NCERT Solutions · Class 8 Maths · Ganita Prakash Part 1 · Chapter 3
Chapter 3: A Story of Numbers (Number Systems)
Step-by-step answers to every "Figure it Out" and in-text question of Chapter 3, A Story of Numbers (NCERT Class 8 Maths, Ganita Prakash Part 1, 2026-27): counting with sticks, letters and symbols, the Gumulgal system, Roman numerals and their arithmetic, Egyptian numerals, base-n systems, the abacus, Mesopotamian, Mayan and Chinese place value systems, and the Hindu numerals with zero. All 34 questions are answered, with the key answer highlighted.
How many numbers can you represent using the sounds of the letters of your language? Can the Roman method (Table 1) be extended to bigger numbers?
Solution
As many as there are letters: English gives 26; Hindi (Devanagari) has about 13 vowels and 33 consonants, so about 46.
Yes: the Roman method continues with new symbols for bigger groups: L for 50, C for 100, D for 500 and M for 1000 (e.g. 40 = XL, 90 = XC, 1999 = MCMXCIX). But it needs more and more new symbols for bigger numbers.
About as many numbers as letters (26 in English, ~46 in Hindi); the Roman system extends with L, C, D, M, but needs ever more symbols.
Using sticks (Method 1), without number names or Hindu numerals, give methods for adding, subtracting, multiplying and dividing two collections.
Solution
Add: put the two bundles together into one bundle.
Subtract: pair the sticks of the smaller bundle one-to-one with sticks of the larger one and remove these pairs; the unpaired sticks left over are the difference.
Multiply: for each stick in the first bundle, put a whole copy of the second bundle into a new heap.
Divide: deal the sticks out one at a time into as many heaps as there are sticks in the divisor (sharing), or keep taking away copies of the divisor bundle (grouping). The size of a heap (or the number of copies taken) is the quotient; any sticks left over form the remainder.
Add by joining bundles, subtract by pairing and removing, multiply by repeated copies, divide by dealing out into equal heaps.
Extend the letter system (Method 2), e.g. using 'aa' for 27, to represent all numbers.
Solution
One way (like column names in a spreadsheet): a, b, …, z (1–26); then aa, ab, …, az (27–52), ba, …, bz (53–78), …, zz (702); then aaa, aab, … and so on. Every number gets a unique name, and names never run out.
(Other ways: use 'a' as a separator meaning "26 and", or use letter pairs as "tens and units".)
E.g. a–z, then aa–az, ba–bz, …, zz, then aaa, …: every number gets a unique string.
Example: use ● for 1, ▲ for 5 and ■ for 25. A number is written by grouping it into as many 25s, then 5s, then 1s as possible: 38 = 25 + 5 + 5 + 1 + 1 + 1 = ■▲▲●●●. (This already uses the idea of landmark numbers that are powers of 5.)
How are the Gumulgal number names formed? Up to what group size can you see the number of objects at a glance? What could be the difficulties of counting only in groups of a single size? How would you write 1345 counting only by 5s?
Solution
They count in 2s: ukasar (2) is repeated, and urapon (1) is added when needed: 5 = 2 + 2 + 1 = ukasar-ukasar-urapon.
Most people can see up to about 4 objects at a glance; beyond that we need to count.
Counting in groups of one size means a big number still needs very many group symbols. For example, 1345 is 269 fives, so we would need 269 copies of the "five" symbol: very long to write and read.
Names are built from 2s and 1s; we see up to about 4 at a glance; 1345 would need 269 "fives", which is too long.
(a) Collect the symbols: 6 C, 4 X, 5 I. Regroup: 5 C = D, 5 I = V. Sum =DCXXXXV (= DCXLV = 645).
(b) Collect: 2 L, 5 X, 2 V, 5 I. Regroup: 5 I = V (now 3 V), 2 V = X (now 6 X, 1 V), 5 X = L (now 3 L, 1 X), 2 L = C (now 1 C, 1 L). Sum =CLXV (= 165).
Multiply in Roman numerals: V × L, L × D, V × D, VII × IX. Also CCXXXI × MDCCCLII.
Solution
V×L=250=CCL
L×D=25000: there is no symbol beyond M, so it needs 25 Ms (later Romans drew a bar over a numeral to mean "× 1000": XXV).
V×D=2500=MMD
VII×IX=63=LXIII
CCXXXI×MDCCCLII=231×1852=4,27,812, which needs 427 Ms followed by DCCCXII (or CDXXVIIDCCCXII with the bar). Doing this directly means multiplying every symbol of one number by every symbol of the other (6 × 8 = 48 products) and regrouping, which is why multiplication was so hard in this system.
Some Pacific island people use different sequences of number names to count different objects. Why might they do this?
Solution
Their counting grew out of daily life and trade: different things were counted in different natural groups (fish or coconuts in pairs or bunches, canoes one by one, long objects differently from round ones). Each object type had its own customary unit, so its own number words. (Many languages still do something similar: "a pair of shoes", "a dozen eggs", "a ream of paper".)
Different objects were traditionally counted in different groups (pairs, bunches), so each kind of object got its own counting words.
Extend the Gumulgal system beyond 6 by counting in 2s, and use it to compute: (i) (u-u-u-u-urapon) + (u-u-u-urapon) (ii) (u-u-u-u-urapon) − (u-u-u) (iii) (u-u-u-u-urapon) × (u-u) (iv) (eight ukasars) ÷ (u-u) [u = ukasar]
Solution
Rules: add by joining all the words and turning two urapons into one ukasar; subtract by crossing out matching words (breaking an ukasar into two urapons if needed); multiply by writing the first number as many times as the second says; divide by splitting into equal groups.
(i) 4 ukasars + urapon and 3 ukasars + urapon: 7 ukasars and 2 urapons = 8 ukasars (16).
(ii) Cross out 3 ukasars from 4 ukasars + urapon: ukasar-urapon (3).
(iii) "ukasar-ukasar" is 4, so write the first number 4 times: 16 ukasars and 4 urapons = 18 ukasars (36).
(iv) 8 ukasars split into 4 equal groups ("ukasar-ukasar" = 4): ukasar-ukasar in each group, i.e. 4 (2 ukasars).
Refine the number system you made earlier using these ideas.
Solution
Make the landmark numbers powers of one number, e.g. ● = 1, ▲ = 5, ■ = 25, ★ = 125, … (base 5). Then no symbol is ever needed more than 4 times, and multiplying landmark numbers gives landmark numbers. Better still, write only the counts in fixed positions with a zero symbol: 38 = 1 twenty-five, 2 fives, 3 ones → "123" in base 5.
Use powers of one number as landmarks, then positions with a zero symbol (e.g. 38 = 123 in base 5).
3.3 The Idea of a Base: Egyptian Numerals
Egyptian symbols: stroke (1), heel (10), coil of rope (100), lotus (1,000), finger (10,000), tadpole (1,00,000), kneeling god (10,00,000) and sun (1,00,00,000).
Is there a number that cannot be represented in our base-5 system? Why or why not?
Solution
No. The landmark numbers 1, 5, 25, 125, … go on forever (we can always make a new symbol for the next power of 5), and any number can be grouped into them using each at most 4 times. (In practice, very large numbers need symbols beyond 3125, which we must invent.)
No, every number can be grouped into powers of 5, as long as we keep inventing symbols for higher powers.
Does the product of landmark numbers give a landmark number in the base-5 system, and in any system with a base?
Solution
Yes. In base 5, 5a×5b=5a+b, another power of 5 (e.g. square × hexagon = 5 × 25 = 125 = circle). In any base n, na×nb=na+b, so the product of two landmark numbers is always a landmark number.
What happens when we multiply a number by a heel (10) in Egyptian numerals? Find (i) (5 coils, 2 heels, 2 strokes) × heel (ii) (lotus, heel) × heel. Give a simple rule.
Solution
Multiplying by 10 turns every symbol into the next higher symbol (stroke → heel, heel → coil, coil → lotus, …), by the distributive property.
(i) 522 × 10 = 5 lotuses, 2 coils, 2 heels (5220)
(ii) 1010 × 10 = 1 finger, 1 coil (10100)
Rule: replace every symbol by the next landmark symbol.
(i) 5 lotuses, 2 coils, 2 heels (ii) 1 finger, 1 coil; rule: replace each symbol by the next higher one.
How would you use the abacus to find 2907 + 43? What is done if the total on a line exceeds 10?
Solution
Bring the counters of each line together:
Ones: 7 + 3 = 10 counters → remove all ten and add 1 counter to the tens line.
Tens: 0 + 4 + 1 = 5 → replace the five by one counter above the tens line (worth 5 tens).
Hundreds: 9; thousands: 2.
Result: 2950. Whenever a line reaches 10 counters, remove them and put one counter on the next line up (and every 5 on a line can be shown by one counter above it).
2907 + 43 = 2950; ten counters on a line become one counter on the next line.
Can a number's Egyptian numeral have one symbol 10 or more times? Why not?
Solution
No, for symbols below the largest: 10 copies of a symbol can always be replaced by one copy of the next symbol (10 strokes = 1 heel, etc.). The only exception is the largest symbol (the sun, 107): numbers of 108 or more would need 10 or more suns, because there is no bigger symbol. That is the shortcoming of the Egyptian system.
No, ten of a symbol always become one of the next; only the largest symbol could repeat 10+ times (for numbers ≥ 10⁸).
Give a simple rule to multiply a number by 5 in the base-5 system.
Solution
Replace every symbol by the next landmark symbol (triangle → square, square → hexagon, hexagon → circle, circle → wave, …), just like multiplying by a heel in the Egyptian system.
Replace each symbol by the next higher symbol.
3.4 Place Value: The Mesopotamian System
We write Y for the wedge (1) and < for the corner wedge (10); a gap separates the places (ones on the right, then 60s, then 3600s).
3600=1×602+0×60+0: a single Y followed by two empty places. Without a zero symbol it looks exactly the same as 1 and 60, which is the ambiguity of the system.
A single Y (with two blank places), the same as 1 and 60: ambiguous without zero.
Why did the Chinese alternate Zong and Heng symbols? Using only Zong symbols, how would 41 be written? Could it be read differently?
Solution
Alternating vertical (zong) and horizontal (heng) rods makes it clear where one place ends and the next begins, even without spaces.
With only zong rods, 41 would be |||| | (four rods, then one rod). Without a clear gap this looks like ||||| (5), or could be read as 14 or 41: it is ambiguous. With alternation, 4 tens in heng and 1 unit in zong cannot be confused.
To separate the places; 41 in zong only is |||| |, which could be read as 5, 14 or 41.
Form a base-2 place value system using 'ukasar' and 'urapon' as the digits. Compare it with the Gumulgal system.
Solution
Let ukasar = 0 and urapon = 1 (the two digits of base 2). Places are 1, 2, 4, 8, 16, … from the right.
Number
Binary
Place-value name
Gumulgal name
3
11
urapon-urapon
ukasar-urapon
6
110
urapon-urapon-ukasar
ukasar-ukasar-ukasar
16
10000
urapon-ukasar-ukasar-ukasar-ukasar
eight ukasars
The Gumulgal system just adds 2s and 1s, so names grow as long as half the number. The place-value system uses positions, so its names grow very slowly (about one word each time the number doubles) and can name every number.
With ukasar = 0, urapon = 1: 6 = urapon-urapon-ukasar. Place value keeps names short (16 needs 5 words, not 8).
Where in daily life and in which professions do Hindu numerals and 0 play an important role? How might life be different without them?
Solution
Everywhere: prices, bills and banking (₹10,050), phone numbers, dates and clocks, PIN codes, page numbers, scores; professions like accountants, shopkeepers, engineers, scientists, doctors (doses), programmers (computers store everything in binary 0s and 1s), pilots and drivers (speeds, distances).
Without them, writing large numbers would need long strings of special symbols, arithmetic would need an abacus or trained experts, calendars, science, computers and digital payments would be almost impossible, and mistakes would be far more common.
They are used in money, phones, dates, science, engineering and computers; without them big numbers and calculations would be extremely hard.
