 = 19

Why the “100 but not 400” rule does not bite here: the only multiple of 100 nearby
is 2100, and it lies outside our range. (For the record, 2100 will not be a leap year,
because it is divisible by 100 but not by 400 — while 2000 was a leap year, since 2000
÷ 400 = 5.)

Page 58 of 70

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Class 6 Maths Chapter 5 Prime Time
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Find the largest and smallest 4-digit numbers that are divisible by 4 and are also
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palindromes.

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A 4-digit palindrome looks like abba — the first and last digits are the same, and so are the

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middle two. For it to be divisible by 4, the number made by the last two digits, ba, must be
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divisible by 4. Since a multiple of 4 is always even, the digit a must be even (2, 4, 6 or 8 — it

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cannot be 0, as a 4-digit number cannot start with 0).
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large

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Test b8 for b = 9, 8, 7, …
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98 ÷ 4 → remainder 2 ✘ 88 ÷ 4 = 22 ✔
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Smallest: take a = 2 and make b as small as possible.

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Test b2 for b = 0, 1, 2, …

02 ÷ 4 → remainder 2 ✘ 12 ÷ 4 = 3 ✔

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So the number is 2112, and 2112 ÷ 4 = 528 ✔
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Answer: largest = 8888, smallest = 2112.

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Note — the book's answer key is wrong here. It gives 9999 and 1001, but both of

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these are odd numbers, so neither can be divisible by 4 (9999 ÷ 4 = 2499 remainder

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3; 1001 ÷ 4 = 250 remainder 1). The correct answers are 8888 and 2112.

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Explore and find out if each statement is always true, sometimes true or never true.
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Q3

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a. Sometimes true.
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2+6=8=4×2✔
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10 + 14 = 24 = 4 × 6 ✔

but 2 + 4 = 6 ✘
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8 + 10 = 18 ✘

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