Q3: Now find the following products- (i) (999 99 ∩∩ ) × ∩ (ii) (P 𓆼 ∩) × 9

Given: Multiplication problems in Egyptian numerals. Rule: Multiplying by a landmark number (a power of 10) promotes each symbol to a higher landmark symbol. Multiplying by ∩ (10) shifts every symbol to the next higher one. Multiplying by 9 (100) shifts every symbol two places higher. (i) (Three…

A Story of Numbers Class 8 Mathematics NCERT Solutions

Q1Figure it Out (Advantages of a Base-n System)

Add the following Egyptian numerals:

(i)

(2 coiled ropes, 3 hobbles, 4 sticks) + (3 coiled ropes, 8 hobbles, 7 sticks)

(ii)

(2 fingers, 3 coiled ropes, 5 hobbles, 6 sticks) + (4 fingers, 8 coiled ropes, 7 hobbles, 9 sticks)

Solution

Given: Addition problems in Egyptian numerals.

To Find: The sums, expressed in proper Egyptian numeral form.

Rule for Simplification: 10 of any symbol are replaced by 1 of the next higher symbol.

(i) (2 ropes, 3 hobbles, 4 sticks) + (3 ropes, 8 hobbles, 7 sticks)

Combine like symbols:
- Sticks: $4 + 7 = 11$ sticks
- Hobbles: $3 + 8 = 11$ hobbles
- Ropes: $2 + 3 = 5$ ropes
Regroup and simplify, starting from the smallest unit (sticks):
- 11 sticks = 10 sticks + 1 stick = 1 hobble + 1 stick.
- Now we have: 5 ropes, (11+1) hobbles, 1 stick = 5 ropes, 12 hobbles, 1 stick.
- 12 hobbles = 10 hobbles + 2 hobbles = 1 rope + 2 hobbles.
- Now we have: (5+1) ropes, 2 hobbles, 1 stick = 6 ropes, 2 hobbles, 1 stick.

Final Answer (i): 6 coiled ropes, 2 hobbles, 1 stick. (Value: 621)

(ii) (2 fingers, 3 ropes, 5 hobbles, 6 sticks) + (4 fingers, 8 ropes, 7 hobbles, 9 sticks)

Combine like symbols:
- Sticks: $6 + 9 = 15$ sticks
- Hobbles: $5 + 7 = 12$ hobbles
- Ropes: $3 + 8 = 11$ ropes
- Fingers: $2 + 4 = 6$ fingers
Regroup and simplify:
- 15 sticks = 1 hobble + 5 sticks.
- Now we have: 6 fingers, 11 ropes, (12+1) hobbles, 5 sticks = 6 fingers, 11 ropes, 13 hobbles, 5 sticks.
- 13 hobbles = 1 rope + 3 hobbles.
- Now we have: 6 fingers, (11+1) ropes, 3 hobbles, 5 sticks = 6 fingers, 12 ropes, 3 hobbles, 5 sticks.
- 12 ropes = 1 finger + 2 ropes.
- Now we have: (6+1) fingers, 2 ropes, 3 hobbles, 5 sticks = 7 fingers, 2 ropes, 3 hobbles, 5 sticks.

Final Answer (ii): 7 fingers, 2 coiled ropes, 3 hobbles, 5 sticks. (Value: 70,235)

Q2Figure it Out (Advantages of a Base-n System)

Add the following numerals that are in the base-5 system that we created: ◯◯□△△ + ◯◯◯□□□△△

Solution

Given: An addition problem in our custom base-5 system.

Symbols: △=1, ♢=5, □=25, ◯=125.
First number: ◯◯□△△ = $2 \times 125 + 1 \times 25 + 2 \times 1 = 250 + 25 + 2 = 277$ .
Second number: ◯◯◯□□□△△ = $3 \times 125 + 3 \times 25 + 2 \times 1 = 375 + 75 + 2 = 452$ .

To Find: The sum in the base-5 system.

Method: Combine and Regroup Rule: 5 of any symbol are replaced by 1 of the next higher symbol.

Combine like symbols:
- △ (1s): $2 + 2 = 4$ triangles.
- ♢ (5s): $0 + 0 = 0$ diamonds.
- □ (25s): $1 + 3 = 4$ squares.
- ◯ (125s): $2 + 3 = 5$ circles.
Regroup and simplify:
- The 4 triangles (△△△△) are fine (less than 5).
- The 4 squares (□□□□) are fine (less than 5).
- The 5 circles (◯◯◯◯◯) must be regrouped. 5 circles ( $5 \times 125$ ) are equal to 1 of the next landmark number, which is $5^4 = 625$ . Let's represent this with a Star ☆.
Final Result: The sum is one Star, four Squares, and four Triangles.

Final Answer: ☆ □□□□ △△△△

Verification (using Hindu numerals):

277 + 452 = 729

. Our result is

1 \times 625 + 4 \times 25 + 0 \times 5 + 4 \times 1 = 625 + 100 + 4 = 729

. The answer is correct.

Q3Figure it Out (Advantages of a Base-n System)

Now find the following products- (i) (999 99 ∩∩ ||) × ∩ (ii) (P 𓆼 ∩) × 9

Solution

Given: Multiplication problems in Egyptian numerals.

Rule: Multiplying by a landmark number (a power of 10) promotes each symbol to a higher landmark symbol.

Multiplying by ∩ (10) shifts every symbol to the next higher one.
Multiplying by 9 (100) shifts every symbol two places higher.

(i) (Three coiled ropes, two hobbles, two sticks) × (one hobble) This is

(300 + 20 + 2) \times 10

. Using the distributive property:

(3 ropes 999) × (1 hobble ∩) = 3 lotus flowers 𓆼𓆼𓆼 ( $300 \times 10 = 3000$ ).
(2 hobbles ∩∩) × (1 hobble ∩) = 2 coiled ropes 99 ( $20 \times 10 = 200$ ).
(2 sticks ||) × (1 hobble ∩) = 2 hobbles ∩∩ ( $2 \times 10 = 20$ ).

Final Answer (i): 3 lotus flowers, 2 coiled ropes, 2 hobbles.

(ii) (One finger, one lotus flower, one hobble) × (one coiled rope) This is

(10000 + 1000 + 10) \times 100

. Using the distributive property:

(1 finger P) × (1 rope 9) = 1 man ( $10000 \times 100 = 1,000,000$ ).
(1 lotus 𓆼) × (1 rope 9) = 1 tadpole ( $1000 \times 100 = 100,000$ ).
(1 hobble ∩) × (1 rope 9) = 1 lotus flower ( $10 \times 100 = 1000$ ).

Final Answer (ii): 1 man, 1 tadpole, 1 lotus flower.

Q1Figure it Out (End of Chapter)

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?

Solution

Question: Why did the Chinese rod numeral system alternate symbol orientations (Zong/vertical and Heng/horizontal)?

Answer: The alternation between Zong (vertical) and Heng (horizontal) symbols was a clever way to avoid ambiguity in their place value system, which originally lacked a symbol for zero.

Zong symbols were used for the units, hundreds, ten thousands, etc. (powers of $10^0, 10^2, 10^4, ...$ ).
Heng symbols were used for the tens, thousands, etc. (powers of $10^1, 10^3, ...$ ).

Hypothetical Scenario: Using only Zong symbols

How would 41 be represented?
- 41 is four tens and one unit.
- If we used only Zong symbols, 4 would be |||| and 1 would be |.
- The representation would be |||| |.
Could this be misinterpreted?
- Yes, absolutely. Without a clear separation or a different orientation, the numeral |||| | looks identical to the Zong symbol for 5, which is |||||. It would be impossible to distinguish 41 from 5.
Another example: The number 22 would be || ||. This could easily be confused with the symbol for 4, which is ||||.

By alternating, 41 is written with a Heng 4 (----) in the tens place and a Zong 1 (|) in the units place, making it ---- |, which is unambiguous. This alternation serves as a structural separator between place values.

Q2Figure it Out (End of Chapter)

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

Solution

Objective: Create a base-2 system using Gumulgal words and compare it to the original Gumulgal system.

1. A Base-2 Place Value System A base-2 (binary) system requires two digits, one for 0 and one for 1.

Let's assign urapon to be the digit for 1.
For the digit 0, the Gumulgal system had no word. We will introduce a placeholder word, say mara (meaning 'empty' or 'nothing').

The place values, from right to left, would be powers of 2: ...eights, fours, twos, ones.

Examples in this new system:

1: urapon (one 1)
2: urapon mara (one 2, zero 1s)
3: urapon urapon (one 2, one 1)
4: urapon mara mara (one 4, zero 2s, zero 1s)
5: urapon mara urapon (one 4, zero 2s, one 1)
6: urapon urapon mara (one 4, one 2, zero 1s)

2. Comparison with the Original Gumulgal System

Feature	Original Gumulgal System	New Base-2 System
Principle	Additive. Numbers are formed by adding the values of the words.	Positional (Place Value). The value of a word depends on its position.
Base	It is based on grouping by 2s, but it is not a true base system.	It is a true base-2 system.
Symbol for Zero	Did not have a concept or symbol for zero.	Requires a symbol for zero (`mara`) to function as a placeholder.
Efficiency	Becomes very long for larger numbers. 6 is `ukasar-ukasar-ukasar`. 8 would be four `ukasar`s.	More compact for larger numbers. 8 would be `urapon mara mara mara`.
Example: 5	`ukasar-ukasar-urapon` (2+2+1)	`urapon mara urapon` (1x4 + 0x2 + 1x1)

Conclusion: The original Gumulgal system shows the emergence of the idea of counting in groups (by 2s), but it is a simple additive system. The new system we created is a fully-formed place value system, which is far more powerful and efficient for representing numbers and performing calculations.

Q3Figure it Out (End of Chapter)

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?

Solution

1. Role in Daily Life and Professions:

The Hindu numerals (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and the concept of zero are fundamental to modern life. They are virtually everywhere.

Daily Life:
- Time: Reading a clock (10:30 AM).
- Money: Banking, prices ($5.09), paying bills.
- Communication: Phone numbers, addresses.
- Technology: Using computers, smartphones, calculators.
- Measurement: Cooking recipes, distances, temperature.
Professions:
- Science & Engineering: All calculations, from physics formulas to building designs, rely on this system.
- Finance & Accounting: Budgets, stock markets, and financial records would be impossible without it.
- Computer Science: The binary system (base-2), which is the foundation of all computing, is a place value system derived from the same principles, and uses 0 and 1.
- Medicine: Dosages, patient records, medical imaging.
- Commerce: Inventory, sales, logistics.

2. Life Without Our Number System and Zero:

If the Hindu-Arabic number system and zero had not been invented or widely adopted, our world would be vastly different and technologically primitive.

Complex Calculations Would Be for Experts Only: Arithmetic, especially multiplication and division, would be extremely difficult. It would likely be performed by specialists using tools like the abacus, as was done in ancient Rome. Everyday commerce would be much slower and more error-prone.
No Advanced Science or Technology: Without an efficient way to handle large numbers and perform complex calculations, there would be no advanced physics, chemistry, or engineering. This means no electricity grids, no modern medicine, no cars, no computers, and no space travel.
No Modern Economy: The global financial system, banking, and stock markets could not exist. Business and government would be unable to manage complex budgets or large-scale projects.
Stagnation of Mathematics: Key mathematical fields like algebra, calculus, and analysis, which are built upon the properties of zero and the number line, would not have developed. The scientific revolution would not have happened.

In short, the invention of this number system was one of the most critical breakthroughs in human history, paving the way for the scientific, technological, and economic development of the modern world.

Q4Figure it Out (End of Chapter)

The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers, and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base-10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2?

Solution

Scenario: If humans had 8 fingers, we would likely have developed a base-8 (octal) number system.

How would numbers be written in Base-8?

Digits: We would only need 8 digits: 0, 1, 2, 3, 4, 5, 6, 7. The symbols for '8' and '9' would not exist.
Place Value: The place values would be powers of 8: ... $8^3(512)$ , $8^2(64)$ , $8^1(8)$ , $8^0(1)$ .
Counting: Counting would proceed as 1, 2, 3, 4, 5, 6, 7, and then the next number (our 'eight') would be written as '10' (meaning one 8 and zero 1s).

Writing the number 25 (base-10) in other bases:

1. In Base-8:

We need to find how many groups of 8 are in 25.
$25 \div 8 = 3$ with a remainder of $1$ .
This means $25 = (3 \times 8^1) + (1 \times 8^0)$ .
The digits are 3 and 1.
Answer: 25 in base-10 is written as 31 in base-8 (often written as $31_8$ ).

2. In Base-5:

We need to find how many groups of 25 and 5 are in 25.
The place values are powers of 5: ..., 25, 5, 1.
$25 = (1 \times 5^2) + (0 \times 5^1) + (0 \times 5^0)$ .
The digits are 1, 0, and 0.
Answer: 25 in base-10 is written as 100 in base-5 ( $100_5$ ).

3. In Base-2 (Binary):

The place values are powers of 2: ..., 16, 8, 4, 2, 1.
We find the largest power of 2 that fits into 25, which is 16.
$25 = 16 + 9$
Now, break down the remainder 9: $9 = 8 + 1$ .
So, $25 = 16 + 8 + 1$ .
In terms of place values: $25 = (1 \times 16) + (1 \times 8) + (0 \times 4) + (0 \times 2) + (1 \times 1)$ .
$25 = (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)$ .
The digits are 1, 1, 0, 0, 1.
Answer: 25 in base-10 is written as 11001 in base-2 ( $11001_2$ ).

Q1Figure it Out (End of Section 3.2)

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?

Solution

Question: Why might a culture use different number names for counting different types of objects?

Possible Reasons: There could be several cultural, practical, or linguistic reasons for this practice:

Object Classification: The number words might change based on the category of the object being counted (e.g., one set of numbers for living things, another for inanimate objects, another for long objects, another for flat objects). This is similar to how some languages use different classifiers or measure words.
Cultural or Spiritual Significance: Certain objects (e.g., sacred items, people, important food sources) might be considered special and thus require a unique set of number names to show respect or acknowledge their importance.
Practical Grouping: The way objects are naturally bundled or handled could influence the counting words. For example, fish might be counted with words that relate to pairs or dozens, while yams might be counted with words related to piles or baskets.
Historical Remnants: The practice could be a remnant of a time when different counting systems were merged, or when counting was a more abstract concept tied closely to the items being counted rather than a universal, abstract sequence of numbers.

Q2Figure it Out (End of Section 3.2)

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-ukasarukasar)

(iii)

(ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)

(iv)

(ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)

Solution

Given: Gumulgal system: urapon = 1, ukasar = 2. Numbers are formed by adding these, e.g., 5 = ukasar-ukasar-urapon (2+2+1).

Arithmetic Operations:

Addition: Combine all the ukasar and urapon terms from both numbers into one long string.
Subtraction: For each term in the number being subtracted, remove a matching term from the first number.
Multiplication: Use repeated addition. To multiply A by B, add A to itself as many times as the value of B.
Division: Use repeated subtraction. To divide A by B, count how many times B can be subtracted from A.

Evaluation: First, let's decode the given numbers:

A = ukasar-ukasar-ukasar-ukasar-urapon = $4 \times 2 + 1 = 9$
B = ukasar-ukasar-ukasar-urapon = $3 \times 2 + 1 = 7$
C = ukasar-ukasar-ukasar = $3 \times 2 = 6$
D = ukasar-ukasar = $2 \times 2 = 4$
E = ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar = $8 \times 2 = 16$

(i) A + B (9 + 7 = 16)

Combine the terms: (four ukasars, one urapon) + (three ukasars, one urapon) = seven ukasars and two urapons.
We know two urapons make one ukasar (urapon-urapon = ukasar).
So, we have seven ukasars + one ukasar = eight ukasars.
Answer: ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(ii) A - C (9 - 6)

Start with A: ukasar-ukasar-ukasar-ukasar-urapon.
Subtract C by removing three ukasars.
We are left with one ukasar and one urapon.
Answer: ukasar-urapon (which is $2+1=3$ )

(iii) A × D (9 × 4)

We need to add A to itself 4 times. Since D is ukasar-ukasar, we add A to itself twice, then add that result to itself.
A + A = 9 + 9 = 18. This is nine ukasars. (ukasar-ukasar-... 9 times).
(A+A) + (A+A) = 18 + 18 = 36. This is eighteen ukasars.
Answer: ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar (18 ukasars)

(iv) E ÷ D (16 ÷ 4)

How many times can we subtract D (ukasar-ukasar) from E (eight ukasars)?
1st subtraction: eight ukasars - two ukasars = six ukasars remain.
2nd subtraction: six ukasars - two ukasars = four ukasars remain.
3rd subtraction: four ukasars - two ukasars = two ukasars remain.
4th subtraction: two ukasars - two ukasars = zero remains.
We subtracted 4 times. The number 4 is ukasar-ukasar.
Answer: ukasar-ukasar

Q3Figure it Out (End of Section 3.2)

Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.

Solution

Comparison of Hindu and Roman Number Systems:

The Hindu number system is significantly more efficient than the Roman system due to three key features:

Place Value System: In the Hindu system, the value of a digit depends on its position (e.g., in 555, the three 5s represent 500, 50, and 5). This allows any number, no matter how large, to be written compactly using a small set of symbols. The Roman system is largely additive (e.g., VIII = 5+1+1+1) and does not have a true place value structure, making large numbers very long and cumbersome to write (e.g., 3888 is MMMDCCCLXXXVIII).
Use of Zero (0): The Hindu system includes the digit 0, which serves two crucial roles:
- As a placeholder: It allows us to distinguish between numbers like 5, 50, and 500. Without a placeholder, a place value system would be ambiguous. The Roman system lacks a zero, which is a major limitation.
- As a number: Zero is treated as a number in its own right, with defined properties for arithmetic operations, which is fundamental to algebra and higher mathematics.
Fixed Base (Base-10): The Hindu system is a base-10 system, where each place value is a power of 10. This consistent structure makes arithmetic operations (addition, subtraction, multiplication, division) systematic and straightforward. The algorithms we use for these operations rely entirely on this base-10 place value structure. Performing calculations in the Roman system is complex and non-algorithmic, often requiring a physical tool like an abacus.

Q4Figure it Out (End of Section 3.2)

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

Solution

Objective: To refine the 'Shape System' created earlier.

Original System: A base-4 system with symbols O (0), | (1), V (2), Δ (3).

Refinements based on historical evolution:

Problem in original system: Writing large numbers requires many symbols. For example, 15 is ΔΔ ( $3 \times 4 + 3$ ). 63 is ΔΔΔ ( $3 \times 16 + 3 \times 4 + 3$ ). This is like a tally system within each place value.
Refinement 1: Introduce Landmark Numbers (like the Roman System): We could introduce a new symbol for a larger number, for instance, a square □ to represent 16 ( $4^2$ ). This is an intermediate step, but not the most efficient.
- Using this, 20 would be □|O (one 16, one 4, zero 1s). This is an improvement but still requires introducing new symbols for higher powers.
Refinement 2: Adopt a Full Place Value System (like the Hindu System): The original system was already defined as a place value system, which is the most efficient idea. The key is to strictly enforce it. The representation of a number in a base-n system should only use n digits (0 to n-1). My system already does this.

Let's re-evaluate the representation:
- Original system: Base-4 with digits O, |, V, Δ for 0, 1, 2, 3.
- Number 15: $15 = 3 \times 4^1 + 3 \times 4^0$ . Representation: ΔΔ. Correct.
- Number 10: $10 = 2 \times 4^1 + 2 \times 4^0$ . Representation: VV. Correct.
The initial design was already based on the most advanced concept (a base-n place value system). The main 'refinement' is to fully appreciate its power compared to earlier systems. It doesn't need new symbols for landmark numbers because the position of a symbol defines its value (power of 4). It avoids ambiguity because each position has only one of the four defined symbols. This system is already quite efficient.

Q1Figure it Out (End of Section 3.3)

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

Solution

Question: Can a single symbol appear 10 or more times in a standard Egyptian numeral representation?

Answer: No, it cannot.

Reasoning: The Egyptian number system is a base-10 system. This means that whenever you have a group of 10 of a certain symbol, it is replaced by a single symbol of the next higher value.

10 sticks (|) are replaced by 1 hobble (∩).
10 hobbles (∩) are replaced by 1 coiled rope (9).
10 coiled ropes (9) are replaced by 1 lotus flower, and so on.

Because of this rule of grouping and replacement, any valid representation of a number in the Egyptian system will have at most 9 of any single symbol. If there were 10 or more, the number would not be in its simplest, standard form.

Q2Figure it Out (End of Section 3.3)

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

Solution

Objective: To create a base-4 number system and represent numbers 1 to 16.

System Name: Quad-Symbol System

Base: 4
Symbols (Digits):
- • (dot) for 1
- •• for 2
- ••• for 3
- For this example, we will not have a symbol for zero, but will use place value. A blank space or context will denote zero.
Representation: This is an additive system like the Egyptian one. To represent a number, we break it down into powers of 4 ( $1, 4, 16, ...$ ).
- Let's assign landmark symbols: • for 1, — for 4, ☐ for 16.

Numbers from 1 to 16:

1: •
2: ••
3: •••
4: —
5: — • ( $4+1$ )
6: — •• ( $4+2$ )
7: — ••• ( $4+3$ )
8: — — ( $4+4$ )
9: — — • ( $4+4+1$ )
10: — — •• ( $4+4+2$ )
11: — — ••• ( $4+4+3$ )
12: — — — ( $4+4+4$ )
13: — — — • ( $4+4+4+1$ )
14: — — — •• ( $4+4+4+2$ )
15: — — — ••• ( $4+4+4+3$ )
16: ☐

Q3Figure it Out (End of Section 3.3)

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

Solution

Given: Our custom base-5 system with landmark numbers being powers of 5 (△=1, ♢=5, □=25, ◯=125, etc.).

To Find: A simple rule for multiplying any number in this system by 5.

Solution: In a base-n system, multiplying by the base n has a very simple effect. Since each landmark number is n times the previous one, multiplying a number by n effectively 'promotes' each of its component symbols to the next higher landmark symbol.

The Rule: To multiply a number by 5 in our base-5 system, replace every symbol in that number's representation with the symbol for the next higher landmark number.

Every triangle (△, for 1) becomes a diamond (♢, for 5).
Every diamond (♢, for 5) becomes a square (□, for 25).
Every square (□, for 25) becomes a circle (◯, for 125).
...and so on.

Example: Let's multiply the number 32 by 5.

First, represent 32 in the base-5 system: $32 = 1 \times 25 + 1 \times 5 + 2 \times 1$ . The representation is □ ♢ △△.
Now, apply the rule: replace each symbol with the next higher one.
- The □ becomes an ◯.
- The ♢ becomes a □.
- The two △s become two ♢s.
The new representation is ◯ □ ♢♢.
Let's check the value: $1 \times 125 + 1 \times 25 + 2 \times 5 = 125 + 25 + 10 = 160$ .
This is correct, since $32 \times 5 = 160$ .

Q1Figure it Out (The Egyptian Number System)

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

Solution

Given: The Egyptian number system symbols:

| (Stick) = 1
∩ (Hobble) = 10
9 (Coiled rope) = 100
(Lotus flower) = 1000
P (Finger) = 10,000
(Tadpole) = 100,000
(Man) = 1,000,000

To Find: Representation of given numbers in the Egyptian system.

Solution:

10458: $1 \times 10000 + 4 \times 100 + 5 \times 10 + 8 \times 1$
- Answer: One Finger, four Coiled ropes, five Hobbles, eight Sticks.
- Symbolically: P 9999 ∩∩∩∩∩ ||||||||
1023: $1 \times 1000 + 2 \times 10 + 3 \times 1$
- Answer: One Lotus flower, two Hobbles, three Sticks.
- Symbolically: (Lotus flower) ∩∩ |||
2660: $2 \times 1000 + 6 \times 100 + 6 \times 10$
- Answer: Two Lotus flowers, six Coiled ropes, six Hobbles.
- Symbolically: (Two Lotus flowers) 999999 ∩∩∩∩∩∩
784: $7 \times 100 + 8 \times 10 + 4 \times 1$
- Answer: Seven Coiled ropes, eight Hobbles, four Sticks.
- Symbolically: 9999999 ∩∩∩∩∩∩∩∩ ||||
1111: $1 \times 1000 + 1 \times 100 + 1 \times 10 + 1 \times 1$
- Answer: One Lotus flower, one Coiled rope, one Hobble, one Stick.
- Symbolically: (Lotus flower) 9 ∩ |
70707: $7 \times 10000 + 7 \times 100 + 7 \times 1$
- Answer: Seven Fingers, seven Coiled ropes, seven Sticks.
- Symbolically: PPPPPPP 9999999 |||||||

Q2Figure it Out (The Egyptian Number System)

What numbers do these numerals stand for? (a) one tadpole, two fingers, three coiled ropes, four hobbles, five sticks. (b) one man, one tadpole, one finger, one coiled rope, one hobble, one stick.

Solution

Given: Collections of Egyptian numerals.

To Find: The value of these collections in the Hindu-Arabic system.

Solution:

(a)

One tadpole = $1 \times 100,000 = 100,000$
Two fingers = $2 \times 10,000 = 20,000$
Three coiled ropes = $3 \times 100 = 300$
Four hobbles = $4 \times 10 = 40$
Five sticks = $5 \times 1 = 5$
Total = $100,000 + 20,000 + 300 + 40 + 5 = 120,345$

Final Answer (a): 120,345

(b)

One man = $1 \times 1,000,000 = 1,000,000$
One tadpole = $1 \times 100,000 = 100,000$
One finger = $1 \times 10,000 = 10,000$
One coiled rope = $1 \times 100 = 100$
One hobble = $1 \times 10 = 10$
One stick = $1 \times 1 = 1$
Total = $1,000,000 + 100,000 + 10,000 + 100 + 10 + 1 = 1,110,111$

Final Answer (b): 1,110,111

Q1Figure it Out (The Mayan Number System)

Represent the following numbers using the Mayan system:

(i)

77

(ii)

100

(iii)

361

(iv)

721 Solution

Given: The Mayan number system.

Symbols: • (dot) for 1, — (bar) for 5, and a seashell for 0.
Place values are written vertically, from bottom to top:
- Bottom level: 1s ( $1$ )
- Second level: 20s ( $20$ )
- Third level: 360s ( $20 \times 18$ )
- Fourth level: 7200s ( $20^2 \times 18$ )

To Find: Representation of given numbers.

Solution:

(i) 77

$77 = (3 \times 20) + 17$
20s place (second level): 3 (•••)
1s place (bottom level): 17 (three bars and two dots ••———)
Answer (written top to bottom): ••• ••———

(ii) 100

$100 = (5 \times 20) + 0$
20s place (second level): 5 (—)
1s place (bottom level): 0 (seashell)
Answer (written top to bottom): — (seashell symbol)

(iii) 361

$361 = (1 \times 360) + (0 \times 20) + 1$
360s place (third level): 1 (•)
20s place (second level): 0 (seashell)
1s place (bottom level): 1 (•)
Answer (written top to bottom): • (seashell symbol) •

(iv) 721

$721 = (2 \times 360) + (0 \times 20) + 1$
360s place (third level): 2 (••)
20s place (second level): 0 (seashell)
1s place (bottom level): 1 (•)
Answer (written top to bottom): •• (seashell symbol) •

Q1Figure it Out (The Mechanism of Counting)

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.

Solution

Given: A number system where numbers are represented by collections of sticks (e.g., 3 is |||, 5 is |||||).

To Find: Methods for arithmetic operations (addition, subtraction, multiplication, division) using only sticks.

Solution: Let us have two collections of sticks, Collection A and Collection B.

Addition (A + B): To add the two collections, simply combine all the sticks from Collection A and Collection B into a single new collection. The total number of sticks in the new collection is the sum.
- Example: To add ||| and ||, we put them together to get |||||.
Subtraction (A - B, where A has more sticks than B): To subtract Collection B from Collection A, take the two collections and for every stick in Collection B, remove one stick from Collection A. The number of sticks remaining in Collection A is the result.
- Example: To subtract || from |||||, we remove two sticks from the larger group, leaving |||.
Multiplication (A × B): To multiply two collections, take one stick from Collection B. Create a new collection that has the same number of sticks as Collection A. Repeat this process for every stick in Collection B. Finally, combine all the newly created collections. The total number of sticks is the product.
- Example: To multiply ||| by ||, we take the first stick from || and create a group of |||. Then we take the second stick from || and create another group of |||. Combining these two groups gives ||| |||, which is ||||||.
Division (A ÷ B): To divide Collection A by Collection B, start with Collection A. From it, repeatedly remove a group of sticks equal in number to the sticks in Collection B. Count how many times you can do this before you can no longer remove a full group. This count is the quotient. The sticks left over, if any, form the remainder.
- Example: To divide ||||||| by |||, we can remove a group of ||| once, leaving ||||. We can remove another group of ||| again, leaving |. We were able to remove the group 2 times, and 1 stick is left over. So, the quotient is 2 (represented as ||) and the remainder is 1 (represented as |).

Q2Figure it Out (The Mechanism of Counting)

One way of extending the number system in Method 2 is by using strings with more than one letter-for example, we could use 'aa' for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!

Solution

Given: A number system using English letters 'a' to 'z' to represent numbers 1 to 26.

To Find: A method to extend this system to represent all numbers.

Solution: This can be achieved by creating a base-26 place value system. The letters 'a' through 'z' can be thought of as 26 distinct digits.

Single-digit numbers (1-26): Use the letters 'a' through 'z' as they are.
- a = 1, b = 2, ..., z = 26.
Two-digit numbers (27 onwards): Use two-letter combinations. The rightmost letter represents the 'units' place (values 1-26), and the letter to its left represents the '26s' place.
- After z (26), the next number is 27. We can represent this as aa. Here, the right 'a' is 1 and the left 'a' represents one group of 26. So, aa = $1 \times 26 + 1 = 27$ . (Note: This is one possible interpretation. A more standard base-26 would map a-z to 0-25, but the question implies a=1). Let's follow the prompt's aa=27 logic.
- If we map a=1, b=2, ..., z=26, then aa could be seen as starting a new cycle. The number 27 would be the first number in the two-letter sequence. So, aa = 27, ab = 28, ..., az = $27 + 25 = 52$ .
- Then, ba = 53, bb = 54, ..., zz = $27 + (26 \times 26 - 1) = 702$ .
General Extension: This system can be extended indefinitely:
- Use three-letter strings (e.g., 'aaa', 'aab', ...) after all two-letter strings are used.
- Use four-letter strings after all three-letter strings, and so on.

This method is analogous to our Hindu-Arabic system, which uses 10 digits and adds more places (tens, hundreds) to represent larger numbers. Here, we are using 26 symbols ('a' through 'z') and creating a base-26 system.

Q3Figure it Out (The Mechanism of Counting)

Try making your own number system.

Solution

Objective: To create a new number system.

My Number System: The Shape System

This will be a base-4 place value system.

Base: The system is base-4, meaning we group things in fours.
Digits (Numerals): There are four symbols for the numbers 0, 1, 2, and 3.
- O (Circle) represents 0.
- | (Line) represents 1.
- V (V-shape) represents 2.
- Δ (Triangle) represents 3.
Place Value: The system uses place value. The rightmost position is the units ( $4^0$ ) place, the next to the left is the fours ( $4^1$ ) place, the next is the sixteens ( $4^2$ ) place, and so on.

Examples of Representing Numbers:

3: Δ
4: This is one group of 4 and 0 units. So, it is represented as |O.
7: This is one group of 4 and 3 units ( $1 \times 4 + 3$ ). So, it is represented as |Δ.
10: This is two groups of 4 and 2 units ( $2 \times 4 + 2$ ). So, it is represented as VV.
16: This is one group of 16, zero groups of 4, and zero units. So, it is represented as |OO.

This system can represent any number using only four basic symbols and the concept of place value.

Q1Figure it Out (The Mesopotamian Number System)

Represent the following numbers in the Mesopotamian system -

(i)

63

(ii)

132

(iii)

200

(iv)

60

(v)

3605

Solution

Given: The Mesopotamian base-60 (sexagesimal) system.

Symbols: ▼ for 1, < for 10.
Place values (right to left): 1s ( $60^0$ ), 60s ( $60^1$ ), 3600s ( $60^2$ ), etc.

To Find: Representation of given numbers.

Solution: We will represent the number of units in each place value, separated by a space for clarity.

(i) 63

$63 = (1 \times 60) + (3 \times 1)$
60s place: 1 (▼)
1s place: 3 (▼▼▼)
Answer: ▼ ▼▼▼

(ii) 132

$132 = (2 \times 60) + (12 \times 1)$
60s place: 2 (▼▼)
1s place: 12 (<▼▼)
Answer: ▼▼ <▼▼

(iii) 200

$200 = (3 \times 60) + (20 \times 1)$
60s place: 3 (▼▼▼)
1s place: 20 (<<)
Answer: ▼▼▼ <<

(iv) 60

$60 = (1 \times 60) + (0 \times 1)$
60s place: 1 (▼)
1s place: 0. This created ambiguity as there was no symbol for zero. A space was used.
Answer: ▼ (This could also be read as 1, which was a flaw in the early system.)

(v) 3605

$3605 = (1 \times 3600) + (0 \times 60) + (5 \times 1)$
3600s place: 1 (▼)
60s place: 0 (represented by a space)
1s place: 5 (▼▼▼▼▼)
Answer: ▼ (space) ▼▼▼▼▼

Q1Figure it Out (The Roman Numerals)

Represent the following numbers in the Roman system.

(i)

1222

(ii)

2999

(iii)

302

(iv)

715 Solution

Given: Numbers in the Hindu-Arabic system.

To Find: Their representation in the Roman numeral system.

Solution: The basic Roman numerals are: I=1, V=5, X=10, L=50, C=100, D=500, M=1000.

(i) 1222

$1222 = 1000 + 200 + 20 + 2$
$1000 = M$
$200 = CC$
$20 = XX$
$2 = II$
So, $1222 = \text{MCCXXII}$

(ii) 2999

$2999 = 2000 + 900 + 90 + 9$
$2000 = MM$
$900 = CM$ (100 less than 1000)
$90 = XC$ (10 less than 100)
$9 = IX$ (1 less than 10)
So, $2999 = \text{MMCMXCIX}$

(iii) 302

$302 = 300 + 2$
$300 = CCC$
$2 = II$
So, $302 = \text{CCCII}$

(iv) 715

$715 = 700 + 10 + 5$
$700 = D + CC = DCC$
$10 = X$
$5 = V$
So, $715 = \text{DCCXV}$

Q2Figure it Out (The Roman Numerals)

Do it yourself now: (b) LXXXVII + LXXVIII

Solution

Given: An addition problem in Roman numerals: LXXXVII + LXXVIII.

To Find: The sum in Roman numerals, without converting to Hindu numerals first.

Solution: We will group the symbols by type (L, X, V, I) and then simplify.

Combine the numerals: LXXXVII + LXXVIII = L XXX V II + L XX V III
Group like symbols:
- L's: L L
- X's: XXX XX
- V's: V V
- I's: II III
Simplify each group:
- L L = C (since $50 + 50 = 100$ )
- XXX XX = XXXXX = L (since $5 \times 10 = 50$ )
- V V = X (since $5 + 5 = 10$ )
- II III = IIIII = V (since $5 \times 1 = 5$ )
Combine the simplified groups: We have C, L, X, and V. Arranging them from largest to smallest: C L X V

Final Answer: LXXXVII + LXXVIII = CLXV

Verification (using Hindu numerals): LXXXVII =

50 + 30 + 7 = 87

LXXVIII =

50 + 20 + 8 = 78

87 + 78 = 165

CLXV =

100 + 50 + 10 + 5 = 165

. The answer is correct.

Q3Figure it Out (The Roman Numerals)

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.

Solution

Objective: To multiply Roman numerals without converting them to the Hindu-Arabic system. This is generally difficult and was done using an abacus historically. For simple cases, we can use the distributive property and knowledge of landmark number products.

Solution:

1. Product of Landmark Numbers:

V × L: This is $5 \times 50$ . We can think of this as five L's: L + L + L + L + L. We know L+L=C, so this is C + C + L = CC L. (Value: 250)
L × D: This is $50 \times 500$ . This is very large. In Hindu-Arabic, $50 \times 500 = 25000$ . This would be written as $\overline{XXV}$ (25 with a bar over it, indicating multiplication by 1000).
V × D: This is $5 \times 500$ . Five D's: D + D + D + D + D. We know D+D=M (1000). So, this is M + M + D = MMD. (Value: 2500)

2. Product of Compound Numerals:

VII × IX: This is $7 \times 9$ $7 \times 9$ . We can use the distributive property. IX can be thought of as (X - I).
- $VII \times (X - I) = (VII \times X) - (VII \times I)$
- $VII \times X = (V+I+I) \times X = (V \times X) + (I \times X) + (I \times X)$ $V II \times X = (V + I + I) \times X = (V \times X) + (I \times X) + (I \times X)$ .
  - $V \times X$ (five X's) = L (50)
  - $I \times X$ = X (10)
  - So, $VII \times X = L + X + X = LXX$ (70).
- $VII \times I = VII$ (7).
- Now subtract: $LXX - VII = 70 - 7 = 63$ . In Roman numerals, 63 is LXIII.

Final Answers:

V × L = CCL
L × D = $\overline{XXV}$
V × D = MMD
VII × IX = LXIII

Q1Figure it Out (Variations on the Egyptian System)

Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50, 137, 293, 651.

Solution

Given: A base-5 system with the following symbols for landmark numbers (powers of 5):

△ (Triangle) = $5^0 = 1$
♢ (Diamond) = $5^1 = 5$
□ (Square) = $5^2 = 25$
◯ (Circle) = $5^3 = 125$

To Find: Representation of the given numbers in this system.

Solution:

15:
- $15 = 3 \times 5 = 3 \times 5^1$
- Answer: ♢♢♢
50:
- $50 = 2 \times 25 = 2 \times 5^2$
- Answer: □□
137:
- $137 = 1 \times 125 + 0 \times 25 + 2 \times 5 + 2 \times 1$
- Answer: ◯ ♢♢ △△
293:
- $293 = 2 \times 125 + 1 \times 25 + 3 \times 5 + 3 \times 1$
- Answer: ◯◯ □ ♢♢♢ △△△
651:
- The next landmark number is $5^4 = 625$ . Let's use a new symbol, say a Star ☆ for 625.
- $651 = 1 \times 625 + 1 \times 25 + 0 \times 5 + 1 \times 1$
- Answer: ☆ □ △

Q2Figure it Out (Variations on the Egyptian System)

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

Solution

Question: Can all positive integers be represented in the described base-5 system?

Answer: No, there is no positive integer that cannot be represented in this system.

Reasoning: This is a base-5 system. Any positive integer can be uniquely expressed as a sum of multiples of powers of 5. This is known as the base-5 representation (or any base-n representation) of a number.

To represent any number N, we follow these steps:

Find the largest power of 5, say $5^k$ , that is less than or equal to N.
Find how many times $5^k$ goes into N. This will be a number from 1 to 4. We write down that many symbols for $5^k$ .
Subtract this value from N to get a remainder.
Repeat the process for the remainder with the next smaller power of 5, $5^{k-1}$ .
Continue until the remainder is zero.

Since this process can be applied to any positive integer, every number has a representation in this base-5 system. The only limitation is that for very large numbers, we would need to invent new symbols for higher powers of 5 (

5^4, 5^5

, etc.), just as the Egyptians did for powers of 10.

Q3Figure it Out (Variations on the Egyptian System)

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

Solution

Given: The concept of a base-n number system.

To Find: The landmark numbers for a base-7 system and a general base-n system.

Solution:

1. Landmark Numbers of a Base-7 System: In a base-n system, the landmark numbers are the powers of n. For a base-7 system, the landmark numbers are the powers of 7.

$7^0 = 1$
$7^1 = 7$
$7^2 = 49$
$7^3 = 343$
$7^4 = 2401$
and so on.

2. Landmark Numbers of a General Base-n System: For any integer n > 1, the landmark numbers of a base-n system are the powers of n.

$n^0 = 1$
$n^1 = n$
$n^2$
$n^3$
$n^4$
and so on, continuing with all non-negative integer powers of n.

A Story of NumbersClass 8 Mathematics NCERT Solutions

Add the following Egyptian numerals:

(2 coiled ropes, 3 hobbles, 4 sticks) + (3 coiled ropes, 8 hobbles, 7 sticks)

(2 fingers, 3 coiled ropes, 5 hobbles, 6 sticks) + (4 fingers, 8 coiled ropes, 7 hobbles, 9 sticks)

Solution

Add the following numerals that are in the base-5 system that we created: ◯◯□△△ + ◯◯◯□□□△△

Solution

Now find the following products- (i) (999 99 ∩∩ ||) × ∩ (ii) (P 𓆼 ∩) × 9

Solution

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?

Solution

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

Solution

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?

Solution

Solution

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?

Solution

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:

(ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon)

(ukasar-ukasar-ukasar-ukasar-urapon) - (ukasar-ukasarukasar)

(ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)

(ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)

Solution

Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.

Solution

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

Solution

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

Solution

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

Solution

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

Solution

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

Solution

What numbers do these numerals stand for? (a) one tadpole, two fingers, three coiled ropes, four hobbles, five sticks. (b) one man, one tadpole, one finger, one coiled rope, one hobble, one stick.

Solution

Represent the following numbers using the Mayan system:

77

100

361

721

Solution

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.

Solution

One way of extending the number system in Method 2 is by using strings with more than one letter-for example, we could use 'aa' for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!

Solution

Try making your own number system.

Solution

Represent the following numbers in the Mesopotamian system -

63

132

200

60

3605

Solution

Represent the following numbers in the Roman system.

1222

2999

302

715

Solution

Do it yourself now: (b) LXXXVII + LXXVIII

Solution

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.

Solution

Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50, 137, 293, 651.

Solution

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

Solution

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

Solution