NCERT Solutions Class 8 Maths Chapter 3 a Story of Numbers

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 3: A Story of Numbers

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

48 – 81 24 55 English

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 3: A Story of Numbers
Chapter 3 of Ganita Prakash Part I follows Reema and her father back through four thousand years of
counting — sticks and tally marks, the Gumulgal’s twos, Roman letters, Egyptian symbols for powers of 10,
the Mesopotamian base-60 tablets, Mayan dots and bars, Chinese rods — to see how four big ideas
(grouping, landmark numbers, a base, and place value with 0) finally produced the Hindu number system
we write with today.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) 48 – 81

SECTIONS QUESTIONS

24 55

MEDIUM

English

In-text Questions — Page 48
3.1 Reema’s Curiosity

Q1 Since when have humans been counting?

Since long before writing existed — at least from the Stone Age, about ten thousand years ago,
and on the evidence of marked bones very much earlier still.

Lebombo bone (South Africa) — 29 notches — about 44,000 years old

Ishango bone (Democratic Republic of Congo) — notches in columns — 20,000 to 35,000

years old

Stone Age herders counting cattle — about 10,000 years ago

Why it happens: counting does not need writing, or even number names. A herder
who keeps one stick for each cow already has a number system. So counting is far
older than any of the numerals in this chapter — the bones only tell us how far back
the record of counting goes, not when counting began.

Page 1 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q2 What was their need for counting? What were they counting?

They counted whatever they had to keep track of, and whatever they had to predict.

Food: the quantity of grain, fruit and vegetables in store.
Livestock: the number of animals in the herd, so a missing animal could be spotted.
Trade: details of goods exchanged with other groups.
Rituals: the number of offerings given.
Time: the passing days, so that the new moon, the full moon or the onset of a season could
be known and predicted in advance.

Did you know? The Lebombo bone has 29 notches — very close to the 29 or 30 days
of a lunar month. That is why many historians think it was used as a lunar calendar.
Counting days was as important as counting cattle.

Q3 Since when have people been writing numbers in the modern form?

About 2000 years ago, in India.

WHEN WHAT HAPPENED

Ancient (Yajurveda Names of numbers built on powers of 10 — eka, dasha, shata, sahasra, āyuta … up
Samhita) to 1012 and beyond, almost as we say them today

c. 3rd century CE Bakhshali manuscript — the first known writing of numbers with ten digits
including 0, then notated as a dot

c. 499 CE Aryabhata first fully explains the ten-symbol Indian system and computes with it

c. 628 CE Brahmagupta codifies 0 as a number, with negative numbers

c. 800 – 830 CE Transmitted to the Arab world; Al-Khwārizmī and Al-Kindi popularise it

c. 1100 – 1200 CE Reaches Europe; Fibonacci argues for adopting it

By the 17th century Adopted throughout Europe — not adopting it would have blocked scientific
progress

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: notice that the spoken form came first. Indian texts had names
based on powers of 10 thousands of years ago; the written form that matches those
names — one digit per power of 10, with 0 for an empty power — took much longer
to appear, because it needed the invention of 0.

Q4 How would the Mesopotamians have written 20? 50? 100?

The Mesopotamian system is base 60. It has just two marks: a wedge for 1 (we write Y) and a
corner wedge for 10 (we write <). Numbers up to 59 sit in a single position; from 60 onwards a
second position to the left is used.

20 = 10 + 10 → <<

50 = 10 + 10 + 10 + 10 + 10 → <<<<<

100 = (1) × 60 + 40 → Y | <<<<

So 20 and 50 need only the ones position. But 100 is bigger than 60, so it needs two positions:
one 60, and 40 left over.

Why it happens: in a base-60 system no power of 60 may occur 60 or more times —
if it did, sixty of them would be regrouped into the next power. So the number of 1s
is always between 0 and 59, which is exactly why marks for 1 and 10 are enough to
fill one position.

Tip: you meet base 60 every day. 1 hour = 60 minutes and 1 minute = 60 seconds are
the Mesopotamian system still in use.

Math Talk — Page 51

Page 3 of 51

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Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

The Mechanism of Counting
co m
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MATH TALK

.co a g l
a s emdo we ensure that all cows have returned safely after grazing?
gl
How
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Q1

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Keep one stick for every cow, and match them .
a s e sticks against the cows as they come back.

agl drop one stick into a pot. Nothing is counted and
In the morning, as each cow leaves,
nothing is named — one cow, one stick.
In the evening, as each cow returns, take one stick out of the pot.
. c om
s e m over, that
If the pot is empty when the last cow is in, every cow is back. If sticks are left
leftover o
.
m
c collection is the number of missing cows.
a gla
a s em
a gl Why it happens: the sticks and the cows have been put into a one-to-one mapping
— no two cows share a stick and no stick is left unused. Two collections that can be

om a s
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matched one to one have the same size, whether or not anyone can name that size.

s e m and it works without a single number
This is the whole mechanism of counting,
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Do we have fewer cows than our neighbour?
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Yes/no can be settled without knowing how many either of us has.

com g l a
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Make our pile of sticks — one per cow. Let the neighbour make his.
m a
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Now pair the piles off: take one stick from ours and one from his, again and again.

agl
Whichever pile runs out first belongs to the person with fewer cows. If both run out together,
the herds are equal.

. c om
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Why it happens: pairing off tests whether a one-to-one mapping
e m between the two
om is possible. If ours can be matched into his with somea of his left over, ours is
.cherds agl
m the smaller herd. Comparing sizes is therefore an easier problem than measuring
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ag them — you can answer “fewer?” long before you can answer “how many?”.
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a

co m
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m ase
.co


a g l Page 4 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q3 If there are fewer, how many more cows would we need so that we have the same
number of cows as our neighbour?

The sticks left over in the neighbour’s pile after the pairing are the answer.

our sticks ↔ his sticks (pair them off)

ours run out first

the sticks still lying in his pile = the number of extra cows we need

Carry that leftover collection of sticks to the market, and buy one cow for each stick. Now the
two herds can be matched exactly.

Why it happens: this is subtraction, done entirely by matching. The unmatched part
of the bigger collection is the difference between the two collections — the same
idea we later write as 15 − 11 = 4, but here the answer is not a word or a numeral, it
is a heap of sticks that can be used directly.

In-text Questions — Page 52 – 53
The Mechanism of Counting — Methods 1 to 3

MATH TALK

Q1 How will you use such sticks to answer the other two questions (Q2 and Q3)?

By pairing one collection against the other and looking at what is left.

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

STEP WHAT YOU DO WITH THE STICKS WHAT IT TELLS YOU

1 One stick for each of our cows; one stick for each of Two collections that stand for the
the neighbour’s cows two herds

2 Remove one stick from each pile at the same time, A one-to-one pairing is being built
again and again

3 Our pile empties first We have fewer cows (Q2)

4 Count nothing — just keep the sticks still lying in his That collection is how many more cows
pile we need (Q3)

Why it happens: the sticks are a faithful copy of the herd, so any question about the
herd can be turned into a question about the sticks — and sticks can be moved,
paired and set aside, which cows cannot. This is why representing a number by
objects is useful even before number names exist.

Q2 How many numbers can you represent in this way using the sounds of the letters of
your language?

Exactly as many as your language has letters — no more.

English: 26 letters a, b, c, …, z → numbers 1 to 26

Devanagari (Hindi): 11 vowels + 33 consonants = 44 letters → numbers 1 to 44

Tamil: 12 vowels + 18 consonants = 30 basic letters → numbers 1 to 30

Whichever language you take, the answer is a fixed, finite number. Count the letters of your own
alphabet and that is your limit.

Why it happens: in this method each number is matched to one letter, following the
letter-order. A one-to-one mapping can never reach further than the collection it
maps into. So the alphabet, being finite, gives a finite standard sequence — and
numbers are unending. That is the flaw: the method is convenient to count with but
it stops.

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Tip: compare this with the stick system. Sticks never run out of numbers but are
clumsy for large collections; letters are easy to say but run out. The Hindu system is
the one that manages both at once.

Q3 Do you see a way of extending this method to represent bigger numbers as well?
How?

Yes — the way the Roman system itself did it. Keep repeating the symbols, and bring in a fresh
symbol whenever the repetition gets too long.

After XX, carry on: XXI, XXII, … XXIX, XXX (30), … XXXX or XL (40)

Instead of writing XXXXX for 50, give it a new symbol: L

Then LX, LXX, LXXX, XC … and again a new symbol for 100: C

and further: D = 500, M = 1000

So the extension works in two moves: (a) repeat the symbol you already have, and (b) when
about five repetitions have piled up, replace that whole group by a brand-new symbol.

Why it happens: repetition alone is just tally marks — 1000 would need a thousand
strokes. Introducing a new symbol at 5, 10, 50, 100, … keeps every numeral short.
But the price is that each new landmark needs a new symbol, so however many
symbols you invent, some number will still be out of reach. Section 3.4 shows the
way out: use the position of a symbol instead of a new symbol.

Figure it Out — Page 54

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

The Mechanism of Counting

MATH TALK TRY THIS

Q1 Suppose you are using the number system that uses sticks to represent numbers,
as in Method 1. Without using either the number names or the numerals of the
Hindu number system, give a method for adding, subtracting, multiplying and
dividing two numbers or two collections of sticks.

Every one of the four operations can be done by moving sticks about. Call the two collections A
and B.

OPERATION WHAT TO DO WITH THE STICKS WHAT THE ANSWER IS

A+B Push the two heaps together into one heap. The combined heap

A−B Pair each stick of B with a stick of A and take The sticks of A still unpaired
both away. Stop when B is exhausted.

A×B For every single stick of B, lay out one fresh copy The pooled heap
of the whole heap A. Then pool all those copies
together.

A÷B From A, keep pulling out bunches, each bunch The markers are the quotient; the
matched one-to-one with B. Beside each sticks left in A that cannot fill a
completed bunch place one marker stick. bunch are the remainder

Why it happens: each rule is the meaning of the operation, stripped of names.
Addition is putting collections together; subtraction is removing a matched part;
multiplication is repeated addition of equal collections; division is repeated
subtraction of equal collections, and the quotient is how many times you could do it
— which is itself a number, so it too gets its own collection of markers. Nothing here
needs the word “seven”.

Check it yourself: take A = 7 sticks and B = 3 sticks. Pull out bunches of 3: you get
two bunches (two markers) and 1 stick left over. That is 7 ÷ 3 = 2 remainder 1 —
obtained without saying a single number.

Page 8 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
One way of extending the number system in Method 2 is by using strings with more
se
Q2

o m l a
than one letter — for example, we could use ‘aa’ for 27. How can you extend this

m
system.c to represent all the numbers? There are many ways ofagdoing it!
l a se
a g

m
Use all the strings of a given length before moving to the next length, in dictionary order.
. co ag
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LENGTH OF STRING STRINGS
g l as NUMBERS COVERED HOW MANY

1 letter
a
a, b, c, …, z 1 – 26 26

. c26om= 676
m
2 letters aa, ab, ac, …, az, ba, …, zz 27 – 702 2

m as e
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3 letters aaa, aab, …, zzz 703 – 18278 263 = 17576

a s
a gl4 letters aaaa, …, zzzz 18279 – 475254 264 = 456976

m a s
m.co agl
se
26 + 676 = 702

g l a
702 + 17576 = 18278
a
18278 + 456976 = 475254
co m
m .
m ase
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The first two-letter string is aa, and it lands on 27 — exactly as the book suggests.
a g l
a s em
a gl Why it happens: there are 26 strings of length k, and the total 26 + 26 + 26 + …
k 2 3

m
grows without limit. So every number gets a string, and no string is wasted — each

a se
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number has exactly one name and each name belongs to exactly one number. That
. a g
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ase
makes this an unending standard sequence, which the plain 26-letter version was

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Try This: a simpler extension is a, b, …, z, then aa, bb, …, zz, then aaa, bbb, …, zzz. It

s e m instead of 26k, so
om grow very long. Both are correct — the first aisgfar lamore efficient.
also never runs out, but it uses only 26 new names at each stage
. cnumerals
a s em
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m a s e
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g l as
a

co m
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m as e
.co


a g l Page 9 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q3 Try making your own number system.

Here is one worked example; yours may look completely different, and that is fine. Choose three
symbols and let each stand for a group ten times the one before.

●=1 ■ = 10 ★ = 100

NUMBER GROUPED AS MY NUMERAL

7 1+1+1+1+1+1+1 ●●●●●●●

23 10 + 10 + 1 + 1 + 1 ■■●●●

140 100 + 10 + 10 + 10 + 10 ★■■■■
306 100 + 100 + 100 + 1+1+1+1+1+1 ★★★●●●●●●

Before you settle on a system, test it against the three demands the chapter has already made:

Is it unending? Mine is not yet — it stops just short of 1000. I must either add a symbol for
1000 or switch to positions.
Is it easy to count with? Yes — the symbols have a fixed order.
Is it easy to compute with? Yes, because each landmark is 10 times the last, so ten ●
become one ■ and ten ■ become one ★.

Why it happens: choosing landmark numbers that are all powers of one number is
what makes regrouping mechanical. If you had chosen ● = 1, ■ = 7, ★ = 30, the
system would still work for writing numbers, but you would have to remember two
different exchange rates instead of one.

In-text Questions — Page 57 – 58

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

3.2 III. Number Names Obtained by Counting in Twos

MATH TALK

Q1 Quickly count the number of objects in each of the following boxes:

Reading the nine boxes on page 57, row by row:

BOX OBJECTS COULD YOU SEE IT AT A GLANCE?

Hens 2 Yes, instantly

Bunch of yellow flowers too many to say No — you would have to pull the bunch apart

Nesting dolls 4 Yes, just about

Steps of the staircase about 10 No — you have to run your eye up them

Dog 1 Yes, instantly

Bunch of grapes too many to say No

Apples 6 No — most people have to count

Pencils about 8 No — you count them one by one

Pyramids 3 Yes, instantly

Why it happens: the boxes have been chosen deliberately. The ones you can read
off — 1 dog, 2 hens, 3 pyramids, 4 dolls — are all small. The moment a box holds 5
or more you stop seeing the number and start counting it. The grapes and flowers
make the point at the other extreme.

Q2 Up to what group size could you immediately see the number of objects without
counting?

Up to about 4. Most humans find it difficult to take in a group of 5 or more objects in a single
glance.

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: this limit of perception is not a mathematical fact but a fact about
human eyes and brains — and it has left its fingerprints all over the history of
numerals. Once a tally reached five marks nobody could read it at sight, so people
replaced every group of five marks by a single new symbol. That is exactly the
Roman V standing in place of IIIII, and X in place of two Vs.

Did you know? This is why the group sizes that keep reappearing in number
systems are small and human-sized: 2 (the Gumulgal), 5 (the Roman V), 10 (fingers of
two hands — the Egyptian and Hindu systems) and 20 (fingers and toes — the
Mayan system).

Q3 What could be the difficulties with using a number system that counts only in
groups of a single particular size? How would you represent a number like 1345 in a
system that counts only by 5s?

The difficulty is that the numeral stays almost as long as the number itself.

1345 ÷ 5 = 269 (since 5 × 269 = 1345)
So 1345 = 5 + 5 + 5 + … + 5 (269 times)

You would have to write the “five” symbol 269 times. Grouping by 5 has shortened the tally from
1345 marks to 269 — a saving, but not nearly enough. Writing it, reading it and comparing two
such numerals are all still impossible in practice.
The cure is not a bigger group size but a sequence of group sizes, each one 5 times the last:

1 → 5 → 25 → 125 → 625 → 3125

1345 = (2 × 625) + (0 × 125) + (3 × 25) + (4 × 5) + (0 × 1)

= 1250 + 0 + 75 + 20 + 0 = 1345 ✓

as a base-5 numeral: 20340

Now the same number needs only 9 symbols in the picture-form (∿∿ ⬡⬡⬡ □□□□) instead of
269.

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Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: with one group size, the length of a numeral grows in proportion
to the number — double the number and you double the writing. With landmark
numbers that are powers of 5, each new symbol you add multiplies the range by 5,
so the length grows only as fast as the number of powers needed. That is the whole
reason a base is worth having.

Figure it Out — Page 59
3.2 IV. The Roman Numerals

Q1 Represent the following numbers in the Roman system. (i) 1222 (ii) 2999 (iii) 302 (iv)
715

Use the book’s own rule: take as many 1000s as possible, then as many 500s, then 100s, 50s,
10s, 5s and finally 1s.

(i) 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1

= MCCXXII

(ii) 2999 = 1000 + 1000 + 900 + 90 + 9

= MM + CM + XC + IX = MMCMXCIX

(iii) 302 = 100 + 100 + 100 + 1 + 1

= CCCII

(iv) 715 = 500 + 100 + 100 + 10 + 5

= DCCXV

Check by reading back: MCCXXII = 1000 + 200 + 20 + 2 = 1222 ✓ MMCMXCIX = 2000 + 900 + 90
+ 9 = 2999 ✓ CCCII = 300 + 2 = 302 ✓ DCCXV = 500 + 200 + 15 = 715 ✓

Page 13 of 51

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Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: 2999 is the interesting one. The plain grouping rule would give 900

m l a
as DCCCC and 90 as LXXXX and 9 as VIIII, producing MMDCCCCLXXXXVIIII — 17
o se
c the “one less than” shortcut (IV for 4, XL for 40) gives
symbols. .Using a g CM, XC, IX and
m
s8esymbols. The book itself notes that people using this system were not always
g l
onlya
aconsistent about it, so both forms occur in old inscriptions; the short form is the one
in use today.
co m
e m . ag
g l as
a
Tip: a symbol placed before a larger one is subtracted (IX = 10 − 1), and after a larger
one is added (XI = 10 + 1). Read a Roman numeral left to right and watch for a small
letter sitting in front of a big one.
co m
em.
m l as
m .co a g
ase Questions — Page 60
agl
In-text
3.2 IV. The Roman Numerals

m a s
.co agl
TRY THIS

a s em
Q1
agl + LXXVIII
Do it yourself now: (b) LXXXVII

co m
.

e m
m exchange rates are not all the same.
coRoman l as
Pool the symbols of both numerals, then regroup from the smallest upwards — remembering
g
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that the

a s
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LXXXVII = L + X + X + X + V + I + I

a se
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LXXVIII = L + X + X + V + I + I + I
m a
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Pool: L L X X X X X V V I I I I I agl

co m
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Now exchange, one landmark at a time:

com g l as
.EXCHANGE a
a s em BECAUSE LEFT WITH

agl 5 ones make a five L L, X X X X X, V V V
.c
IIIII→V

s e m
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VV→X 2 fives make a ten L L, X X X X X X, V

se m
XXXXX→L
g l a5 tens make a fifty L L L, X, V

LL→C
a 2 fifties make a hundred C, L, X, V

co m
m .
m ase
.co


a g l Page 14 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Sum = C + L + X + V = CLXV

Check: 87 + 78 = 165 = 100 + 50 + 10 + 5 ✓

Why it happens: notice how much care this needs. Going from I to V you exchange
five; from V to X you exchange two; from X to L again five; from L to C again two. In
the Egyptian or Hindu systems every single exchange is “ten of these make one of
those”, so the same regrouping can be done without thinking. That single difference
is why Roman arithmetic needed an abacus and trained specialists.

Q2 How will you multiply two numbers given in Roman numerals, without converting
them to Hindu numerals? Try to find the product of the following pairs of landmark
numbers: V × L, L × D, V × D, VII × IX.

There is no general shortcut, because the Roman landmark numbers are not the powers of any
one number. Each product has to be worked out separately by repeated addition or by the
distributive law.

PRODUCT WORKING ROMAN NUMERAL

V×L 5 fifties = L+L+L+L+L = 250 CCL
= C+C+L

L×D 50 five-hundreds = 25,000 MMMMMMMMMMMMMMMMMMMMMMMMM (M
= 25 thousands written 25 times)

V×D 5 five-hundreds = 2500 = MMD
M+M+D

VII × IX VII × (X − I) = LXX − VII = 70 LXIII
− 7 = 63 = L + X + III

Why it happens: in the Egyptian system, landmark × landmark is always another
landmark (102 × 103 = 105), so a product can be written down the moment you know
the two factors. Here, V × L = CCL is not a landmark at all, and L × D = 25,000 has no
symbol whatever — you are forced to write M twenty-five times. The Roman
landmarks jump ×5, ×2, ×5, ×2, ×5, ×2, so there is no rule to remember and no
pattern to exploit.

Page 15 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Tip: VII × IX is easiest done as VII × IX = VII × (X − I). Since a smaller symbol before a
larger one already means subtraction in this system, the trick is very natural here.

Q3 Multiply CCXXXI and MDCCCLII

First read the two numerals.

CCXXXI = 100 + 100 + 10 + 10 + 10 + 1 = 231
MDCCCLII = 1000 + 500 + 300 + 50 + 2 = 1852

Break the second numeral into its landmark parts and use the distributive law:

231 × 1852 = 231 × (1000 + 800 + 50 + 2)

= 231 × 1000 + 231 × 800 + 231 × 50 + 231 × 2

= 231000 + 184800 + 11550 + 462

= 427812

Check: 231000 + 184800 = 415800; 415800 + 11550 = 427350; 427350 + 462 = 427812 ✓
Now try to write 427812 in Roman numerals. The largest Roman symbol is M = 1000, so you
would need 427 M’s followed by DCCCXII. Later Romans got round this with a bar over a
numeral to mean “times 1000”:

427812 = 427 thousands + 812

427 = CDXXVII, 812 = DCCCXII

so 427812 = C̄ D̄ X̄ X̄ V̄ Ī Ī DCCCXII

Why it happens: this is the joke in the cartoon. Being asked to multiply CCXXXI by
MDCCCLII really was about as pleasant as fighting a lion — not because the
arithmetic is deep, but because the notation gives you no help. In Hindu numerals
you would set out 231 × 1852 in columns and be done in a minute, because every
carry is “ten of these make one of those”.

Page 16 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Figure it Out — Page 60 – 61
3.2 IV. The Roman Numerals

MATH TALK TRY THIS

Q1 A group of indigenous people in a Pacific island use different sequences of number
names to count different objects. Why do you think they do this?

Because in such languages the number word is still tied to the thing being counted — the count
has not yet been separated from what is counted.

The name carries extra information. One series for coconuts, another for canoes, another
for fish: the word tells the listener both how many and of what, so nothing has to be said
twice.
Different goods are handled in different natural groups. Fish may come in pairs, coconuts
in bunches, yams by the basket. A counting series that grew out of the way a good is actually
stacked and traded is the most convenient one for that good.
Status and ceremony. A separate series was often kept for people, or for objects used in
rituals, marking them out as different from ordinary trade goods.

Why it happens: the mathematically important point is what such a system cannot
do. If “three” for fish and “three” for coconuts are different words, then 3 fish + 3
coconuts cannot be added inside the language — the system has no idea of “three”
by itself. Realising that the threeness of three fish and three coconuts is one and the
same thing is the step that makes arithmetic possible. Our own language still keeps
a trace of the older habit: we say a pair of shoes, a brace of birds, a dozen eggs.

Page 17 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q2 Consider the extension of the Gumulgal number system beyond 6 in the same way
of counting by 2s. Come up with ways of performing the different arithmetic
operations (+, –, ×, ÷) for numbers occurring in this system, without using Hindu
numerals. Use this to evaluate the following: (i) (ukasar-ukasar-ukasar-ukasar-
urapon) + (ukasar-ukasar-ukasar-urapon) (ii) (ukasar-ukasar-ukasar-ukasar-urapon)
– (ukasar-ukasar-ukasar) (iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-
ukasar) (iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-
ukasar)

In this system urapon = 1 and ukasar = 2, and a number is simply a string: as many ukasar as
there are 2s in it, followed by one urapon if it is odd. Extending past 6 needs no new word at all
— just longer strings.
The four rules, stated inside the system:

Add: write the two strings one after the other; if two urapon now appear, replace them by
one ukasar. (A string never needs more than one urapon.)
Subtract: cancel ukasar against ukasar and urapon against urapon. If the larger number has
no urapon to cancel, first split one of its ukasar into two urapon.
Multiply: replace each ukasar of the second number by a full copy of the first number, and
each urapon by half a copy — or more simply, write the first number out as many times as
the second number says, then tidy up the urapons.
Divide: break the string into equal bunches, each bunch matching the divisor; the number of
bunches is the answer.

PART IN 2S AND 1S WORKING ANSWER

(i) (2+2+2+2+1) + 7 ukasar and 2 urapon; the two urapon ukasar-ukasar-ukasar-
(2+2+2+1) make one more ukasar → 8 ukasar ukasar-ukasar-ukasar-
=9+7 ukasar-ukasar (16)

(ii) (2+2+2+2+1) − Cancel three ukasar; one ukasar and one ukasar-urapon (3)
(2+2+2) urapon remain
=9−6

(iii) (2+2+2+2+1) × Write the 9-string four times: 16 ukasar ukasar repeated 18 times
(2+2) and 4 urapon; the four urapon make 2 (36)
=9×4 more ukasar → 18 ukasar

(iv) (8 ukasar) ÷ (2 Break the 8 ukasar into bunches of 2 ukasar-ukasar (4)
ukasar) ukasar: 4 bunches
= 16 ÷ 4

Page 18 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: every rule here is just the tally rule with one extra exchange — “two

m l a se
urapon make one ukasar”. That single exchange halves the length of a numeral,
o
which is a.creal gain, but no further exchange is ever available: 100
a gstill needs fifty
m
se Counting in a single group size takes you exactly one step beyond tally marks
l a
g no further.
ukasar.
aand

com
e m . ag
g l as
a
Q3 Identify the features of the Hindu number system that make it efficient when
compared to the Roman number system.

co m
e m.
m l as
.co a g
Five features, and each of them fixes a specific Roman weakness.

a s em
a glFEATURE HINDU NUMBER SYSTEM ROMAN NUMBER SYSTEM

a s
com
Place value — the 3 in 375 means 3 × Fixed value — X is 10 wherever it stands

agl
Value of a
102, the 3 in 32 means 3 × 10
.
symbol
m
ase
agl
A symbol for 0, used as a digit and as a number No symbol for zero at all
nothing

How many Ten symbols write every number,
. c om
A new symbol needed for each new
however large
s e mD M …
landmark: I V X L C
m a
symbols

m . co agl×5, ×2, ×5, ×2 — no pattern,
asenumbers
Landmark All powers of 10, so landmark × Ratios jump

agl
landmark is again a landmark so no multiplication rule

Length of a Grows very slowly — 1888 needs 4 Grows fast — 1888 = MDCCCLXXXVIII, 13
se m
com g l a
.
numeral digits symbols

m a
ase
agl
Why it happens: the deepest of these is place value together with 0. Because each
position already announces which power of 10 it counts, no symbol has to be

. com
invented for 108 or 1015; and because 0 can hold an empty position, 305 cannot be

m a s em long
misread as 35. Everything else — short numerals, column addition,

. o
cmultiplication a gl
em
and division — follows from those two ideas.

as
agl c
Check it yourself: add 1888 and 1976 in Roman numerals, then in Hindu numerals.
m .
m a s e
co agl
The second takes a few seconds; the first takes several exchanges and a lot of care.

m .
ase
a g l

com
m .
m ase
.co


a g l Page 19 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q4 Using the ideas discussed in this section, try refining the number system you might
have made earlier.

Take the system from page 54 (● = 1, ■ = 10, ★ = 100) and put it through the two refinements the
chapter has introduced so far.
Refinement 1 — make the landmark numbers powers of one number. Ours already are: 1,
10, 100. So extend the same way instead of inventing unrelated jumps: the next landmarks must
be 1000, 10000, 100000, and so on.
Refinement 2 — check the two tests the chapter has set.

TEST BEFORE AFTER REFINING

Is the sequence unending? No — stops below 1000 Yes, once every power of 10 has its
own symbol

Is regrouping simple? Yes — ten ● make one ■ Yes, and now the same rule holds at
every level

Is a landmark × a landmark a Yes — ■ × ■ = ★ Yes, so multiplication has a rule
landmark?

Does a symbol ever repeat 10 No — ten of any symbol become No — so at most 9 of each
times? one of the next

One problem survives: an unending supply of new symbols is still needed. That is exactly the
Egyptian shortcoming of page 69, and the fix is on page 73 — stop drawing the landmark
symbol and let the position say which landmark you mean. Since no symbol ever repeats more
than 9 times, exactly nine marks plus a mark for “none” — ten symbols in all — are enough.
Refined all the way, your system has turned into the Hindu number system.

Why it happens: this is the whole argument of the chapter in miniature. Grouping
shortens numerals; a base makes regrouping and multiplying mechanical; place
value removes the need for endless symbols; and 0 is what place value needs to be
unambiguous.

Figure it Out — Page 62

Page 20 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

3.3 I. The Egyptian Number System

Q1 Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784,
1111, 70707.

The Egyptian symbols run through the powers of 10:

|=1 ∩ = 10 coil = 102 lotus = 103 finger = 104 tadpole = 105 kneeling man = 106

sun = 107

Group each number into powers of 10 and draw that many of each symbol.

NUMBER GROUPED AS SYMBOLS NEEDED

10458 10000 + 400 + 50 + 8 1 finger, 4 coils, 5 arches, 8 strokes

1023 1000 + 20 + 3 1 lotus, 2 arches, 3 strokes

2660 2000 + 600 + 60 2 lotus, 6 coils, 6 arches

784 700 + 80 + 4 7 coils, 8 arches, 4 strokes

1111 1000 + 100 + 10 + 1 1 lotus, 1 coil, 1 arch, 1 stroke

70707 70000 + 700 + 7 7 fingers, 7 coils, 7 strokes

Why it happens: compare the last two lines with the Hindu numerals 1111 and
70707. The Egyptians wrote as many copies as the digit says, in any order, and had no
way to show a zero — but they did not need one, because a missing power of 10
simply means that symbol is absent. In 1023, no coil is drawn; in 70707, no lotus and
no arch. The digit 0 becomes necessary only when the symbols are dropped and the
position has to do the work.

Tip: the number of symbols you draw is the sum of the digits. 70707 needs 7 + 0 + 7
+ 0 + 7 = 21 symbols, while the Hindu numeral needs 5. That is the price the Egyptian
system pays.

Page 21 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q2 What numbers do these numerals stand for?

Count each kind of symbol on page 62 and multiply by its landmark value.

SYMBOLS SHOWN VALUE NUMBER

(i) 2 coils, 7 arches, 6 strokes 2 × 100 + 7 × 10 + 6 × 1 276
= 200 + 70 + 6

(ii) 4 lotus, 3 coils, 2 arches, 2 strokes 4 × 1000 + 3 × 100 + 2 × 10 + 2 × 1 4322
= 4000 + 300 + 20 + 2

Why it happens: reading an Egyptian numeral is pure addition — you never have to
worry about where a symbol sits, only how many of each there are. That makes it
foolproof to read but wasteful to write: 4322 takes 11 symbols here and 4 digits in
our system. Notice also how naturally the counts 4, 3, 2, 2 line up with the digits of
4322 — the Egyptian system is base 10 without place value, and the Hindu system is
base 10 with it.

In-text Questions — Page 62
3.3 II. Variations on the Egyptian System and the Notion of Base

Q1 Instead of grouping together 10 collections of size equal to the previous landmark
number (as in the case of the Egyptian system), can we get a number system by
grouping together 5 collections of size equal to the previous landmark number? Can
this 5 be replaced by any positive integer?

Yes to the first question, and almost yes to the second.
Grouping in 5s. Start with 1 as the first landmark and multiply by 5 each time:

50 = 1 51 = 5 52 = 25 53 = 125 54 = 625 55 = 3125

These are all powers of 5, so we get a perfectly good base-5 system, with symbols △ □ ⬡ ○ ∿ ↑
for the six landmarks shown.

Replacing 5 by any n. The same construction works for any n: landmarks 1, n, n2, n3, … This is
exactly the definition of a base-n number system, and the Egyptian system is the case n = 10.

Page 22 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: one positive integer must be ruled out — n = 1. Then the landmark
numbers are 1, 1 × 1, 1 × 1 × 1, … which are all just 1. Nothing new is ever created, so
a “base-1 system” is nothing but tally marks. So the correct statement is: 5 may be
replaced by any integer greater than 1.

Did you know? Every computer you use works in base 2, with landmark numbers 1,
2, 4, 8, 16, … and only two digits. A smaller base means fewer symbols to remember
but longer numerals — the number 25 is “25” in base 10 and “11001” in base 2.

In-text Questions — Page 63
3.3 II. Variations on the Egyptian System and the Notion of Base

Q1 Express the number 143 in this new system.

Start from the largest landmark number that is smaller than 143. That is 53 = 125, since the next
one, 625, is too big.

143 − 125 = 18

18 = 5 + 5 + 5 + 3 (three 5s, then 3 ones)

So 143 = 125 + 5 + 5 + 5 + 1 + 1 + 1

Numeral: ○ □□□ △△△

Check: 125 + 15 + 3 = 143 ✓

Written as base-5 digits, the same number is

143 = (1)×125 + (0)×25 + (3)×5 + (3)×1 = 1033 in base 5

Why it happens: the hexagon ⬡ (25) does not appear at all, because after taking
away 125 only 18 is left and 18 < 25. In the picture form its absence is simply a
symbol not drawn; in the digit form the same absence must be written as the digit 0.
That one difference is the gap between the Egyptian idea and place value.

Page 23 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: no symbol appears five or more times here. If it ever did, five of

as e
om 0 and 4. l
them would be swapped for one of the next symbol up — which is why each base-5
digit stays.cbetween a g
a s em
agl
Figure it Out — Page 63
. com ag
a s emNotion of Base
3.3 II. Variations on the Egyptian System and the

agl
Write the following numbers in the above base-5 system using the symbols in Table
m
Q1

co
m.
2: 15, 50, 137, 293, 651.

m as e
.co a g l
se m
g l a 5⁰ = 1 5¹ = 5 5² = 25 5³ = 125 5⁴ = 625 5⁵ = 3125

a
m a s
m.co agl
l a se
a g
The base-5 symbols built on page 63 — one symbol for each landmark number.

co m
m .
ase

m l
.co a g
The symbols are △ = 1, □ = 5, ⬡ = 25, ○ = 125, ∿ = 625, ↑ = 3125. Each time, take as many of the
m
l a se
largest possible landmark as you can, then move down.

ag
se m
NUMBER GROUPING NUMERAL BASE-5 DIGITS

com g l a
m. a
5+5+5 30

ase
15 □□□

50 25 + 25
agl ⬡⬡ 200

○ □□ △△

com
137 125 + 5 + 5 + 1 + 1 1022

.
em
⬡ □□□ △△△
m las
293 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1 ○○ 2133

. co ∿ ⬡a△g
e m
las
651 625 + 25 + 1 10101

ag c
Checks: 15 = 3 × 5 ✓ 50 = 2 × 25 ✓ 125 + 10 + 2 = 137 ✓ 250 + 25 + 15 + 3 = 293 ✓ 625 + 25 +
m .
m a s e
agl
1 = 651 ✓
. co
e m
g l as
a

co m
m .
m as e
.co


a g l Page 24 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: taking the largest landmark first is not just habit — it is what
guarantees that no symbol ever has to be drawn five times. If for 651 you began
with 125s you would need five of them, and five ○ are one ∿, so you would be forced
to regroup anyway. Being greedy from the top gets the shortest numeral straight
away.

Tip: 651 is striking — a number in the six hundreds written with just three symbols,
because 651 = 625 + 25 + 1 sits almost exactly on the powers of 5.

Q2 Is there a number that cannot be represented in our base-5 system above? Why or
why not?

Yes — zero. There is no symbol for it, and there is no way to build it, because a numeral here is a
collection of symbols and zero would have to be an empty collection, which cannot be seen on
the page.
Every counting number 1, 2, 3, … can be written, and here is the argument:

Take any number N.

Pick the largest landmark 5k with 5k ≤ N. Draw that symbol and subtract.
The leftover is smaller than before, so the process must stop.

When it stops, nothing is left over — every part has been drawn.

Provided you never use a symbol five or more times, that numeral is also the only one for N.

Why it happens: notice that the same objection applies to the Egyptian and Roman
systems — none of them can write zero either, and none of them needed to, because
they only ever recorded quantities that were actually there. Zero becomes
indispensable only in a place value system, where an empty position must be shown
so that 305 is not read as 35. That is why the invention of 0 belongs to the last stage
of the story, not the first.

Did you know? Negative numbers and fractions cannot be written in this system
either. Brahmagupta’s work of 628 CE, which gave 0 and the negative numbers the
full status of numbers, is what opened all of that up.

Page 25 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q3 Compute the landmark numbers of a base-7 system. In general, what are the
landmark numbers of a base-n system?

Start at 1 and multiply by 7 each time.

70 = 1

71 = 7

72 = 49

73 = 343

74 = 2401

75 = 16807 …

So the landmark numbers of a base-7 system are 1, 7, 49, 343, 2401, 16807, …
In general, the landmark numbers of a base-n system are the powers of n:

n0 = 1, n, n2, n3, n4, …

Why it happens: the definition on page 63 says the first landmark is 1 and each next
one is the current one multiplied by n. Starting at 1 and multiplying by n repeatedly
is exactly what taking powers of n means. So “base-n” and “landmarks are the
powers of n” are two ways of saying the same thing — and this is the property that
will make landmark × landmark = landmark on page 67.

Check it yourself: 7 × 343 = 2401 and 49 × 49 = 2401 as well. Two different pairs of
landmarks give the same landmark, because 71 × 73 and 72 × 72 both come to 74.

Figure it Out — Page 65

Page 26 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

3.3 Advantages of a Base-n System

Q1 Add the following Egyptian numerals: (i) … and … (ii) … and …

(i) Read the two numerals on page 65 first.

First: 9 lotus, 6 coils, 8 strokes = 9000 + 600 + 8 = 9608

Second: 5 coils, 7 strokes = 500 + 7 = 507

Pool the symbols and regroup from the smallest upwards — ten of any symbol make one of the
next.

SYMBOL TOTAL REGROUPING KEPT

strokes (1) 8 + 7 = 15 10 strokes → 1 arch 5 strokes, carry 1 arch

arches (10) 0+0+1=1 — 1 arch

coils (100) 6 + 5 = 11 10 coils → 1 lotus 1 coil, carry 1 lotus

lotus (1000) 9 + 1 = 10 10 lotus → 1 finger 0 lotus, carry 1 finger

finger (10000) 1 — 1 finger

Sum = 1 finger + 1 coil + 1 arch + 5 strokes

= 10000 + 100 + 10 + 5 = 10115
Check: 9608 + 507 = 10115 ✓

(ii)

First: 1 lotus, 8 arches = 1000 + 80 = 1080

Second: 4 arches, 6 strokes = 40 + 6 = 46

Page 27 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

strokes: 0 + 6 = 6

arches: 8 + 4 = 12 → 1 coil and 2 arches

coils: 0 + 1 (carried) = 1
lotus: 1

Sum = 1 lotus + 1 coil + 2 arches + 6 strokes = 1000 + 100 + 20 + 6 = 1126

Check: 1080 + 46 = 1126 ✓

Why it happens: this is column addition, and the girl on page 66 is right to notice
the resemblance. “Ten strokes make an arch, ten arches make a coil” is exactly “carry
1 from the ones column to the tens column”. It works because every landmark is 10
times the one below, so the same carrying rule serves at every level — which is
precisely what fails in the Roman system.

Q2 Add the following numerals that are in the base-5 system that we created.
Remember that in this system, 5 times a landmark number gives the next one!

Read the two numerals from page 65:

○ ⬡⬡ □ △△ = 125 + 25 + 25 + 5 + 1 + 1 = 182

○○○ ⬡ □□ △△ = 375 + 25 + 5 + 5 + 1 + 1 = 412

Now pool them, symbol by symbol:

SYMBOL VALUE FIRST + SECOND TOTAL REGROUP?

△ 1 2+2 4 No, 4 < 5

□ 5 1+2 3 No

⬡ 25 2+1 3 No

○ 125 1+3 4 No

Page 28 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
⬡⬡⬡ □□□ △△△△
e
Sum = ○○○○
m l as
.co
= 4 × 125 + 3 × 25 + 3 × 5 + 4 × 1
m a g
l a se
g
= 500 + 75 + 15 + 4 = 594
aCheck: 182 + 412 = 594 ✓

com
m . ag
l a se
Why it happens: this sum happens to need no regrouping at all — every column
ag reached 5, five of that symbol would have been
came to 4 or less. Had any column
swapped for one of the next symbol up, exactly as ten strokes became an arch in Q1.

co m
m.
The reminder in the book is a warning to check every column, not a hint that a carry

m as e
l
is hiding here.
.co a g
a s em
a gl Try This: add ○○○ □□□ △△△ (375 + 15 + 3 = 393) to ○○ □□ △△△ (250 + 10 + 3 = 263).
Now the triangles come to 6, so five of them become a □. The answer is 656; check

m a s
.co agl
that your symbols give 5×125 + 0×25 + 6×5 + 1, regrouped to ∿ ⬡ □ △ = 625 + 25 + 5
+ 1 = 656.
se m
g l a
a

co m
In-text Questions — Page 66
m .
m as e
.co l
3.3 Advantages of a Base-n System

a g
a s em How to multiply two numbers in Egyptian numerals?
a gl Q1
se m
com g l a
. a

m
ase
Split each numeral into the landmark numbers it is made of, multiply every landmark of the first

agl
by every landmark of the second, then pool the results and regroup.

co m
.
(a + b) × (c + d) = ac + ad + bc + bd (the distributive law)

se m
o m l a
g pages:
m .c whole method rests on one fact, established on the nextatwo
se
The

a
agl c
10a × 10b = 10a+b — the product of two landmarks is again a landmark
m .
s e
. c om a g la
So each of the small products iseam
a s single symbol, and the only work left is counting how many of

agl up with.
each symbol you have ended

co m
m .
m ase
.co


a g l Page 29 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: in the Roman system the same plan collapses at the first step,
because V × L = 250 is not a landmark and L × D = 25000 has no symbol at all. Having
landmarks that are powers of one number is what turns multiplication from a special
problem into a routine.

Q2 What is any landmark number multiplied by ∩ (that is 10)? Find the following
products — (i) ∩ × ∩ (ii) coil × ∩ (iii) lotus × ∩ (iv) finger × ∩

Multiplying a landmark by 10 gives the next landmark, because it raises the power of 10 by one.

10a × 10 = 10a+1

PRODUCT IN POWERS OF 10 ANSWER

(i) ∩×∩ 10 × 10 = 102 100 — the coil

(ii) coil × ∩ 102 × 10 = 103 1000 — the lotus

(iii) lotus × ∩ 103 × 10 = 104 10,000 — the finger

(iv) finger × ∩ 104 × 10 = 105 1,00,000 — the tadpole

Why it happens: 10a means 10 multiplied by itself a times. Multiplying by one more
10 makes it a + 1 tens multiplied together, which is 10a+1. So the rule is not a
coincidence of Egyptian drawing — it is the law of exponents, and it is the reason the
symbol list is ordered the way it is.

Q3 What is any landmark number multiplied by the coil (10²)? Find the following
products — (i) ∩ × coil (ii) coil × coil (iii) lotus × coil (iv) finger × coil

Multiplying by 102 moves a landmark up by two steps.

10a × 102 = 10a+2

Page 30 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

PRODUCT IN POWERS OF 10 ANSWER

(i) ∩ × coil 10 × 102 = 103 1000 — the lotus

(ii) coil × coil 102 × 102 = 104 10,000 — the finger

(iii) lotus × coil 103 × 102 = 105 1,00,000 — the tadpole

(iv) finger × coil 104 × 102 = 106 10,00,000 — the kneeling man

Why it happens: multiplying by 102 is multiplying by 10 twice over, so each answer
here is two symbols further along the list than the answers in the previous question.
In Hindu numerals the same rule reads “multiplying by 100 puts two zeroes at the
end”.

In-text Questions — Page 67
3.3 Advantages of a Base-n System

MATH TALK

Q1 Find the following products — (i) ∩ × tadpole (ii) coil × lotus (iii) lotus × lotus (iv)
finger × kneeling man

Add the powers of 10 each time.

PRODUCT IN POWERS OF 10 ANSWER

(i) ∩ × tadpole 101 × 105 = 101+5 106 — the kneeling man

(ii) coil × lotus 102 × 103 = 102+3 105 — the tadpole

(iii) lotus × lotus 103 × 103 = 103+3 106 — the kneeling man

(iv) finger × kneeling man 104 × 106 = 104+6 1010

Thus the product of any two landmark numbers is another landmark number.

Page 31 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: part (iv) is worth pausing over. 1010 is a landmark number all right,
but the Egyptian symbol list stops at 107 (the sun). So the answer exists as a number
but cannot be written down in Egyptian numerals. This is precisely the shortcoming
set out on page 69: a system that gives each landmark its own symbol needs an
unending supply of symbols.

Tip: 1010 is one thousand crore, or 1 followed by ten zeroes. In our system it is no
trouble at all — the same ten digits handle it.

Q2 Does this property hold true in the base-5 system that we created? Does this hold
for any number system with a base?

Yes in the base-5 system, and yes in every base-n system.

In base 5: ⬡ × □ = 25 × 5 = 125 = ○ (52 × 51 = 53)

○ × □ = 125 × 5 = 625 = ∿ (53 × 51 = 54)

⬡ × ⬡ = 25 × 25 = 625 = ∿ (52 × 52 = 54)

In general: na × nb = na+b

Why it happens: na is n written down as a factor a times, and nb is n written down b
times. Multiply them and you have n written down a + b times — which is na+b, again
a power of n, and therefore again a landmark number. Nothing about 10 or 5 was
used; the argument is about counting factors, so it holds for every base.

Check it yourself: now compare V × L = CCL in the Roman system. Roman
landmarks are 1, 5, 10, 50, 100, 500, 1000 — the ratios are 5, 2, 5, 2, 5, 2, so they are
not the powers of any one number, and the property fails. That single failure is the
whole reason Roman multiplication is hard.

Page 32 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q3 What can we conclude about the product of a number and ∩ (10), in the Egyptian
system?

Every symbol in the numeral is replaced by the next symbol up, and the number of each kind
stays the same.
The book shows why with two examples.

(i) coil coil × ∩

coil coil is the same as coil + coil, so

(coil + coil) × ∩ = (coil × ∩) + (coil × ∩) (distributive law)

= lotus + lotus = lotus lotus
In numbers: (100 + 100) × 10 = 1000 + 1000 = 2000 ✓

(ii) coil ∩∩ | × ∩

coil ∩∩ | is the same as coil + ∩∩ + |, so

(coil + ∩∩ + |) × ∩ = (coil × ∩) + (∩∩ × ∩) + (| × ∩)
= lotus + coil coil + ∩

= lotus coil coil ∩

In numbers: 121 × 10 = 1210 ✓

Why it happens: the distributive law lets you multiply each part separately, and
each part is a landmark, so each part simply steps up one place in the symbol list.
The counts are untouched — two coils become two lotus, two arches become two
coils. In Hindu numerals this same statement reads: “to multiply by 10, write a 0 at
the end.” 121 becomes 1210, and the digits 1, 2, 1 have not changed at all, they have
only moved one place left.

In-text Questions — Page 68

Page 33 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

3.3 Advantages of a Base-n System
co m
e m.
m l as
Q1
.co a g
Now find the following products — (i) (coil coil coil coil coil coil ∩∩ ||) × ∩ (ii) (lotus
m
l a se
∩) × ∩

a g

com
. ag
(i) Read the numeral in the bracket first: 6 coils, 2 arches, 2 strokes.
e m
g l as
= 600 + 20 + 2 = 622 a
622 × 10 = 6220
co m
em.
m l as
.co
Symbol by symbol, each one steps up:
m a g
l a se
a g WAS BECOMES VALUE

a s
com agl
6 coils (600) 6 lotus 6000

m .
ase
2 arches (20) 2 coils 200

2 strokes (2) agl 2 arches 20

co m
m .
m
Product = 6 lotus, 2 coils, 2 arches = 6000 + 200 + 20 = 6220 ✓
as e
.co a g l
a s em
gl
(ii) lotus ∩ = 1000 + 10 = 1010
a
se m
1010 × 10 = 10100
com g l a
m . a
ase
lotus → finger (1000 → 10,000)

∩ → coil (10 → 100) agl
Product = 1 finger, 1 coil = 10000 + 100 = 10100 ✓
co m
m .
o m l a se
ag has an empty hundreds
.cWhy it happens: in part (ii) notice that the numeral 1010
m place and an empty ones place, and after multiplying, the product 10100 has empty
l a se
ag thousands, tens and ones places. In Egyptian writing those gaps cost nothing — the
.c
s e m
m a
symbol is simply not drawn. It is only when the symbols are dropped in favour of

m . co
positions that these gaps must be filled with the digit 0.
e agl
g l as
a

co m
m .
m ase
.co


a g l Page 34 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q2 What would be a simple rule to multiply a number with ∩?

Replace every symbol by the next symbol in the list, keeping the count of each symbol
exactly as it was.

stroke → arch → coil → lotus → finger → tadpole → kneeling man → sun

In Hindu numerals the very same rule reads: write a 0 at the end.

4 coils + 3 arches + 7 strokes = 437

↓ each symbol steps up
4 lotus + 3 coils + 7 arches = 4370 = 437 × 10 ✓

Why it happens: by the distributive law the number splits into its landmark parts,
and by the law of exponents each landmark multiplied by 10 becomes the next
landmark. Nothing merges and nothing is created, so the counts survive untouched
— which is why the numeral looks the same, just shifted one place along.

Tip: the rule has one limit. The Egyptian list stops at the sun, 107. Multiply a numeral
containing suns by 10 and there is no next symbol to step up to. In our system there
is no such wall, because the next position is always available.

In-text Questions — Page 69
3.3 Abacus that Makes Use of the Decimal System

Q1 How would you use this to find the sum?

Slide the counters from the right of the vertical partition across onto the matching lines on the
left, then tidy up each line.
On page 69 the board holds 2907 on the left and 43 on the right:

Page 35 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

LINE 2907 HAS 43 HAS AFTER BRINGING TOGETHER VALUE

1000 2 counters on the line — 2 on the line 2000

100 1 above (500) + 4 on the — 1 above + 4 on the line 900
line (400)

10 — 4 4 + 1 carried = 5 → replaced by 1 counter 50
counters above the line

1 1 above (5) + 2 on the line 3 7 + 3 = 10 → cleared, 1 counter added to 0
(2) counters the 10s line

Sum = 2000 + 900 + 50 + 0 = 2950

Check: 2907 + 43 = 2950 ✓

Why it happens: the abacus is a decimal place value machine made of wood. Each
line is a power of 10 and the counters on it are the digit — so pushing the two sets of
counters together is addition, and the exchanges keep every digit below 10. Its
whole design carries the base-10 idea, which is why people who wrote in Roman
numerals still calculated in base 10.

Q2 The counters along each line were brought together. What is to be done if the total
in a line exceeded 10?

Two exchanges keep the board readable:

Ten counters on a line are taken off and replaced by one counter on the next line up —
the next power of 10.
Five counters on a line are taken off and replaced by one counter just above that line,
since a counter above a line stands for 5.

In 2907 + 43: 7 ones + 3 ones = 10 ones

→ clear the 1s line, put 1 counter on the 10s line

10s line: 4 + 1 = 5 counters → replaced by 1 counter above the 10s line (= 50)

Page 36 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Why it happens: this is exactly the carrying you do in column addition, made visible.
“Ten of these make one of those” is possible on every line because the lines are
successive powers of 10 — the very property that defines a base. Try the same board
with Roman exchanges (five I make a V, two V make an X, five X make an L) and you
would need a different rule on almost every line.

Figure it Out — Page 69 – 70
3.3 III. Shortcomings of the Egyptian System

MATH TALK

Q1 Can there be a number whose representation in Egyptian numerals has one of the
symbols occurring 10 or more times? Why not?

No — not in the tidied-up form that the Egyptians actually wrote.

10 copies of a landmark 10a = 10 × 10a = 10a+1

and 10a+1 has a symbol of its very own.

So ten identical symbols would always be swapped for one symbol of the next kind. Each symbol
can therefore appear at most 9 times.

Why it happens: this is the same fact that fixes the range of our digits. A digit in the
Hindu system counts how many of that power of 10 there are, and the answer can
never be 10 or more — so the digits run 0 to 9 and no further. Nine is not an
arbitrary stopping point; it is one less than the base.

Tip: there is one place where the argument breaks. The topmost symbol is the sun,
107. Ten suns would be 108, and no symbol exists for that — so for numbers of ten
crore and above, the sun really would have to be drawn ten or more times. That is
exactly the shortcoming this section is about.

Page 37 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q2 Create your own number system of base 4, and represent numbers from 1 to 16.

Take three symbols for the first three landmarks of base 4:

∟ = 40 = 1 △ = 41 = 4 □ = 42 = 16

NUMBER GROUPING NUMERAL BASE-4 DIGITS

1 1 ∟ 1

2 1+1 ∟∟ 2

3 1+1+1 ∟∟∟ 3

4 4 △ 10

5 4+1 △∟ 11

6 4+2 △∟∟ 12

7 4+3 △∟∟∟ 13

8 4+4 △△ 20

9 8+1 △△∟ 21

10 8+2 △△∟∟ 22

11 8+3 △△∟∟∟ 23

12 4+4+4 △△△ 30

13 12 + 1 △△△∟ 31

14 12 + 2 △△△∟∟ 32

15 12 + 3 △△△∟∟∟ 33

16 16 □ 100

Page 38 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: read down the numeral column and you can see the base at work

m l a se
— no symbol ever appears four or more times, because four ∟ are exchanged for
o
m
one △ and.cfour △ for one □. That is why the base-4 digits are onlyag0, 1, 2, 3. It also
l a se why 16 is written with a single symbol while 15 needs six: 16 is a landmark
g 15 is just below one.
explains
aand
o m
e
. c
mand 20 = □△, and check that 16 + 4 = 20. ag
s
Check it yourself: continue to 17 = □∟
a
agl

co m
m.
Q3 Give a simple rule to multiply a given number by 5 in the base-5 system that we

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

a Replace every symbol by the next symbol up, keeping the counts unchanged.
m a s
△→□→⬡→○→∿→↑
em
.co agl
a s
a l
(1 → 5 → 25 → 125 → 625 →g3125)

co m
.
Worked example. Take ○ □□ △ = 125 + 5 + 5 + 1 = 136.

e m
co∿m □□ → ⬡⬡ △ → □ g l as
. a
em
○→
a s
agl Result: ∿ ⬡⬡ □ = 625 + 25 + 25 + 5 = 680
Check: 136 × 5 = 680 ✓
se m
com g l a
m. a
gl ase
a
Why it happens: break the number into its landmark parts (distributive law), and
multiply each part by 5. Since 5 × 5a = 5a+1, every part becomes the next landmark,
and nothing has to be regrouped because the counts have not grown. Written as
co m
m .
as e
base-5 digits, the rule is the familiar one: put a 0 at the end — 136 is 1021 in base 5,
m
.co
and 680 is 10210.
a g l
sem
g l a
a c
.
Tip: the same rule works in every base — multiplying by the base itself shifts

s e m
a
everything up one place. That is why “multiply by 10, add a zero” works in base 10
m
. co
and “multiply by 2, add a zero” works in binary.
e m agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

In-text Questions — Page 72
3.4 I. The Mesopotamian Number System

Q1 Can we represent this more compactly?

Yes — and the way we do it is the birth of place value.
The number on page 71 was 640, grouped as

640 = (10) × 60 + 40

Following the Egyptian idea we would draw the 60-symbol ten times over and then four 10-
wedges — fourteen symbols in all. Instead, write the numeral for ten in the 60s slot and the
numeral for forty in the ones slot:

< | <<<<

read as “ten 60s and one 40” — exactly what the equation says

The same for 7530:

7530 = (2) × 3600 + (5) × 60 + 30

Check: 7200 + 300 + 30 = 7530 ✓

Numeral: YY | YYYYY | <<<

Once each slot has its own place, the symbols marking which power of 60 it is are no longer
needed at all — the position says it. Drop them, and the numeral is as compact as it can be.

Why it happens: the compression works because no power of 60 can ever occur 60
or more times. If it did, sixty of them would be regrouped into one of the next power
— the book shows this with (1) × 3600 + (70) × 60 + 2 = (2) × 602 + (10) × 60 + 2. So
each slot holds a number between 0 and 59, which a handful of wedges can always
write. A bounded slot is what makes place value possible.

Figure it Out — Page 73

Page 40 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

3.4 I. The Mesopotamian Number System

Q1 Represent the following numbers in the Mesopotamian system — (i) 63 (ii) 132 (iii)
200 (iv) 60 (v) 3605

Write Y for the 1-wedge and < for the 10-wedge. Reading right to left, the slots stand for 1s, 60s,
3600s.

GROUPED AS 3600S 60S 1S

(i) 63 1 × 60 + 3 — Y YYY

(ii) 132 2 × 60 + 12 — YY < YY

(iii) 200 3 × 60 + 20 — YYY <<

(iv) 60 1 × 60 + 0 — Y (blank)

(v) 3605 1 × 3600 + 0 × 60 + 5 Y (blank) YYYYY

Checks: 60 + 3 = 63 ✓ 120 + 12 = 132 ✓ 180 + 20 = 200 ✓ 3600 + 0 + 5 = 3605 ✓

Why it happens: parts (iv) and (v) are the ones that expose the flaw. In (iv) the single
wedge stands in the 60s slot with an empty ones slot — but the emptiness is
invisible, so the numeral looks exactly like the numeral for 1. In (v) two consecutive
slots are empty, and no amount of careful spacing tells a reader whether one blank
was intended or two. This is what the later Mesopotamians fixed with a placeholder
symbol, and what our 0 fixes completely.

Tip: to convert any number, divide repeatedly by 60. For 3605: 3605 ÷ 60 = 60
remainder 5, and 60 ÷ 60 = 1 remainder 0. Reading the remainders upwards gives
the slots 1, 0, 5 — that is 1 × 3600 + 0 × 60 + 5.

In-text Questions — Page 73

Page 41 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

3.4 I. The Mesopotamian Number System

Q1 Look at the representation of 60. What will be the representation for 3,600?

A single wedge Y again — this time standing in the 3600s slot, with two empty slots to its right.

NUMBER 3600S 60S 1S WHAT IS ACTUALLY WRITTEN

1 — — Y Y

60 — Y (blank) Y

3600 Y (blank) (blank) Y

All three look the same on the tablet. Only the context could tell a reader which was meant.

Why it happens: in a place value system the position carries the meaning — but a
position that is empty carries no ink, and an empty position at the end of a numeral
cannot even be seen. The later Mesopotamians invented a placeholder mark for a
blank in the middle of a numeral, which is a remarkable idea and very close to our 0.
But they did not use it at the end, so 1, 60 and 3600 stayed ambiguous. The Hindu
system removed the ambiguity entirely by treating 0 as a digit like any other, which is
why we can write 1, 60 and 3600 and never confuse them.

Check it yourself: the same page shows 12, 602 and 36002 all written as < YY, with
only the spacing to tell them apart. In our system those are three plainly different
numerals, because every empty position is filled by a 0.

In-text Questions — Page 76
3.4 II. The Mayan Number System

Q1 Represent the following numbers using the Mayan system: (i) 77 (ii) 100 (iii) 361 (iv)
721

The Mayan landmark numbers are 1, 20, 360, 7200, 144000 — note that the third is 20 × 18 =
360, not 400. A dot is 1, a bar is 5, and a shell is 0. The sets of symbols are written one below the
other, with the lowest set counting the 1s, the next the 20s, the next the 360s.

Page 42 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

GROUPED AS 360S ROW 20S ROW 1S ROW

(i) 77 3 × 20 + 17 — 3 dots 3 bars + 2 dots

(ii) 100 5 × 20 + 0 — 1 bar shell

(iii) 361 1 × 360 + 0 × 20 + 1 1 dot shell 1 dot

(iv) 721 2 × 360 + 0 × 20 + 1 2 dots shell 1 dot

Checks: 60 + 17 = 77 ✓ 100 + 0 = 100 ✓ 360 + 0 + 1 = 361 ✓ 720 + 0 + 1 = 721 ✓

77 100 361 721

360s

20s

1s

Each numeral is read from the bottom up: the lowest set counts 1s, the next counts 20s, the next
counts 360s. The orange shell is the Mayan placeholder for 0.

Why it happens: in (iii) and (iv) the shell is doing real work. Without it, 361 would be
a dot, a gap and a dot — and no reader could tell how wide the gap was meant to
be. That is the same trouble the Mesopotamians had, and the Maya solved it
independently, on the other side of the world.

Did you know? Because the third landmark is 360 and not 400, this is almost but not
quite a base-20 system. That is why the chapter says it lacks the computing
advantages of a true base: 20 × 20 = 400 is not a landmark number, so landmark ×
landmark is not always a landmark. Scholars think 360 was chosen to suit the Mayan
calendar.

In-text Questions — Page 78

Page 43 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

3.4 IV. The Hindu Number System
co m
e m.
m l as
.co
Where does the Hindu/Indian number system figure in the evolution of ideas of
a g
em representation? What are its landmark numbers? And does it use a place
Q1

a s
number

a gl value system?

. com ag

a s em — it is the one system that carries every idea in
Where it figures: right at the end of the story

aglone of its own.
the chapter at once, and adds the last

IDEA FIRST SEEN IN
co
PRESENT IN THE HINDU
m
se m.
SYSTEM?

m group size
coa fixed Yes g l a
. a— groups of 10
em
Counting in Gumulgal (2s)

a s
aglA sequence of landmark numbers Roman Yes

a s
com agl
Landmarks that are powers of one number Egyptian (base 10) Yes — base 10, decimal
— a base
m .
e
asMesopotamian, Mayan,
agl Chinese
Position instead of a landmark symbol Yes — place value

co m
.
0 as a placeholder and as a number Indian mathematics Yes — and only here

em
m l as
.co
Its landmark numbers are the powers of 10:
m a g
l a se
ag 1, 10, 102 = 100, 103 = 1000, 104, 105, … — without end

se m
com g l a
m . a
ase
Yes, it is a place value system. The book’s own example:

375 = (3) × 102 + (7) × 10 + (5) × 1
agl

com
= 300 + 70 + 5 = 375 ✓
m .
m as e
.co a g l
se m Why it happens: what makes this system unambiguous, where the Mesopotamian
g l a
a one was not, is two rules working together: one digit in each position, and a digit 0
c
m .
e
for an empty position. Together they mean that a numeral can be read in only one

om la s
. c
way — 305, 35 and 350 are three different strings for three different numbers, with
no reliance on spacing. Andm a g
s e because the landmarks are powers of 10, landmark ×

a gla
landmark is again a landmark, so multiplication and division have simple general
algorithms.

co m
m .
m ase
.co


a g l Page 44 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Did you know? Zero was not just a placeholder in Indian mathematics. Aryabhata
used its arithmetic properties in 499 CE, and Brahmagupta in 628 CE gave 0 and the
negative numbers the full standing of numbers — creating what we now call a ring, a
set closed under addition, subtraction and multiplication. Those ideas are the
foundation of modern algebra and analysis.

Figure it Out — Page 80
3.4 The Hindu Number System

MATH TALK

Q1 Why do you think the Chinese alternated between the Zong and Heng symbols? If
only the Zong symbols were to be used, how would 41 be represented? Could this
numeral be interpreted in any other way if there is no significant space between
two successive positions?

Why they alternated: because the rod numerals were actual sticks laid out on a counting
board, and a blank space between two places is invisible. Turning the rods through a right angle
at every place makes the boundary between places visible instead.

Zong (upright rods) — units, hundreds, ten-thousands …
Heng (flat rods) — tens, thousands, hundred-thousands …

So in the numeral for 2634 on page 77 you read 2 (Heng), 6 (Zong), 3 (Heng), 4 (Zong), and there
is never any doubt where one place ends and the next begins.
41 with Zong only: 41 = 4 tens and 1 one, so you would put down Zong-4 followed by Zong-1:

| | | | | — five upright rods in a row

Yes, it can be read in many other ways if the gaps are not clear. Five rods can be split up as:

Page 45 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

HOW THE RODS ARE SPLIT READS AS

| | | | | (no split) 5

| |||| 14

|| ||| 23

||| || 32

|||| | 41

| || || 122

| | ||| 113

|| | || 212

||| | | 311

Why it happens: alternating the orientation is a clever partial cure, but only a partial
one. It marks where the places begin, yet it still cannot say that a place is empty, so a
blank had to be left for a missing place and blanks are hard to count. As the book
notes, a symbol for zero would have turned the rod numerals into a fully developed
place value system. It is exactly the step the Indian system took.

Q2 Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare
this system with that of the Gumulgal’s.

A base-2 system needs exactly two digits, one of which must stand for zero. Let

urapon = 0 ukasar = 1

landmark numbers: 1, 2, 4, 8, 16, 32, …

Page 46 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

NUMBER GROUPED INTO POWERS BASE-2 PLACE VALUE GUMULGAL SYSTEM
OF 2 SYSTEM

1 1 ukasar urapon

2 2 ukasar urapon ukasar

3 2+1 ukasar ukasar ukasar-urapon

4 4 ukasar urapon urapon ukasar-ukasar

5 4+1 ukasar urapon ukasar ukasar-ukasar-urapon

6 4+2 ukasar ukasar urapon ukasar-ukasar-ukasar

7 4+2+1 ukasar ukasar ukasar ukasar-ukasar-ukasar-
urapon

8 8 ukasar urapon urapon ukasar-ukasar-ukasar-
urapon ukasar

Comparison.

In the Gumulgal system the position of a word means nothing — you simply add: each
ukasar contributes 2 and each urapon contributes 1. Its only landmark numbers are 1 and 2.
In the base-2 place value system the position is everything: the same word ukasar means 1,
2, 4 or 8 depending on where it stands. Its landmark numbers go on for ever: 1, 2, 4, 8, 16, …
Length: to write N, the Gumulgal need about N ÷ 2 words; the place value system needs only
about log2N digits. For 1024 that is 512 words against 11 digits.
The place value version needs a digit for zero; the Gumulgal system has no use for one,
because a place that is empty is simply not spoken.

Why it happens: both systems count in 2s, and yet they behave completely
differently. The Gumulgal use the group size only to make names; the base-2 system
uses it to make places, and each new place doubles the range. Watch out for one
trap: 2 is ukasar in the Gumulgal system but ukasar urapon in the base-2 system —
the same words, read by two different rules.

Page 47 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Q3 Where in your daily lives, and in which professions, do the Hindu numerals, and 0,
play an important role? How might our lives have been different if our number
system and 0 hadn’t been invented or conceived of?

In daily life: prices and change at a shop, dates on a calendar, the time on a clock, phone
numbers, PIN codes, bus and train numbers, house numbers, cricket scores and run rates,
marks in a report card, the reading on a weighing scale, page numbers in this book.
In professions:

Shopkeepers and accountants — bills, ledgers, GST calculations.
Bankers — interest, instalments, account balances.
Engineers, architects and surveyors — measurements, scale drawings, load calculations.
Doctors and chemists — dosages in milligrams, blood reports.
Farmers — land area in hectares, yields, market rates.
Scientists — very large and very small numbers written as 1023 or 10-9, which is place value
taken to its natural conclusion.
Programmers — every computer works in base 2, using only 0 and 1.

Without them, life would look very different:

Arithmetic would need a device. Recall page 60: people using Roman numerals had to use an
abacus, and only specially trained people could work it. Ordinary shopkeepers could not
have checked their own bills.
There would be no column addition, no long multiplication, no long division on paper.
Without 0 as a number there would be no negative numbers as Brahmagupta defined them,
so no algebra of the modern kind, no coordinate geometry, no equations of motion.
There would be no decimals, no scientific notation, and no binary — so no computers, no
digital payments, no mobile phones.

Why it happens: the Hindu system does not merely record numbers, it makes
calculation possible for everybody. That is why Laplace called its significance so
profound and yet so easily overlooked — it “placed arithmetic foremost among
useful inventions” precisely by making it ordinary.

Page 48 of 51

as e
Class 8 Maths Chapter 3 A Story of Numbers
a g l AglaSem · NCERT Solutions

co m
m.
The ancient Indians likely used base 10 for the Hindu number system because
se
Q4

o m l a
humans have 10 fingers, and so we can use our fingers to count. But what if we had
only 8.cfingers? How would we be writing numbers then? What gwould the Hindu
m a
l a se
numerals look like if we were using base 8 instead? Base 5? Try writing the base-10
ag Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write
it in base-2?

com
m . ag
l a se
ag have settled on base 8. Everything else about the
With 8 fingers we would almost certainly
system would be unchanged — only the base would differ.

co m
e m.
m l as
.co
Landmark numbers: 1, 8, 64, 512, 4096, …
a g
a s em
Digits needed: 0, 1, 2, 3, 4, 5, 6, 7 — eight of them, one fewer than the base

a gl A number N would be written a a k k−1 … a1 a0, meaning

m a s
agl
N = ak·8k + ak−1·8k−1 + … + a1·8 + a0, with 0 ≤ ai ≤ 7

m.co
l a se
Writing 25:
a g
m
.co
BASE GROUPING NUMERAL

se m 31
8
com
25 = 3 × 8 + 1
g l a
5m
. a
ase
agl
25 = 1 × 25 + 0 × 5 + 0 × 1 100

m
2 25 = 16 + 8 + 1 = 1×16 + 1×8 + 0×4 + 0×2 + 1×1 11001

a se
10 25 = 2 × 10 + 5
. com 25
a g l
m
ase
agl
Checks: 3 × 8 + 1 = 25 ✓ 1 × 25 = 25 ✓ 16 + 8 + 0 + 0 + 1 = 25 ✓

c o m
Why it happens: the base decides two things at once. It fixes how many digits you
.
m numerals are. A
s e
need — always the base itself, counting 0 — and it fixes how long

.csmaller gla
om base means fewer symbols to learn but longeranumerals: 25 is “31” in base 8,

se m “100” in base 5 and “11001” in base 2. Base 10 is not mathematically special in any
g l a
a way; ten fingers are the whole reason for it. What is special, and works in every base,
c
m .
e
is the structure — landmarks that are powers of the base, one digit per position, and
m a s
. co agl
a 0 for an empty position.

e m
g l as
a

com
m .
m ase
.co


a g l Page 49 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Check it yourself: 100 in base 5 is 25 in base 10, and 100 in base 8 is 64. In every
base, the numeral “100” means the square of the base — which is exactly what place
value promises.

Chapter at a glance
To count you need a standard sequence of objects, names or written symbols with a fixed
order — that sequence is a number system. Counting a collection means making a one-to-
one mapping onto it.
The written symbols of a number system are called numerals. Numbers that get a symbol
of their own and act as reference points are called landmark numbers.
If the landmark numbers are 1, n, n2, n3, … — the powers of one number — the system is a
base-n system. Base 10 is called decimal.
In a base-n system the product of two landmark numbers is again a landmark number,
which is exactly why multiplication is easy there and hard in the Roman system.
A system that uses the position of a symbol to decide which landmark it counts is a place
value system. Mesopotamia, the Maya, China and India all reached this idea.
The Hindu (Indian) number system — base 10, place value, and 0 treated both as a
placeholder digit and as a number in its own right — writes every number unambiguously
with just ten symbols. It began in India about 2000 years ago and is now used everywhere.

Page 50 of 51

Class 8 Maths Chapter 3 A Story of Numbers AglaSem · NCERT Solutions

Quick revision

NUMBER SYSTEM LANDMARK THE IDEA IT ADDS HOW 27 IS WRITTEN
NUMBERS

Sticks / tally marks 1 only One-to-one mapping: one mark 27 marks
per object

Gumulgal 1, 2 Counting in a fixed group size ukasar repeated 13 times,
(Australia) (2s) then urapon (14 words)

Roman 1, 5, 10, 50, 100, A sequence of landmark numbers XXVII
500, 1000 — but not powers of one number

Egyptian 1, 10, 102, …, 107 Landmarks are powers of 10 — a 2 arches + 7 strokes
base

Our base-5 system 1, 5, 25, 125, 625, The same idea with a different ⬡ △△ (25 + 1 + 1)
3125 base

Mesopotamian 1, 60, 602, 603 Place value; later a placeholder << YYYYYYY (two 10-
(Babylonian) for a blank wedges, seven 1-wedges)

Mayan 1, 20, 360, 7200, Place value written vertically; a 1 dot above; 1 bar + 2
144000 shell for 0 dots below

Chinese rod powers of 10 Place value made readable by Heng-2 above the tens
numerals alternating Zong and Heng place, Zong-7 in the units

Hindu (Indian) 1, 10, 100, 1000, … 0 as a digit and as a number; ten 27
symbols for everything

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