NCERT Solutions Class 8 Maths Chapter 1 a Square and a Cube

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 1: A Square and A Cube

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

1 – 18 18 65 English

Solutions, notes, sample papers & more at 48 pages

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 1: A Square and A Cube
Chapter 1 of Ganita Prakash Part I opens with Queen Ratnamanjuri's will and a room of 100 lockers. Counting
how many times each locker is toggled turns out to be counting factors — and that single idea leads to
square numbers, square roots, cubes and cube roots.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) 1 – 18

SECTIONS QUESTIONS

18 65

MEDIUM

English

In-text Questions — Page 1
Queen Ratnamanjuri's puzzle

Q1 Before the process begins, Khoisnam realises that he already knows which lockers
will be open at the end. How did he figure out the answer? Hint: Find out how many
times each locker is toggled.

He counted factors. Person k touches locker n exactly when k divides n, so locker n is toggled
once for every factor of n.

Locker 6 is touched by persons 1, 2, 3, 6 → 4 toggles

Every locker starts closed, so it ends open only after an odd number of toggles

So the question becomes: which numbers from 1 to 100 have an odd number of factors?

Why it happens: Factors come in partner pairs — if d divides n, then so does n ÷ d,
and the two multiply to n. Pairing off every factor with its partner would make the
count even. The count can only be odd when one factor is its own partner, that is
when d × d = n. That happens exactly for the squares.

So Khoisnam knew the answer before anyone moved: the lockers left open are

Page 1 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Check it yourself: Locker 16 is touched by 1, 2, 4, 8, 16 — five people, an odd
number, so it stays open. Locker 12 is touched by 1, 2, 3, 4, 6, 12 — six people, so it
closes.

In-text Questions — Page 2
Factors and partner factors

Q1 Does every number have an even number of factors?

No. Most numbers do, but the squares do not.

6 : 1 × 6, 2 × 3 → 4 factors (even)

10 : 1 × 10, 2 × 5 → 4 factors (even)

9 : 1 × 9, 3 × 3 → 3 factors (odd)

Why it happens: Writing a number as a product of two factors always pairs a factor
with a partner. As long as the two members of every pair are different, the factors
can be matched up two by two and the total is even. A number breaks that pattern
only when one of its pairs has both members equal — and a pair d × d means the
number is d2.

So: every number except a perfect square has an even number of factors.

Q2 Can you use this insight to find more numbers with an odd number of factors?

Yes — take any number times itself.

1 × 1 = 1, 2 × 2 = 4, 3 × 3 = 9, 4 × 4 = 16, 5 × 5 = 25, …

Page 2 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Their factor counts are 1, 3, 3, 5, 3, … — all odd.

NUMBER FACTORS HOW MANY

1 1 1

4 1, 2, 4 3

9 1, 3, 9 3

16 1, 2, 4, 8, 16 5

25 1, 5, 25 3

36 1, 2, 3, 4, 6, 9, 12, 18, 36 9

Every one of these is a square, and no number that is not a square appears on such a list.

Q3 For instance, 36 has a factor pair 6 × 6 where both numbers are 6. Does this number
have an odd number of factors? If every factor of 36 other than 6 has a different
factor as its partner, then we can be sure that 36 has an odd number of factors.
Check if this is true.

Yes, 36 has 9 factors — an odd number. Here is the check, pair by pair.

1 × 36 → partners 1 and 36, different

2 × 18 → partners 2 and 18, different

3 × 12 → partners 3 and 12, different

4 × 9 → partners 4 and 9, different

6 × 6 → partner of 6 is 6 itself

Four pairs of distinct factors give 4 × 2 = 8 factors, and the lone 6 adds one more:

8 + 1 = 9 factors — 1, 2, 3, 4, 6, 9, 12, 18, 36

Page 3 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: The claim in the question is exactly the general argument. All the

m l a se
factors except the self-partner can be swept into disjoint pairs, contributing an even
o g Only a square
count; the.cself-partner is left over on its own, tipping the total toaodd.
m
se a self-partner, so only a square can have an odd number of factors.
hasasuch
ag l

co m
e m . ag
as
In-text Questions — Page 3
Section 1.1 Square Numbers
a g l

co m
m.
Q1 Write the locker numbers that remain open.

as e
ANSWER .c
om a g l
a s em
agl
The lockers with square numbers between 1 and 100:

m a s
.co agl
1, 4, 9, 16, 25, 36, 49, 64, 81, 100

a s em
That is 10 lockers, namely 12, 2a2g
l
, 32, …, 102. The next square, 112 = 121, is beyond 100.

co m
Tip: The count of open lockers up to 100 is 10 because 102 = 100 — in general the
m .
m as e
l
number of squares from 1 to N is the whole-number part of √N.

m .co a g
l a se
ag
Q2 Which are these five lockers?
se m
com g l a
m . a
ase
agl

The passcode lockers are the ones touched exactly twice, so their numbers have exactly two
factors — the prime numbers.

co m
m .
m as e
.co l
2 → 1, 2
a g
a s em 3 → 1, 3

agl
.c
5 → 1, 5

s e m
m a
agl
7 → 1, 7
. co
em
as
11 → 1, 11

a g l
The first five such lockers are 2, 3, 5, 7 and 11, so the code is 2-3-5-7-11.

co m
m .
m ase
.co


a g l Page 4 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Why it happens: A locker toggled exactly twice is opened by person 1 and closed by
exactly one other person. That means the number has 1 and itself as factors and
nothing else — the definition of a prime. Note that 1 itself is touched only once, so it
is not on the list.

Q3 Can we have a square of sidelength 3/5 or 2.5 units?

Yes. A side length need not be a whole number; squaring works for fractions and decimals in
exactly the same way.

(3/5)2 = (3/5) × (3/5) = 9/25 sq units

(2.5)2 = 2.5 × 2.5 = 6.25 sq units

Why it happens: The area of a square is side × side whatever the side is. Squaring a
fraction squares the numerator and the denominator separately, because (a/b) ×
(a/b) = (a×a)/(b×b).

Tip: 9/25 and 6.25 are squares of fractions, but they are not perfect squares — that
name is kept for squares of natural numbers.

In-text Questions — Page 4

Page 5 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Patterns and Properties of Perfect Squares

MATH TALK

Q1 Find the squares of the first 30 natural numbers and fill in the table below.

1² = 1 11² = 121 21² = 441

2² = 4 12² = 22² =

3² = 9 13² =

4² = 16 14² =

5² = 25 15² =

6² = 16² =

7² = 17² =

8² = 18² =

9² = 19² =

10² = 20² =

The completed table of the first 30 perfect squares:

Page 6 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

12 = 1 112 = 121 212 = 441

22 = 4 122 = 144 222 = 484

32 = 9 132 = 169 232 = 529

42 = 16 142 = 196 242 = 576

52 = 25 152 = 225 252 = 625

62 = 36 162 = 256 262 = 676

72 = 49 172 = 289 272 = 729

82 = 64 182 = 324 282 = 784

92 = 81 192 = 361 292 = 841

102 = 100 202 = 400 302 = 900

Tip: You do not need to multiply each time. Since (n+1)2 = n2 + (2n + 1), each square
is the previous one plus the next odd number: 400 + 41 = 441, 441 + 43 = 484, 484 +
45 = 529, and so on.

Q2 What patterns do you notice? Share your observations and make conjectures.

Several patterns show up at once in that table.

Units digits. Every square ends in 0, 1, 4, 5, 6 or 9. None ends in 2, 3, 7 or 8.
Mirror pairs. Numbers whose units digits add to 10 have squares with the same units digit:
1 and 9 both give 1, 2 and 8 both give 4, 3 and 7 both give 9, 4 and 6 both give 6.
Differences. Consecutive squares differ by consecutive odd numbers: 4 – 1 = 3, 9 – 4 = 5, 16 –
9 = 7, …
Parity. The square of an even number is even, the square of an odd number is odd.
Zeros. 102 = 100, 202 = 400, 302 = 900 — one zero at the end of the number becomes two at
the end of the square.

Page 7 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Why the units digit works: The last digit of a product depends only on the last
digits of the factors. So the last digit of n2 depends only on the last digit of n, and
there are only ten cases to check: 0→0, 1→1, 2→4, 3→9, 4→6, 5→5, 6→6, 7→9, 8→4,
9→1. The set of outcomes is exactly {0, 1, 4, 5, 6, 9}.

Q3 If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?

No. The units digit is a one-way test.

16 = 42 ✓ and 36 = 62 ✓, but 26 ends in 6 and is not a square

Also 20, 21, 24, 30, 31 … all end in an allowed digit but are not squares

Why it happens: Ending in 0, 1, 4, 5, 6 or 9 is a necessary condition, not a sufficient
one. A square must pass it, but passing it does not make a number a square. The
useful direction is the contrapositive: if a number ends in 2, 3, 7 or 8, it is certainly
not a square.

Tip: This is the fastest first filter you have. 1027 ends in 7 → not a square, and you
never need to factorise it.

Q4 Write 5 numbers such that you can determine by looking at their units digit that
they are not squares.

Choose any numbers ending in 2, 3, 7 or 8. For example:

42, 153, 267, 638, 1082

Page 8 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
NUMBER UNITS DIGIT CONCLUSION

m as e
42
.co
2 not a square
a g l
se m
g l a
a
153 3 not a square

267 7 not a square

co m
638 8
e m . not a square ag
g l as
1082 2
a not a square

No square of any natural number can end in these four digits, so a single glance settles all five.
co m
em.
m l as
.co a g
a s em
l
The squares, 1², 9², 11², 19², 21², and 29², all have 1 in their units place. Write the
g
Q5
a next two squares. Notice that if a number has 1 or 9 in the units place, then its

s
square ends in 1.
m a
em
.co agl
a s
agl with the next numbers ending in 1 or 9 — that is 31 and 39.

The list 1, 9, 11, 19, 21, 29 continues

co m
312 = 961
m .
m as e
.co
392 = 1521
a g l
se m
g l a
a Both end in 1, as expected.

se m
com g l a
Why it happens: Only the units digit
. of the number decides the units digit of the
a
a s em
l
square. 1 × 1 = 1 ends in 1, and 9 × 9 = 81 also ends in 1. No other digit gives 1: the
ag1→1, 2→4, 3→9, 4→6, 5→5, 6→6, 7→9, 8→4, 9→1. So a
ten possibilities are 0→0,
square ends in 1 exactly when its root ends in 1 or 9.

co m
m .
m as e
.co a g l
s e m Questions — Page 5
In-text
agla
.c
s e m
m a
em . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Patterns and Properties of Perfect Squares

Q1 Which of the following numbers have the digit 6 in the units place? (i) 38² (ii) 34² (iii)
46² (iv) 56² (v) 74² (vi) 82²

Square only the units digit of each number and look at the last digit of the result.

NUMBER UNITS DIGIT ITS SQUARE UNITS DIGIT OF THE SQUARE

(i) 38 8 64 4

(ii) 34 4 16 6✓

(iii) 46 6 36 6✓

(iv) 56 6 36 6✓

(v) 74 4 16 6✓

(vi) 82 2 4 4

So (ii) 342, (iii) 462, (iv) 562 and (v) 742 have 6 in the units place.

Why it happens: A square ends in 6 exactly when its root ends in 4 or 6, because 4 ×
4 = 16 and 6 × 6 = 36 are the only single-digit products ending in 6.

Check it yourself: 342 = 1156, 462 = 2116, 562 = 3136, 742 = 5476 — all end in 6,
while 382 = 1444 and 822 = 6724 end in 4.

Page 10 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q2 Find more such patterns by observing the numbers and their squares from the
table you filled earlier.

1² = 1 11² = 121 21² = 441

2² = 4 12² = 22² =

3² = 9 13² =

4² = 16 14² =

5² = 25 15² =

6² = 16² =

7² = 17² =

8² = 18² =

9² = 19² =

10² = 20² =

The table of squares from page 4 — the one you filled in.

Here are four more patterns that the table of squares up to 302 makes visible.

Squares of numbers ending in 5. 152 = 225, 252 = 625 — the square always ends in 25, and
the part before it is the tens digit times the next number: 1×2 = 2, 2×3 = 6.
Digit sums. The digit sum of a square, reduced repeatedly, is always 1, 4, 7 or 9. For 576:
5+7+6 = 18 → 9. For 441: 4+4+1 = 9. Never 2, 3, 5, 6 or 8.
Squares of numbers just below a round number. 192 = 361 = 400 – 39, 292 = 841 = 900 – 59
— subtract the two "neighbours" added together.
Nine-times pattern. Squares that are multiples of 3 are multiples of 9: 9, 36, 81, 144, 225,
324, 441, 576, 729, 900.

Why the "ending in 5" rule works: A number ending in 5 is 10a + 5, and (10a + 5)2 =
100a2 + 100a + 25 = 100 × a(a+1) + 25. So the last two digits are always 25, and the
leading part is a × (a + 1). For 25: a = 2, a(a+1) = 6, giving 625.

Page 11 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q3 If a number contains 3 zeros at the end, how many zeros will its square have at the
end?

Six zeros.

2000 = 2 × 1000

20002 = 4 × 1000 × 1000 = 4 × 1000000 = 4000000 → 6 zeros

Why it happens: A number ending in three zeros is m × 103, where m does not end
in 0. Squaring gives m2 × 106, and m2 cannot end in 0, so exactly 6 zeros are left at
the end. Squaring simply doubles the count of terminal zeros.

Q4 What do you notice about the number of zeros at the end of a number and the
number of zeros at the end of its square? Will this always happen? Can we say that
squares can only have an even number of zeros at the end?

The square always has twice as many terminal zeros as the number.

102 = 100 → 1 zero becomes 2

202 = 400 → 1 zero becomes 2

1002 = 10000 → 2 zeros become 4

7002 = 490000 → 2 zeros become 4

Yes, this always happens, and yes, a square can only end in an even number of zeros.

Why it happens: Write the number as m × 10k, where m has no zero at the end.
Then its square is m2 × 102k, and m2 also has no zero at the end. So the count of
terminal zeros of the square is exactly 2k — always even. This gives a second quick
test: a number ending in an odd number of zeros, such as 1000 or 490, cannot be a
perfect square.

Page 12 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q5 What can you say about the parity of a number and its square?

A number and its square always have the same parity — even squares to even, odd squares to
odd.

even : 42 = 16, 102 = 100, 262 = 676 → all even

odd : 32 = 9, 72 = 49, 212 = 441 → all odd

Why it happens: An even number is 2k, so its square is 4k2 = 2(2k2) — a multiple of
2, hence even. An odd number is 2k + 1, so its square is 4k2 + 4k + 1 = 2(2k2 + 2k) + 1
— one more than an even number, hence odd.

Tip: Read this backwards too. If a perfect square is odd, its square root must be odd;
if it is even, the root is even. That halves your search when you are hunting for a
root.

Q6 Let us explore the differences between consecutive squares. What do you notice? 4
– 1 = 3, 9 – 4 = 5, 16 – 9 = 7, 25 – 16 = 9. See if this pattern continues for the next few
square numbers.

The differences are the consecutive odd numbers, and the pattern does continue.

36 – 25 = 11

49 – 36 = 13

64 – 49 = 15

81 – 64 = 17
100 – 81 = 19

Page 13 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: The gap between n2 and (n+1)2 is

o m
(n + 1)2 – n2 = n2 + 2n + 1 – n2 = 2n + 1,
l a se
g another from
.c the (n+1)-th odd number. Adding these gaps oneaafter
se
which is mexactly
a
1 lgives
a1g + 3 + 5 + … + (2n – 1) = n2.

co m
e m . ag
g l as
a

. c om2
5 moreem →3
m a s
e m . co agl
g l as
a
3 more → 22
m a s
m .co agl
l a se
ag
1 12
com
m .
m as e
.co a g l
se m
g l a
a
Each new odd number is an L-shaped band that grows a square into the next square.

se m
com g l a
. a
In-text Questions — Pagea6sem
agl

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

co m
m .
m ase
.co


a g l Page 14 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Perfect Squares and Odd Numbers

Q1 Using the pattern above, find 36², given that 35² = 1225.

1+3 1 + 3 + (3 + 2) 1+3+5 1 + 3 + 5 + (5 + 2) 1+3+5+7

Page 6 — the picture that explains why each subsequent inverted L gives the next odd
number.

1225 is the sum of the first 35 odd numbers, so 362 is that sum plus the 36th odd number.

36th odd number = 2 × 36 – 1 = 71

362 = 1225 + 71 = 1296

Why it happens: Since the sum of the first n odd numbers is n2, going from 352 to
362 means adding just one more term of that series. The same step in algebra: (n +
1)2 = n2 + 2n + 1, and here 2 × 35 + 1 = 71.

Check it yourself: 36 × 36 = 36 × 30 + 36 × 6 = 1080 + 216 = 1296. ✓

Q2 How do we find the 36th odd number?

List the odd numbers with their positions and the rule appears at once.

POSITION N 1 2 3 4 5 6

ODD NUMBER 1 3 5 7 9 11

Each odd number is one less than twice its position: 2 × 6 – 1 = 11. So

Page 15 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

36th odd number = 2 × 36 – 1 = 71

Why it happens: The even numbers in order are 2, 4, 6, …, so the n-th even number
is 2n. Every odd number sits exactly one step below an even number, which gives 2n
– 1.

Q3 What is the nth odd number?

The n-th odd number is 2n – 1.

n = 1 → 2(1) – 1 = 1

n = 5 → 2(5) – 1 = 9
n = 50 → 2(50) – 1 = 99

Why it happens: Any odd number leaves remainder 1 on division by 2, so it can be
written as 2k + 1 for some whole number k ≥ 0. Counting from k = 0 gives 1, 3, 5, …,
and the term in position n has k = n – 1, so the value is 2(n – 1) + 1 = 2n – 1.

Tip: This formula is what makes the odd-number test practical. Adding 1 + 3 + 5 + …
+ (2n – 1) gives n2, so the position at which you land on 0 while subtracting is the
square root.

In-text Questions — Page 7
Perfect Squares and Triangular Numbers; Square Roots

Q1 Find how many numbers lie between two consecutive perfect squares. Do you
notice a pattern?

Between n2 and (n+1)2 there are exactly 2n numbers.

Page 16 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

SQUARES NUMBERS STRICTLY BETWEEN HOW MANY 2N

1 and 4 2, 3 2 2×1

4 and 9 5, 6, 7, 8 4 2×2

9 and 16 10 … 15 6 2×3

16 and 25 17 … 24 8 2×4

Why it happens: The gap itself is (n + 1)2 – n2 = 2n + 1. That count includes one of
the two end points, so the numbers strictly in between number 2n + 1 – 1 = 2n.
Notice the answer is always even, so no two consecutive squares are ever "close
together" once n grows.

Tip: The official answer writes this as q – p – 1 for consecutive squares p and q. That
is the same thing: with p = n2 and q = (n+1)2, q – p – 1 = (2n + 1) – 1 = 2n.

Q2 How many square numbers are there between 1 and 100? How many are between
101 and 200? Using the table of squares you filled earlier, enter the values below,
tabulating the number of squares in each block of 100.

1 – 100 101 – 200 201 – 300 301 – 400 401 – 500

501 – 600 601 – 700 701 – 800 801 – 900 901 – 1000

What is the largest square less than 1000?

Blockwise count of the perfect squares up to 1000:

Page 17 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

BLOCK SQUARES IN IT HOW MANY

1 – 100 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 10

101 – 200 121, 144, 169, 196 4

201 – 300 225, 256, 289 3

301 – 400 324, 361, 400 3

401 – 500 441, 484 2

501 – 600 529, 576 2

601 – 700 625, 676 2

701 – 800 729, 784 2

801 – 900 841, 900 2

901 – 1000 961 1

Total = 10 + 4 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 1 = 31 squares below 1000, and the largest is

312 = 961 (the next square, 322 = 1024, is past 1000)

Why the counts thin out: The gaps between consecutive squares are 2n + 1, which
keeps growing. Early on the gaps are 3, 5, 7 — many squares fit in a block of 100.
Around 900 the gap is already about 60, so only one or two squares can fall in each
block.

Page 18 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Do you remember triangular numbers?
e
Q3

m l as
m .co a g
l a se
a g

co m
m . ag
l a se
1 3 ag 6 10 15

Page 7 — the first five triangular numbers.
co m
e m.
m l as
.co
Can you see any relation between triangular numbers and square numbers? Extend
m a g
l a se the pattern shown and draw the next term.

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

co m
1 + 3 = 4 = 2² 3 + 6 = 9 = 3² 6 + 10 = 16 = 4²

m .
o m l se box is left empty for
Page 7 — each square split into two triangular numbers; thealast

m .c ag
se
you.

g l a
a
se m
com g l a
.

m a
ase
Yes — two consecutive triangular numbers always add to a square.

agl
1 + 3 = 4 = 22

co m
3 + 6 = 9 = 32
m .
m as e
.co
6 + 10 = 16 = 42
a g l
se m
g l a 10 + 15 = 25 = 52 ← the next term
a c
m .
15 + 21 = 36 = 62
m a s e
. co a gl
m
se of dots split by a staircase into a triangle of 10 dots and a
l a
So the next picture is a 5 × 5 square
triangle of 15 dots. ag

co m
m .
m as e
.co


a g l Page 19 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

2
10 (blue) + 15 (red) = 25 = 5
A 5 × 5 square of dots cut into the triangular numbers 10 and 15.

Why it happens: The n-th triangular number is Tn = 1 + 2 + … + n = n(n+1)/2. Then

Tn–1 + Tn = (n–1)n/2 + n(n+1)/2 = n[(n–1) + (n+1)]/2 = n × 2n/2 = n2.
Geometrically, two staircase triangles of heights n – 1 and n slot together into an n ×
n square.

Q4 The area of a square is 49 sq. cm. What is the length of its side?

The side is the number that gives 49 when multiplied by itself.

side × side = 49

7 × 7 = 49, so side = 7 cm

We call 7 the square root of 49, and write √49 = 7.

Page 20 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Why we take only 7: Algebraically both 7 and –7 square to 49, so 49 has two integer
square roots. A side length is a measured length, so only the positive root makes
sense here. That is why this chapter works with the positive square root throughout.

In-text Questions — Page 8
Square Roots

MATH TALK

Q1 What is the square root of 64?

8 × 8 = 64, and (–8) × (–8) = 64 as well.

82 = 64 and (–8)2 = 64
So the square roots of 64 are +8 and –8

√64 = ±8; in this chapter we take √64 = 8

Why there are two: A negative times a negative is positive, so any perfect square
has two integer roots that differ only in sign. In general √(n2) = ±n. The symbol √ by
itself is reserved for the positive one, which is why √100 = 10, not –10.

Q2 Given a number, such as 576 or 327, how do we find out if it is a perfect square? If it
is a perfect square, how can we find its square root?

Use the tests in order — cheapest first.

1. Units digit. 327 ends in 7, so it is certainly not a perfect square. 576 ends in 6, so it survives
this test but is not yet settled.
2. List the squares. 202 = 400, 212 = 441, 222 = 484, 232 = 529, 242 = 576. So 576 is a perfect
square and √576 = 24. This works, but it is slow for large numbers.
3. Subtract odd numbers. Take away 1, 3, 5, … in turn; landing exactly on 0 at step k means the
number is k2.

Page 21 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

4. Prime factorisation. Split the prime factors into two identical groups. 576 = 26 × 32 = (23 × 3)
× (23 × 3) = 24 × 24.

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

= (2 × 2 × 2 × 3) × (2 × 2 × 2 × 3)

= 242, so √576 = 24

Why prime factorisation is the reliable test: If a number is m2, every prime in m
appears twice as often in m2. So a perfect square is exactly a number in which every
prime occurs an even number of times — and halving each exponent reads off the
root directly.

Q3 Can we find the square root of 729 using this method?

Yes, but it is time-consuming — it takes 27 subtractions.

729 – 1 = 728, 728 – 3 = 725, 725 – 5 = 720, …

… and the 27th subtraction, of 53, leaves 0

So 729 = 272 and √729 = 27

Why it is slow: Finding √N this way needs about √N steps. For 81 that is 9 steps,
which is fine; for 729 it is 27, and for a six-digit number it would run into the
hundreds. Prime factorisation is far shorter: 729 = 36 = (33)2 = 272, done in one line.

Check it yourself: The odd numbers 1 + 3 + 5 + … + 53 add to 272 = 729, because 53
= 2 × 27 – 1.

In-text Questions — Page 9

Page 22 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Square Roots by prime factorisation

Q1 We know that a perfect square is obtained by multiplying an integer by itself. Will
looking at a number's prime factorisation help in determining whether it is a
perfect square?

Yes. If the prime factors can be divided into two equal groups, the product of one group is the
square root.

36 = 2 × 2 × 3 × 3 = (2 × 3) × (2 × 3) = 62 → √36 = 6

50 = 2 × 5 × 5 → the 2 has no partner → not a perfect square

Why it works: Squaring m repeats every prime of m exactly twice as often. Reading
that backwards, a number is a perfect square precisely when each of its primes
appears an even number of times, so the whole factorisation can be dealt out into
two identical halves. Halving each exponent gives the root at once.

Q2 Is 324 a perfect square?

Yes, 324 = 182.

324 = 2 × 2 × 3 × 3 × 3 × 3

Two equal groups: (2 × 3 × 3) × (2 × 3 × 3)

= (2 × 3 × 3)2 = 182
Or in pairs: (2 × 2) × (3 × 3) × (3 × 3)

Therefore √324 = 18

Why both ways agree: Pairing the primes shows each prime occurs an even
number of times — 2 twice, 3 four times. Taking one member from each pair, 2 × 3 ×
3 = 18, rebuilds the root. Grouping into two halves and pairing are two views of the
same fact, that every exponent is even.

Page 23 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Is 156 a perfect square?
e
Q3

m l as
m .co a g
ase

No.gl
a

co m
. ag
156 = 2 × 2 × 3 × 13
e m
g l as
The pair (2 × 2) is fine, but 3 and 13 are left unpaired
a
So 156 cannot be split into two identical groups

co m
em.
as
Therefore 156 is not a perfect square.
m l
.co a g
a s em it yourself: 122 = 144 and 132 = 169, so 156 sits between two consecutive
gl
Check
a squares and cannot be one itself.

m a s
m .co agl
l a se
g
Find whether 1156 and 2800 are perfect squares using prime factorisation.
a
Q4

co m
m .
1156 is a perfect square; 2800 is not.
m as e
.co a g l
s m = 2 × 578 = 2 × 2 × 289 = 2 × 2 × 17 × 17
e1156
gl a
a
m
= (2 × 17) × (2 × 17) = 342
a se
√1156 = 34
.com a g l
m
ase
agl
2800 = 2 × 1400 = 2 × 2 × 700 = 2 × 2 × 2 × 350 = 2 × 2 × 2 × 2 × 175

co m
=2×2×2×2×5×5×7
m .
m as e
.co a g l
The four 2s pair up and the two 5s pair up, but the single 7 has no partner

a s em
agl So 2800 is not a perfect square.
.c
s e m
. c om
Why the leftover 7 decides it: In a perfect square every prime must occur an even
a g la
s e
number of times. In 2800 the mprime 7 occurs once, an odd number of times, so no
a
aglare arranged one 7 will always be stranded.
matter how the factors

co m
m .
m ase
.co


a g l Page 24 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Tip: 2800 = 400 × 7 = 202 × 7. Multiplying 2800 by 7 would give 19600 = 1402.

Figure it Out — Page 10
Section 1.1 Square Numbers

Q1 Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027
(iv) 1089

(i) 2032, (ii) 2048 and (iii) 1027 are not perfect squares. 1089 is.

NUMBER REASON VERDICT

(i) 2032 ends in 2 not a square

(ii) 2048 ends in 8 not a square

(iii) 1027 ends in 7 not a square

(iv) 1089 1089 = 3 × 3 × 11 × 11 = 332 a perfect square

Why the digit test suffices for the first three: No square ends in 2, 3, 7 or 8, so
those three are ruled out on sight. The fourth ends in 9, which is allowed, so it needs
a real check — and factorising gives 1089 = 332, so √1089 = 33.

Tip: 2048 = 211. The exponent 11 is odd, which is a second reason it cannot be a
square.

Q2 Which one among 64², 108², 292², 36² has last digit 4?

Only the units digit of the base matters.

Page 25 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

SQUARE UNITS DIGIT OF BASE ITS SQUARE LAST DIGIT

642 4 16 6

1082 8 64 4✓

2922 2 4 4✓

362 6 36 6

Two of them qualify: 1082 and 2922.

Check it yourself: 1082 = 11664 and 2922 = 85264 — both end in 4; 642 = 4096 and
362 = 1296 end in 6. The question says "which one", but the numbers give two
answers, so both must be named.

Why it happens: A square ends in 4 exactly when its root ends in 2 or 8, since 2 × 2 =
4 and 8 × 8 = 64 are the only single-digit products ending in 4.

Q3 Given 125² = 15625, what is the value of 126²? (i) 15625 + 126 (ii) 15625 + 262 (iii) 15625
+ 253 (iv) 15625 + 251 (v) 15625 + 512

(iv) 15625 + 251

1262 = (125 + 1)2 = 1252 + 2 × 125 + 1

= 15625 + 250 + 1 = 15625 + 251

= 15876

Why it happens: Consecutive squares differ by the next odd number, and here that
is the 126th odd number, 2 × 126 – 1 = 251. Equivalently, 1262 – 1252 = (126 + 125)
(126 – 125) = 251 × 1.

Check it yourself: 1262 = 126 × 126 = 15876, and 15625 + 251 = 15876. ✓

Page 26 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q4 Find the length of the side of a square whose area is 441 m².

The side is √441.

441 = 3 × 3 × 7 × 7

= (3 × 7) × (3 × 7) = 212
Side = 21 m

Why prime factorisation is the neat route: Both primes appear twice, so the
factors split into two identical groups of 3 × 7. The product of one group, 21, is the
square root. Estimating would also get you there — 202 = 400 and the number ends
in 1, so the root ends in 1 or 9, and 21 is the only candidate in range.

Q5 Find the smallest square number that is divisible by each of the following numbers:
4, 9, and 10.

First find the LCM, then top it up to a square.

4 = 22, 9 = 32, 10 = 2 × 5

LCM = 22 × 32 × 5 = 180
In 180 the prime 5 appears once — an odd number of times

Multiply by 5: 180 × 5 = 900 = 22 × 32 × 52 = 302

Smallest such square = 900

Why 900 is smallest: Any number divisible by 4, 9 and 10 must be a multiple of their
LCM, 180. A multiple of 180 that is a square must contain every prime of 180 an even
number of times, so it needs at least one more 5. The least such multiple is 180 × 5 =
900, and 900 = 302 is indeed divisible by 4, 9 and 10.

Page 27 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q6 Find the smallest number by which 9408 must be multiplied so that the product is a
perfect square. Find the square root of the product.

Factorise 9408 and look for the prime with an odd exponent.

9408 = 2 × 4704 = 2 × 2 × 2352 = 2 × 2 × 2 × 1176

= 2 × 2 × 2 × 2 × 588 = 2 × 2 × 2 × 2 × 2 × 294 = 2 × 2 × 2 × 2 × 2 × 2 × 147

147 = 3 × 49 = 3 × 7 × 7

So 9408 = 26 × 3 × 72

The 2s and the 7s pair up; only the single 3 is stranded. So multiply by 3.

9408 × 3 = 28224 = 26 × 32 × 72

= (23 × 3 × 7)2 = 1682
√28224 = 168

Why 3 is the smallest choice: To make every exponent even, each prime with an
odd exponent must be supplied once more. Here that is only the prime 3, so the
least multiplier is 3 itself; any smaller factor would leave the 3 unpaired.

Q7 How many numbers lie between the squares of the following numbers? (i) 16 and 17
(ii) 99 and 100

Between n2 and (n+1)2 there are 2n numbers.

(i) n = 16 → 2 × 16 = 32 numbers (from 257 to 288, since 162 = 256, 172 = 289)

(ii) n = 99 → 2 × 99 = 198 numbers (from 9802 to 9999, since 992 = 9801, 1002 = 10000)

Why it happens: (n + 1)2 – n2 = 2n + 1 counts the whole step from one square to the
next. Excluding the square at the top end leaves 2n numbers strictly in between.

Page 28 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: 289 – 256 – 1 = 32 and 10000 – 9801 – 1 = 198. ✓

m as e
.co a g l
a s em
aQ8gl In the following pattern, fill in the missing numbers: 1² + 2² + 2² = 3², 2² + 3² + 6² = 7²,
3² + 4² + 12² = 13², 4² + 5² + 20² = (___)², 9² + 10² + (___)² = (___)²

com
e m . ag
as

Fourth row first:
a g l

co m
m.
42 + 52 + 202 = 16 + 25 + 400 = 441 = 212

o m l a se
g is the product of the
Now read.cthe rule from the completed rows. In each one the third base
a
m
e and the answer is one more than that product.
stwo,
l a
ag
first

a s
com
N N+1 N(N + 1) N(N + 1) + 1

m . agl
ase
1 2 2 3

2 3 agl6 7

co m
.
3 4 12 13

e m
m l as
.co g
4 5 20 21

e9m a
a s
gl
10 90 91

a
se m
92 + 102 + 902 = 81 + 100 + 8100 = 8281 = 912
com g l a
m . a
ase
agl
Why the rule holds: Put k = n(n + 1). Then
n2 + (n+1)2 + k2 = n2 + n2 + 2n + 1 + k2 = 2n2 + 2n + 1 + k2 = 2k + 1 + k2 = (k + 1)2,
co m
m .
m as e
using 2n2 + 2n = 2n(n + 1) = 2k. So the identity works for every n, not just the rows

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

co m
m .
m as e
.co


a g l Page 29 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q9 How many tiny squares are there in the following picture? Write the prime
factorisation of the number of tiny squares.

Page 11 — the tiled square.

Count in two stages — blocks, then tiny squares inside a block.

The picture is a 9 × 9 arrangement of blocks → 81 blocks

Each block (upright or tilted) is a 5 × 5 grid → 25 tiny squares

Total = 81 × 25 = 2025 tiny squares

Prime factorisation:

Page 30 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

81 = 34, 25 = 52

2025 = 34 × 52
=3×3×3×3×5×5

Why the answer is a square: Every prime here has an even exponent, so 2025 is a
perfect square: 2025 = (32 × 5)2 = 452. The picture shows this directly — a 9 × 9
square of blocks, each block a 5 × 5 square of tiny squares, so the whole picture is a
(9 × 5) × (9 × 5) = 45 × 45 square.

Tip: Tilting a block does not change how many tiny squares it holds, so the
diamonds count exactly like the upright ones.

In-text Questions — Pages 11–12
Section 1.2 Cubic Numbers

MATH TALK

Q1 How many cubes of side 1 cm make a cube of side 2 cm?

8 cubes.

Each layer is 2 × 2 = 4 unit cubes

There are 2 such layers

Total = 2 × 2 × 2 = 23 = 8

Why we multiply three times: A cube has length, breadth and height all equal.
Filling it with unit cubes means choosing 2 positions along each of the three
directions, so the count is 2 × 2 × 2. This is exactly why n × n × n is called "n cubed".

Page 31 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q2 How many cubes of side 1 cm will make a cube of side 3 cm?

27 cubes.

Each layer is 3 × 3 = 9 unit cubes

There are 3 such layers

Total = 3 × 3 × 3 = 33 = 27

Tip: Doubling the side does not double the number of unit cubes — it multiplies it by
8, because each of the three directions doubles. Going from side 2 to side 3 raises
the count from 8 to 27, not from 8 to 12.

Q3 These numbers are called perfect cubes. Can you see why they are named so?

Because each of them counts the unit cubes that fill a cube.

1 = 1 × 1 × 1 → a cube of side 1

8 = 2 × 2 × 2 → a cube of side 2

27 = 3 × 3 × 3 → a cube of side 3

64 = 4 × 4 × 4 → a cube of side 4

Why the name carries over: The same thing happened with squares: 1, 4, 9, 16 are
called squares because they count the unit squares in a square. Here the solid figure
lends its name to the number. Sanskrit made the same choice — ghana means both
the solid cube and the third power.

Q4 Is 9 a cube?

No.

Page 32 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

2×2×2=8

3 × 3 × 3 = 27

9 lies between 8 and 27, with no whole number in between

So 9 is not a perfect cube — and neither is any number from 10 to 26.

Why the gap is so wide: Cubes grow much faster than squares. Between 8 and 27
there is no room for another cube because there is no whole number between 2 and
3. That is why perfect cubes are far rarer than perfect squares: below 100 there are
ten squares but only four cubes (1, 8, 27, 64).

Q5 Can you estimate the number of unit cubes in a cube with an edge length of 4
units?

64 unit cubes.

Each square layer holds 4 × 4 = 16 unit cubes

There are 4 such layers

Total = 16 × 4 = 4 × 4 × 4 = 43 = 64

Tip: Thinking in layers is worth keeping. It turns a three-dimensional count into "area
of one layer × number of layers", which is how volume is measured throughout your
later work.

Page 33 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Complete the table below.
e
Q6

1³ = .1co
m g l as
a
sem
11³ = 1331

a 2³ = 8
agl 12³ =

o m
3³ = 27 13³ = 2197

14³ = 2744 em
. c ag
as
4³ = 64

g l
5³ = 125 15³a=

co m
m.
6³ = 16³ =

m as e
.co
7³ = 17³ = 4913
a g l
a s em8³ =
gl
18³ = 5832
a
9³ = 19³ = 6859

m a s
m.co agl
se
10³ = 20³ =

g l a
a
What patterns do you notice in the table above?

co m
m .
e

m l as
.co g
The completed table of cubes:

e1m= a
a s
agl
3 1 113 = 1331

se m
com a
23 = 8 123 = 1728

. a g l
m
ase
33 = 27 133 = 2197

43 =
agl
64 143 = 2744

m
co3375
53 = 125 153 =
m .
m as e
.co a g l
em
63 = 216 163 = 4096

a s
agl 3 343 173 = 4913

c
7 =

m .
m a s e
co agl
83 = 512 183 = 5832

m .
93 =
l a
729
se 193 = 6859
g
a 1000
103 = 203 = 8000

co m
m .
m ase
.co


a g l Page 34 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Patterns worth naming:

Parity carries over. Odd numbers have odd cubes, even numbers have even cubes.
Every last digit occurs. Unlike squares, cubes end in all ten digits: 1, 8, 7, 4, 5, 6, 3, 2, 9, 0 for
bases 1 to 10.
The units digit determines the base's units digit. A cube ending in 8 has a base ending in
2; ending in 7 → base ends in 3; ending in 3 → base ends in 7; ending in 2 → base ends in 8.
The digits 0, 1, 4, 5, 6, 9 stay put.
Zeros triple. 103 = 1000, 203 = 8000 — one zero in the number becomes three in the cube.
Cubes are sparse. Only 20 cubes reach 8000, while there are 89 squares in the same range.

Why every digit appears: Cubing the digits 0 to 9 gives 0, 1, 8, 27, 64, 125, 216, 343,
512, 729, whose last digits are 0, 1, 8, 7, 4, 5, 6, 3, 2, 9 — all ten, each exactly once. So
a cube's last digit tells you the last digit of its cube root uniquely, which is not true
for squares.

In-text Questions — Page 13
Section 1.2 Cubic Numbers

Q1 We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are
the possible last digits of cubes?

All ten digits, 0 to 9, are possible.

LAST DIGIT OF THE NUMBER 0 1 2 3 4 5 6 7 8 9

LAST DIGIT OF ITS CUBE 0 1 8 7 4 5 6 3 2 9

Why cubes differ from squares: Each of the ten digits gives a different last digit
when cubed, so the map is one-to-one and covers all ten. With squares, 2 and 8 both
give 4, 3 and 7 both give 9, and so on — the map folds two digits onto one and only
six outcomes survive. That is why the units digit rules out many non-squares but
rules out no cube at all.

Tip: This makes guessing cube roots easy. 4913 ends in 3, so its cube root ends in 7;
the root is between 10 and 20 because 103 = 1000 and 203 = 8000, so it must be 17.

Page 35 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q2 Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3
digits? What do you observe?

Count them directly from 13, 23, 33, … and compare with the squares.

DIGITS CUBES HOW SQUARES, FOR HOW
MANY COMPARISON MANY

1 1, 8 2 1, 4, 9 3

2 27, 64 2 16, 25, 36, 49, 64, 81 6

3 125, 216, 343, 512, 5 100, 121, …, 961 22
729

Observation: cubes are much rarer than squares. There are only 9 cubes below 1000, against
31 squares.

Why the thinning happens: The gap between consecutive cubes is (n+1)3 – n3 = 3n2
+ 3n + 1, which grows like n2, while the gap between consecutive squares is only 2n
+ 1. Since the cube gaps widen far faster, fewer of them fit into any fixed stretch of
numbers.

Q3 Can a cube end with exactly two zeroes (00)? Explain.

No. A cube can end with 0 zeros, 3 zeros, 6 zeros … but never exactly 2.

103 = 1000 → 3 zeros

203 = 8000 → 3 zeros

1003 = 1000000 → 6 zeros

Why the count must be a multiple of 3: Write the number as m × 10k where m
does not end in 0. Cubing gives m3 × 103k, and m3 cannot end in 0 either — if it did,
10 would divide m3, forcing both 2 and 5 into m. So the cube ends in exactly 3k
zeros, always a multiple of 3. Two zeros is impossible.

Page 36 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Tip: Compare with squares, where the terminal zeros always come in an even count.
In each case the count of zeros is multiplied by the power you are taking.

Try This — Page 13
Taxicab Numbers

TRY THIS

Q1 The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in
which each of these can be expressed as the sum of two positive cubes.

Search the cubes below each number and look for two that add up.

4104 = 23 + 163 = 8 + 4096

4104 = 93 + 153 = 729 + 3375

13832 = 23 + 243 = 8 + 13824

13832 = 183 + 203 = 5832 + 8000

How to search efficiently: For 4104, the largest possible cube is 163 = 4096, since
173 = 4913 is already too big. So run a from 1 to 16, subtract a3 from 4104, and check
whether what is left is a cube. Only a = 2 and a = 9 work. The same sweep on 13832
with a up to 24 gives a = 2 and a = 18.

Did you know? Hardy's remark that 1729 was "rather a dull number" is what drew
Ramanujan's reply that it is the smallest number expressible as a sum of two cubes
in two different ways: 1729 = 13 + 123 = 93 + 103. Notice that 4104 = 23 × 513 and
13832 = 23 × 1729 — the third taxicab number is just 8 times the first.

In-text Questions — Page 14

Page 37 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Perfect Cubes and Consecutive Odd Numbers; Cube Roots

Q1 Later in this series, we get the following set of consecutive numbers: 91 + 93 + 95 +
97 + 99 + 101 + 103 + 105 + 107 + 109. Can you tell what this sum is without doing the
calculation?

1000, which is 103.

Count the terms: 91, 93, …, 109 → 10 consecutive odd numbers

In the pattern, a run of n consecutive odd numbers adds to n3

So the sum = 103 = 1000

Why it works: The rows use 1, then 2, then 3, … consecutive odd numbers, giving 13,
23, 33, …. A run of 10 must therefore give 103. You can also see it from the middle:
the terms pair off around the centre — 91 + 109 = 200, 93 + 107 = 200, and so on,
five pairs of 200, giving 5 × 200 = 1000.

Tip: Each row starts where the previous one stopped, so the runs never overlap. That
is why adding all the rows up to n gives 13 + 23 + … + n3 as a single unbroken block
of odd numbers.

Q2 Let us check if 3375 is a perfect cube.

Yes, 3375 = 153.

3375 = 3 × 3 × 3 × 5 × 5 × 5

Three identical groups: (3 × 5) × (3 × 5) × (3 × 5)

= (3 × 5)3 = 153

Or as triplets: (3 × 3 × 3) × (5 × 5 × 5) = 33 × 53
Therefore ∛3375 = 15

Page 38 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
Why three groups is the right test: Cubing m repeats every prime of m exactly

m l a se
three times as often. So a perfect cube is a number in which every prime occurs a
o g deals out
.c 3 times, which is the same as saying the whole factorisation
a
m
multiple of
a e identical groups. Dividing each exponent by 3 reads off the cube root.
sthree
ag l
into

co m
e m . ag
as
Q3 Is 500 a perfect cube?

a g l

co m
m.
No.

m as e
500 = 2.c×o2 × 5 × 5 × 5 a g l
a s em
agl The three 5s form a triplet, but only two 2s are available
s
So the factors cannot be split into three identical groups
m a
m .co agl
Therefore 500 is not a perfect cube.
l a se
ag
Why one short exponent is enough to decide: In 500 = 22 × 53, the exponent of 2 is
co m
2, which is not a multiple of 3. No rearrangement can fix that.
m .
m as e
.co a g l
s e m it yourself: 73 = 343 and 83 = 512, so 500 falls between two consecutive
la
Check
ag cubes. Multiplying 500 by 2 gives 1000 = 103, which is the smallest multiplier that
makes it a cube.
se m
com g l a
m . a
gl ase
a
In-text Questions — Page 15
Cube Roots; Successive Differences
co m
m .
m a e
s(iii)
.c o Find the cube roots of these numbers: (i) ∛64 = (ii) ∛
ag
512l = ∛729 =
m
Q1

l a se
ag ANSWER
.c
s e m
m a
co agl
Factorise each into triplets.

m .
as e
a g l

co m
m .
m ase
.co


a g l Page 39 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

(i) 64 = 2 × 2 × 2 × 2 × 2 × 2 = (2 × 2)3 = 43 → ∛64 = 4

(ii) 512 = 29 = (23)3 = 83 → ∛512 = 8

(iii) 729 = 36 = (32)3 = 93 → ∛729 = 9

Why exponents make this quick: Each of these is a power of a single prime, so
taking the cube root just divides the exponent by 3: 26 → 22, 29 → 23, 36 → 32.

Did you know? 64 is both a perfect square (82) and a perfect cube (43), because 64 =
26 and 6 is divisible by both 2 and 3. The next such number is 729 = 36 = 272 = 93.

Q2 Compute successive differences over levels for perfect cubes until all the
differences at a level are the same. What do you notice?

Start from the cubes and keep differencing.

PERFECT CUBES 1 8 27 64 125 216 343

LEVEL 1 7 19 37 61 91 127

LEVEL 2 12 18 24 30 36

LEVEL 3 6 6 6 6

The differences become constant at level 3, and the constant is 6.

Why level 3, and why 6: For squares, one round of differencing turns n2 into 2n + 1
(degree 1) and a second round gives the constant 2. For cubes,
level 1: (n+1)3 – n3 = 3n2 + 3n + 1, a degree-2 expression;
level 2: differencing that gives 6n + 6, a degree-1 expression;
level 3: differencing again gives the constant 6.
Each round of differencing lowers the degree by one, so a third power needs three
rounds. The constant left behind is 3 × 2 × 1 = 6, just as squares leave 2 × 1 = 2.

Page 40 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Tip: Predict before you compute — fourth powers would flatten out at level 4, with
the constant 4 × 3 × 2 × 1 = 24. Try it on 1, 16, 81, 256, 625, 1296.

In-text Questions — Page 16
Section 1.3 A Pinch of History

Q1 Why is the word ‘root’ (the root of a plant) used for the mathematical operation √
(square root, cube root, etc.)?

Because the Sanskrit word mula — the root of a plant, and also basis, cause, origin — was the
term ancient Indian mathematicians used for this operation.

varga = square (the figure, its area, and the second power); varga-mula = square root.
ghana = cube (the solid and the third power); ghana-mula = cube root.
pada (foot, basis, origin) was another word for root. Brahmagupta (628 CE) writes, "The pada
(root) of a krti (square) is that of which it is a square."

Why the image fits: A root is what a thing grows out of. The number 3 is what 9
"grows from" when it is squared, so calling 3 the root of 9 is the same picture as a
plant growing from its root. The word travelled: Arabic took jidhr and Latin took
radix, both meaning the root of a plant, and English "root" and "radical" come from
the Latin.

Did you know? Mula in this mathematical sense has been used in India since at least
the first century BCE, and varga and ghana since at least the third century BCE. The
fourth power had its own name, varga-varga — square of the square.

Figure it Out — Page 16
Section 1.3 A Pinch of History

Q1 Find the cube roots of 27000 and 10648.

Factorise into triplets.

Page 41 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

27000 = 27 × 1000 = 33 × 103 = (3 × 10)3 = 303

∛27000 = 30

10648 = 2 × 5324 = 2 × 2 × 2662 = 2 × 2 × 2 × 1331

1331 = 11 × 11 × 11

10648 = 23 × 113 = (2 × 11)3 = 223
∛10648 = 22

Why you could also guess these: 10648 ends in 8, so its cube root ends in 2. It lies
between 8000 = 203 and 27000 = 303, so the root is between 20 and 30 — and 22 is
the only number in that range ending in 2.

Q2 What number will you multiply by 1323 to make it a cube number?

Factorise and see which prime falls short of a multiple of 3.

1323 = 3 × 441 = 3 × 3 × 147 = 3 × 3 × 3 × 49

= 3 3 × 72

The 3s already form a triplet; the 7s are one short. So multiply by 7.

1323 × 7 = 9261 = 33 × 73 = (3 × 7)3 = 213
∛9261 = 21

Why 7 is the smallest such multiplier: To make a cube, every exponent must
become a multiple of 3. Here 7 has exponent 2, so it needs exactly one more 7 to
reach 3. Nothing smaller can do it, and any larger multiplier would introduce fresh
primes to fix.

Page 42 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Q3 State true or false. Explain your reasoning. (i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may
be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more
digits. (v) Cube numbers have an odd number of factors.

All five are false. Here is why in each case.

STATEMENT VERDICT REASON

(i) Cube of any odd number is even False Odd × odd × odd is odd: 33 = 27, 73 =
343

(ii) No perfect cube ends with 8 False 23 = 8, 123 = 1728, 223 = 10648

(iii) Cube of a 2-digit number may have 3 digits False The smallest is 103 = 1000, already 4
digits

(iv) Cube of a 2-digit number may have 7 or more False The largest is 993 = 970299, only 6 digits
digits

(v) Cube numbers have an odd number of factors False 8 has factors 1, 2, 4, 8 — four of them

Why (i) and (ii) fail: An odd number is 2k + 1, and any product of odd numbers is
odd, so no cube of an odd number can be even. For (ii), the last digit of a cube is
decided by the last digit of the base, and 2 cubed ends in 8 — in fact every number
ending in 2 has a cube ending in 8.

Why (iii) and (iv) fail: Two-digit numbers run from 10 to 99, so their cubes run from
103 = 1000 to 993 = 970299. Every cube in that range has 4, 5 or 6 digits — never 3,
never 7.

Why (v) fails: An odd number of factors is the mark of a square, not a cube. A cube
n3 has an odd number of factors only when it is also a square, as with 64 = 43 = 82
(factors 1, 2, 4, 8, 16, 32, 64 — seven of them). For 8 or 27 the count is even.

Page 43 of 48

as e
Class 8 Maths Chapter 1 A Square and A Cube
a g l AglaSem · NCERT Solutions

co m
m.
You are told that 1331 is a perfect cube. Can you guess without factorisation what
e
Q4

m l as
.co
its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

a g
se m
g l a
a

Yes — use the last digit to fix the units digit, and the size to fix the tens digit.

m
coRANGE
CUBE LAST DIGIT → UNITS DIGIT OF
e m . CUBE
ag
ROOT
g l as ROOT

1331 1→1
a 1000 = 103, 8000 = 203 → root in 10– 11
20
co m
e m.
m l as
.co
4913 3→7 root in 10–20 17

a g
se m 7→3
l a 8000 = 203, 27000 = 303 → root in 20–
12167 23

a g 30

m a s
o agl
32768 8→2 3 3 32

c 30–40
27000 = 30 , 64000 = 40 → root in

m .
gl ase
a
Why the guess is certain, not lucky: Cubing sends each digit to a different last
digit, so the last digit of a cube pins down the units digit of its root uniquely (1→1,
co m
m .
e
3→7, 7→3, 8→2). The size of the number then pins down the tens digit. Together
m l as
.co g
these leave exactly one candidate.
m a
l a se
ag Check it yourself: 113 = 1331, 173 = 4913, 233 = 12167, 323 = 32768. ✓

se m
com g l a
m . a
Which of the followinglis
e
asthe greatest? Explain your reasoning. (i) 67³ – 66³ (ii) 43³ –
Q5
42³ (iii) 67² – 66² (iv) a
g
43² – 42²

com
m .
e

m l as
.co a g
(i) 673 – 663 is the greatest. Use the two difference formulas rather than the full powers.
m
l a se
ag a3 – (a – 1)3 = 3a2 – 3a + 1
.c
s e m
m a
co agl
a2 – (a – 1)2 = 2a – 1
m .
as e
a g l

co m
m .
m as e
.co


a g l Page 44 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

EXPRESSION WORKING VALUE

(i) 673 – 663 3(672) – 3(67) + 1 = 13467 – 201 + 1 13267

(ii) 433 – 423 3(432) – 3(43) + 1 = 5547 – 129 + 1 5419

(iii) 672 – 662 2(67) – 1 = 134 – 1 133

(iv) 432 – 422 2(43) – 1 = 86 – 1 85

Why you can see the answer before computing: A difference of consecutive cubes
grows like 3a2, while a difference of consecutive squares grows only like 2a. So (i)
and (ii) are far bigger than (iii) and (iv), and between the two cube differences the
one with the larger a wins. Since 67 > 43, option (i) is greatest.

Check it yourself: 673 = 300763 and 663 = 287496, and 300763 – 287496 = 13267. ✓

Puzzle Time — Square Pairs! — Page 18
Square Pairs!

TRY THIS

Q1 Look at the following numbers: 3 6 10 15 1. They are arranged such that each pair of
adjacent numbers adds up to a square. 3 + 6 = 9, 6 + 10 = 16, 10 + 15 = 25, 15 + 1 = 16.
Try arranging the numbers 1 to 17 (without repetition) in a row in a similar way —
the sum of every adjacent pair of numbers should be a square.

Here is an arrangement that works:

16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17

Every adjacent pair sums to 9, 16 or 25:

16+9 = 25 9+7 = 16 7+2 = 9 2+14 = 16 14+11 = 25 11+5 = 16

5+4 = 9 4+12 = 16 12+13 = 25 13+3 = 16 3+6 = 9 6+10 = 16

10+15 = 25 15+1 = 16 1+8 = 9 8+17 = 25

Page 45 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

How to find it rather than stumble on it: Two numbers from 1 to 17 add to at least
3 and at most 33, so the only possible square sums are 9, 16 and 25. For each
number, list its allowed neighbours — 1 goes with 8 and 15; 2 goes with 7 and 14; 16
goes with 9 only; 17 goes with 8 only. Then join the numbers up like a chain, starting
from the most restricted ones.

Q2 Can you arrange them in more than one way? If not, can you explain why?

No — apart from reading the same row backwards, the arrangement is the only one.

16 + p is a square only for p = 9 (16 + 9 = 25)

17 + p is a square only for p = 8 (17 + 8 = 25)

Why that forces everything: A number with only one allowed neighbour cannot sit
in the middle of the row, since a middle position needs two neighbours. So 16 and
17 must be the two ends. That fixes the start as 16, 9, … and the finish as …, 8, 17.
From there each step is forced too: after 9 the untried partners are 7 (16) and 16 (25),
but 16 is already used, so 7 must follow — and the same reasoning at every step
leaves exactly one chain. Reversing it gives the same row read the other way, not a
genuinely new arrangement.

Tip: Check the possible partners of a few numbers: 4 pairs with 5, 12; 6 pairs with 3,
10; 13 pairs with 3, 12. Numbers with two choices sit inside the row; the two with one
choice must be the ends.

Q3 Can you do the same with numbers from 1 to 32 (again, without repetition), but this
time arranging all the numbers in a circle?

Yes. Here is one circular arrangement of 1 to 32 in which every neighbouring pair adds to a
square:

Page 46 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2,

23, 26, 10, 15 — and then back to 1

The sums going round are all 9, 16, 25, 36 or 49:

1+8 = 9 8+28 = 36 28+21 = 49 21+4 = 25 4+32 = 36 32+17 = 49

17+19 = 36 19+30 = 49 30+6 = 36 6+3 = 9 3+13 = 16 13+12 = 25

12+24 = 36 24+25 = 49 25+11 = 36 11+5 = 16 5+31 = 36 31+18 = 49

18+7 = 25 7+29 = 36 29+20 = 49 20+16 = 36 16+9 = 25 9+27 = 36

27+22 = 49 22+14 = 36 14+2 = 16 2+23 = 25 23+26 = 49 26+10 = 36

10+15 = 25 15+1 = 16

Why a circle is harder than a row: In a row the two ends need only one neighbour
each, so a number with a single partner can be parked at an end. In a circle every
number needs two neighbours, so no number may have only one allowed partner.
That is why 1 to 17 cannot be closed into a circle — 16 and 17 have one partner each
— while 32 is large enough for every number to have at least two.

Try This: Check that a circle is impossible for 1 to 31: 31 pairs only with 5 and 18,
which is fine, but 30 pairs only with 6 and 19 — trace further and see where the
chain is forced to break. Circular arrangements exist for 32 and for every number
above it.

Chapter at a glance
A locker is toggled once for each factor of its number, so it ends up open only when the
factor count is odd. Factors come in partner pairs, and a pair repeats itself only in a square
— so exactly the square-numbered lockers stay open.
A perfect square is n × n = n2 for a natural number n. Perfect squares end only in 0, 1, 4, 5, 6
or 9, and can carry only an even number of terminal zeros.
The sum of the first n odd numbers is n2. Subtracting 1, 3, 5, … in turn is therefore a test: if
you land exactly on 0 at the k-th step, the number is k2.

Page 47 of 48

Class 8 Maths Chapter 1 A Square and A Cube AglaSem · NCERT Solutions

Prime factorisation settles both questions. A number is a perfect square when its prime
factors split into two identical groups, and a perfect cube when they split into three.
Square root is the inverse of squaring; every perfect square has two integer roots, + and –,
and this chapter uses only the positive one, written √. Cube root is written ∛ and a cube has
just one real cube root.
Successive differences of the squares become constant at level 2 and of the cubes at level 3,
a first taste of how the degree of a pattern shows itself — and Sanskrit named all of this
long ago: varga for square, ghana for cube, mula (root of a plant) for the root operation, the
ancestor of the Arabic jidhr and the Latin radix.

Quick revision

IDEA MEANING TEST / RULE EXAMPLE

Square number n × n, written n2 Factors split into two identical 182 = 324 = (2×3×3)2
groups

Units digit Last digit of a square Only 0, 1, 4, 5, 6, 9 possible 1027 ends in 7 → not a
square

Terminal zeros Zeros at the end of a square Always an even count 4002 = 160000 (4 zeros)

Odd-number 1 + 3 + 5 + … + (2n–1) Equals n2 1+3+5+7+9 = 25 = 52
sum

n-th odd number 2n – 1 36th odd number = 71 352 + 71 = 362 = 1296

Gap between Count of numbers between 2n Between 162 and 172:
squares n2 and (n+1)2 32 numbers

Triangular link Tn–1 + Tn Equals n2 10 + 15 = 25 = 52

Square root x with x2 = y √y; two integer roots, ± √441 = 21

Cube number n × n × n, written n3 Factors split into three 3375 = (3×5)3 = 153
identical groups

Cube units digit Last digit of a cube All ten digits 0–9 occur 23 = 8, 83 = 512

Cube root x with x3 = y ∛y ∛10648 = 22

Successive Repeated differencing Constant at level 2 for squares, Cubes: 6, 6, 6, …
differences level 3 for cubes

Page 48 of 48