Friday, August 29, 2025

Class 8 – Ganita Prakash – CHAPTER 3 A STORY OF NUMBERS

 Class 8 – Ganita Prakash – CHAPTER 3 A STORY OF NUMBERS


SUMMARY 

  •  To represent numbers, we need a standard sequence of objects, names, or written symbols that have a fixed order. This standard sequence is called a number system. 
  •  The symbols representing numbers in a written number system are called numerals. 
  •  In a number system, landmark numbers are numbers that are easily recognisable and used as reference points for understanding and working with other numbers. They serve as anchors within the number system, helping people to orient themselves and make sense of quantities, particularly larger ones. 
  •  A number system whose landmark numbers are the powers of a number n is referred to as a base-n number system. 
  •  Number systems having a base that make use of the position of a symbol in determining the landmark number that it is associated with are called positional number systems or place value systems. 
  •  Place value representations were used in the Mesopotamian (Babylonian), Mayan, Chinese and Indian civilisations. 
  •  The system of numerals that we use throughout the world today is the Hindu number system (also sometimes called the Indian number system, or the Hindu-Arabic number system). It is a place value system with (usually) 10 digits, including the digit 0 which is treated on par with other digits. Due to its use of 0 as a number, the system enables the writing of all numbers unambiguously using just finitely many symbols, and also enables efficient computation. The system originated in India around 2000 years ago, and then spread across the world, and is considered one of human history’s greatest inventions.

Page 51

 Q1. How do we ensure that all cows have returned safely after grazing?

 Q2. Do we have fewer cows than our neighbour? 

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

    • Answer: To compare, you would create a separate pile of sticks for your neighbour's herd (e.g., one stick per cow they have). Then, you would try to pair up each stick from your pile with a stick from your neighbour's pile.

    • If your pile runs out of sticks first, you have fewer cows.

    • If your neighbour's pile runs out first, they have fewer cows.

    • If both piles run out at the same time, you both have the same number.

  • Q3: If there are fewer, how many more cows would we need?

    • Answer: After pairing the sticks from the two piles (as in Q2), the number of unpaired sticks remaining in your neighbour's pile directly tells you how many more cows you need to have the same number as them.

Question 2 & 3: These are part of the table asking you to fill in the representations for numbers 1 through 5 and beyond using sticks.

NumberIts representation (using sticks)
1**** (1 stick)
2**** (2 sticks)
3**** (3 sticks)
4**** (4 sticks)
5**** (5 sticks)
.. (The pattern continues)



Page 53

 How many numbers can you represent in this way using the sounds of the letters of your language?
  • Answer: This depends entirely on how many distinct sounds or names you have in your standard sequence.

    • If you use the English alphabet, you can represent 26 numbers (from a=1 to z=26).

    • If you use the Hindi vowels and consonants (เคตเคฐ्เคฃเคฎाเคฒा), you could represent around 52 numbers (13 vowels + 33 consonants + 2 special consonants?).

    • The key point is: The size of your collection of sounds limits how high you can count. This is the major drawback of this method.

Question 2: This is the second "?" in the table, asking to continue the pattern of representing numbers with letters.
Do you see a way of extending this method to represent bigger numbers as well? How?

NumberIts representation (using sounds or names)
1a
2b
3c
4d
5e
..
..
..
26z
27? (This is the question. You need to extend the system.)
  • Answer: The text suggests one way to extend it: use strings of more than one letter. For example:

    • 27 could be aa

    • 28 could be ab

    • ...

    • 52 could be az

    • 53 could be ba

    • This pattern could continue indefinitely, similar to how Excel labels its columns. This creates an unending sequence, solving the limitation of the first method.

 Figure it Out 

Page 54


 1. 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.

  • Adding: Simply combine the two piles of sticks into one large pile. The new pile represents the sum.

    • Example: |||| (4) + ||| (3) = ||||||| (7)

  • Subtracting: Take the smaller pile away from the larger pile. The sticks left in the larger pile represent the difference.

    • Example: |||||| (6) - ||| (3) = ||| (3)

  • Multiplying: Multiplication is repeated addition. For example, to calculate 4 × 3, you would create 4 separate groups, each containing 3 sticks. Then, you combine all groups into one pile.

    • Example: 4 × 3 = ||| + ||| + ||| + ||| = |||||||||||| (12)

  • Dividing: Division is equal sharing. For example, to calculate 12 ÷ 3, you start with a pile of 12 sticks and distribute them one by one into 3 new groups until all sticks are gone. The number of sticks in each final group is the answer.

    • Example: 12 ÷ 3: Group 1: ||||, Group 2: ||||, Group 3: ||||. Answer is 4.

 

 2. 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! 
  • Answer: You can treat the letters like digits in a base-26 number system. Just like we use 0-9 and then go to 10, you can use a-z and then go to aa.

    • Proposed System:

      • a = 1, b = 2, c = 3, ... , z = 26

      • aa = 27, ab = 28, ... , az = 52

      • ba = 53, bb = 54, ... , bz = 78

      • ...

      • za = (26*26 + 1) = 677, ... , zz = (26*26 + 26) = 702

      • aaa = 703, and so on.

    • This is efficient because it uses a finite set of symbols (26 letters) to represent an infinite set of numbers, which was the main drawback of the original Method 2.

 3. Try making your own number system.

  • Answer: This is a creative exercise. A good number system should:

    1. Have a base (a grouping number).

    2. Have a unique symbol for every number from 1 up to (but not including) the base.

    3. Use place value (the position of a symbol changes its value).

  • Example: A Base-4 System

    • Symbols: Let's say your symbols are: 1 = , 2 = :, 3 = 

    • How it works: The rightmost digit is the 1's place. The next digit to the left is the 4's place. The next is the 16's place (4²), and so on.

    • Representing Numbers:

      •  = 1

      • : = 2

      •  = 3

      • ∙∙ = (1×4) + (1×1) = 5

      • :∙ = (2×4) + (1×1) = 9

      • ∴∙ = (3×4) + (1×1) = 13

      • ∙∴ = (1×4) + (3×1) = 7

      • :: = (2×4) + (2×1) = 10

Page 56 (Bottom) - Two ? Marks

Context: The Gumulgal number names are given:

  • 1 = urapon

  • 2 = ukasar

  • 3 = ukasar-urapon (2 + 1)

  • 4 = ukasar-ukasar (2 + 2)

  • 5 = ukasar-ukasar-urapon (2 + 2 + 1)

  • 6 = ukasar-ukasar-ukasar (2 + 2 + 2)

Question 1: Can you see how their number names are formed?

  • Answer: Yes. The Gumulgal system is additive and based on grouping by twos. The name for the number 2 (ukasar) is used as a building block. Larger numbers are formed by combining this word for "two" with the word for "one" (urapon).

    • An odd number ends with -urapon.

    • An even number is made by repeating ukasar.

Question 2: Can you see how the names of the other numbers are formed?

  • Answer: The pattern is consistent. Each number is simply the sum of 2s and 1s, and its name is the corresponding sequence of words:

    • 3 (ukasar-urapon): Formed by combining the word for 2 (ukasar) and the word for 1 (urapon), meaning "two and one".

    • 4 (ukasar-ukasar): Formed by repeating the word for 2 twice, meaning "two and two".

    • 5 (ukasar-ukasar-urapon): Formed by saying the word for 2 twice and then the word for 1, meaning "two and two and one".

    • 6 (ukasar-ukasar-ukasar): Formed by repeating the word for 2 three times, meaning "two and two and two".

This system is efficient for small numbers but becomes very long and cumbersome for larger numbers (e.g., 10 would be ukasar-ukasar-ukasar-ukasar-ukasar). This limitation is why more advanced systems with larger grouping numbers (like base-10) evolved.

Page 57 (Bottom) - One ? Mark

Context: This page discusses the human limit for instantly recognizing small quantities (subitizing) and how this led to grouping tally marks (e.g., grouping by 5s).

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

  • Answer: For most humans, the immediate, accurate perception of quantity without counting (a process called subitizing) is limited to about 4 or 5 objects. Beyond that, you have to consciously count them. This is why tally marks are often grouped in sets of 5 (~~////~~ )—it's much easier to quickly see three groups of 5 (15) than to count 15 individual marks.


Page 58 (Bottom) - One ? Mark

Context: This page introduces the Roman numeral system and its "landmark numbers" (I=1, V=5, X=10, L=50, C=100, D=500, M=1000). It explains how to form numbers by grouping into these landmarks.

Question: Example: Let us take the number 27. ... So, 27 in Roman numerals is XXVII.

  • Answer: The process is shown correctly.

    1. Group into 10s: 27 has two 10s. 10 = X, so XX = 20.

    2. Group the remainder: 27 - 20 = 7.

    3. Group into 5s: 7 has one 5. 5 = V, so now we have XXV = 25.

    4. Group the final remainder: 7 - 5 = 2. 2 = II, so the final numeral is XXV + II = **XXVII**.


Page 59 (Bottom) - Figure it Out & ? Marks

This page has a "Figure it Out" box and a "Math Talk" box with question marks.

FIGURE IT OUT:

1. Represent the following numbers in the Roman system.
(i) 1222
Answer: 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1
* Breakdown: M (1000) + CC (200) + XX (20) + II (2)
Numeral: MCCXXII

(ii) 2999
Answer: This is a tricky one. 2999 = 1000 + 1000 + (1000 - 100) + (100 - 10) + (10 - 1)
* Modern standard form: MM (2000) + CM (900) + XC (90) + IX (9)
Numeral: MMCMXCIX

(iii) 302
Answer: 302 = 100 + 100 + 100 + 1 + 1
* Breakdown: CCC (300) + II (2)
Numeral: CCCII

(iv) 715
Answer: 715 = 500 + 100 + 100 + 10 + 5
* Breakdown: D (500) + CC (200) + X (10) + V (5)
Numeral: DCCXV

MATH TALK (The ? ? Mark):

Question: How will you multiply two numbers given in Roman numerals, without converting them to Hindu numerals?

  • Answer: This is meant to show the extreme difficulty of the task. There is no simple algorithm like in the Hindu system. Methods were incredibly cumbersome:

    1. Repeated Addition: The most basic method. To calculate V × III, you would add V + V + V.

    2. Using an Abacus: Romans used a special counting board called an abacus with grooves for different values (I, X, C, M, etc.). They would place counters on it to represent numbers and then physically manipulate them according to rules for doubling, halving, etc., to perform multiplication and division. This required specialized training.

    3. Doubling and Halving (Egyptian Method): This was a common ancient method that didn't rely on place value. You would create two columns. In one column, you would start with one number and repeatedly double it. In the other, you would start with 1 and repeatedly double it. You would then add the doubled values from the first column that correspond to odd values in the second column after halving.

Trying to multiply large numbers like CCXXXI (231) and MDCCCIII (1803) using only pen and paper with Roman numerals would be a "Daredevil Contest" or a "Fight With a Lion" because it was so impractical. This vividly illustrates the main weakness of non-place-value systems.

Page 60 (Bottom) - One ? Mark

Context: This is in the "Math Talk" box, continuing the discussion on the difficulty of arithmetic with Roman numerals.

Question: Try to find the product of the following pairs of landmark numbers: V × L, L × D, V × D, VII × IX.

  • Answer: The key here is to convert the Roman numerals to their values, multiply, and then see if the product is itself a landmark number.

    • V × L: 5 × 50 = 250. 250 is not a Roman landmark number (the landmarks are I=1, V=5, X=10, L=50, C=100, D=500, M=1000). This shows a problem: the product of two landmark numbers isn't always a landmark number.

    • L × D: 50 × 500 = 25,000. This is far larger than M (1000) and is not a standard landmark number.

    • V × D: 5 × 500 = 2,500. Not a standard landmark number.

    • VII × IX: First, convert: VII = 7, IX = 9. 7 × 9 = 63. 63 is not a landmark number. It would be written as LXIII (50+10+3).

This exercise demonstrates why multiplication was so difficult in the Roman system—it didn't have the elegant property where digit × base simply shifts the value, a key feature of place-value systems.


Page 60-61 - Figure it Out (4 Questions)

1. 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?

  • Answer: This likely relates to cultural or linguistic traditions where certain words are considered appropriate or lucky for specific types of objects (e.g., counting people vs. counting boats vs. counting coconuts). It might also be a way to classify nouns, similar to grammatical gender in other languages, where the counter word depends on the class of the object being counted.

2. Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s... Use this to evaluate the following:

  • First, we need to extend the system based on the pattern:

    • 7 = 6 + 1 = (ukasar-ukasar-ukasar) + urapon = ukasar-ukasar-ukasar-urapon

    • 8 = 6 + 2 = (ukasar-ukasar-ukasar) + ukasar = ukasar-ukasar-ukasar-ukasar

    • 9 = 6 + 2 + 1 = ukasar-ukasar-ukasar-ukasar-urapon

    • 10 = 6 + 2 + 2 = ukasar-ukasar-ukasar-ukasar-ukasar

Now for the operations. The easiest way is to convert the names to numerical values, perform the operation, and then convert back to Gumulgal names.

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

  • Convert: ukasar-ukasar-ukasar-ukasar-urapon = 4+4+1? Wait, let's use the pattern from 5 and 6. The pattern is additive. So:

    • ukasar-ukasar-ukasar-ukasar-urapon = 2+2+2+2+1 = 9

    • ukasar-ukasar-ukasar-urapon = 2+2+2+1 = 7

  • Add: 9 + 7 = 16

  • Convert back: 16 is eight groups of 2. So the name would be ukasar repeated 8 times.

  • Answer: ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar-ukasar)

  • Convert: 9 - 6 = 3

  • Convert back: ukasar-urapon

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

  • Convert: 9 × 4 = 36

  • Convert back: 36 is eighteen groups of 2. This would be an extremely long name: ukasar repeated 18 times.

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

  • Convert: ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar = 14 (7 groups of 2)
    ukasar-ukasar = 4 (2 groups of 2)

  • Divide: 14 ÷ 4 = 3.5

  • This is a problem. This system, designed for counting whole objects, likely has no way to represent fractions. You could say the answer is 3 (ukasar-urapon) with a remainder of 1 (urapon).

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

  • Place Value: The value of a digit depends on its position (e.g., the '3' in 35 is 30, but in 53 it is 3). Roman numerals are additive and don't have this property.

  • A Base (10): All numbers are built from powers of 10, making the system consistent and predictable.

  • The Digit Zero (0): Acts as a placeholder, allowing us to distinguish between numbers like 205 and 25. Romans had no symbol for zero.

  • Only Ten Symbols: Any number can be written using just 0,1,2,3,4,5,6,7,8,9. Romans needed new symbols (I, V, X, L, C, D, M) for larger values.

  • Ease of Calculation: The properties above make algorithms for addition, subtraction, multiplication, and division (like the ones you learn in school) simple and efficient. Performing these operations with Roman numerals is very difficult.

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

  • Answer: This is a creative prompt. A refined system should incorporate the breakthroughs discussed:

    1. Choose a Base: Instead of just tally marks, decide on a grouping number (e.g., 5, 10, 20).

    2. Create Landmark Numbers: Have unique symbols for the base and its powers (e.g., for base-5, have symbols for 1, 5, 25, 125...).

    3. Use an Additive Principle: Represent numbers by adding the symbols for their components (e.g., 17 in a base-5 system is 15 + 2, or 3x5 + 2).

    4. (Advanced) Add Place Value: The most efficient refinement would be to use the position of a symbol to indicate if it represents 1s, 5s, 25s, etc., and then you would only need symbols for 0, 1, 2, 3, and 4.


Page 62 (Bottom) - Figure it Out (2 Questions) & One ? Mark

FIGURE IT OUT:

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

  • Method: Break the number down into the largest possible powers of 10 (the Egyptian landmarks), using the symbols:

    • | = 1

    • n = 10

    • 9 = 100

    •  = 1,000

    • 8 = 10,000

    • ? = 100,000

    • ? = 1,000,000

  • Example for 10458:

    • 10458 = 10,000 + 400 + 50 + 8

    • = (1 x 10,000) + (4 x 100) + (5 x 10) + (8 x 1)

    • Numeral: 8 9999 nnnnn ||||||||

2. What numbers do these numerals stand for?
(Since the specific numerals aren't provided here, the general method is:)

  • Identify each symbol and its value.

  • Add all the values together.

  • Example: If the numeral is ∩∩ 99 nnn ||||, it equals (1000+1000) + (100+100) + (10+10+10) + (1+1+1+1) = 2000 + 200 + 30 + 4 = 2234.

ONE ? MARK (Math Talk):

Question: Will the distributive property hold here? (Referring to multiplying Egyptian numerals).

  • Answer: Yes, the distributive property holds. The text gives the example: ( + + |) × = ( × ) + ( × ) + (| × )

  • This works because multiplication is distributive over addition, regardless of the symbols used to represent the numbers. The Egyptian number + + | is just another way of writing the number 121 (100+20+1). Multiplying it by 10 (n) is the same as (100×10) + (20×10) + (1×10) = 1000 + 200 + 10 = 1210. The property is a fundamental rule of arithmetic, not dependent on the numeral system.

    Page 63 (Bottom) - One ? Mark & Figure it Out

    Context: This page introduces the concept of a base-n number system, using a base-5 system as an example.

    The ? Mark Question:
    Express the number 143 in this new system.

    • Answer: The new system is base-5. The landmark numbers are powers of 5: 1, 5, 25, 125, 625...

    • Step 1: Find the largest landmark number ≤ 143. 125 (5³) fits.

      • 143 ÷ 125 = 1, remainder 18. So we have 1 group of 125.

    • Step 2: Take the remainder (18). The next landmark is 25 (5²). 18 ÷ 25 = 0. So we have 0 groups of 25.

    • Step 3: Next landmark is 5 (5¹). 18 ÷ 5 = 3, remainder 3. So we have 3 groups of 5.

    • Step 4: The remainder is 3. The last landmark is 1 (5⁰). 3 ÷ 1 = 3. So we have 3 groups of 1.

    • Conclusion: 143 = (1 × 125) + (0 × 25) + (3 × 5) + (3 × 1)

    • Using the symbols from the book (○=125, □=25, △=5, ∇=1), the representation is: ○ □□□ △△△ (Note: The book might use a different symbol for 1, like a dot. The three □ symbols represent the zero groups of 25? Wait, let's check the pattern. The text says: "So the number 143 in the new system is ○□□□△△△". This seems to use a positional idea where the number of symbols shows the count, and the absence of a symbol for 25 might be implied. A more precise representation would be 1 symbol for 125, 0 for 25, 3 for 5, and 3 for 1:  △△△ ∇∇∇)

    FIGURE IT OUT:

    1. Write the following numbers in the above base-5 system...: 15, 50, 137, 293, 651.

    • Method: Decompose each number into powers of 5.

    • 15: 15 = (3 × 5) + (0 × 1) -> 3 groups of 5, 0 groups of 1 -> △△△

    • 50: 50 = (1 × 25) + (0 × 5) + (0 × 1) -> 1 group of 25, 0 of 5, 0 of 1 ->  (But to show the places, it might be written with placeholders)

    • 137: 137 = (1 × 125) + (0 × 25) + (2 × 5) + (2 × 1) -> 1, 0, 2, 2 ->  △△ ∇∇

    • 293: 293 = (2 × 125) + (1 × 25) + (3 × 5) + (3 × 1) -> 250 + 25 + 15 + 3 -> 2, 1, 3, 3 -> ○○  △△△ ∇∇∇

    • 651: 651 = (1 × 625) + (0 × 125) + (1 × 25) + (0 × 5) + (1 × 1) -> 625 + 0 + 25 + 0 + 1 -> 1, 0, 1, 0, 1 ->    (assuming ⊕=625)

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

    • Answer: No. A key property of a complete place-value system (even an additive one like this Egyptian-style base-5) is that every positive integer can be represented. The process of repeatedly dividing by the base (5) guarantees a unique representation for any number. This system can represent any number, though the representations for large numbers get very long.

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

    • Answer: The landmark numbers are the powers of the base.

    • Base-7: 7⁰ = 1, 7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401, ...

    • Base-n: n0=1n1=nn2n3n4, ...


    Page 64 (Bottom) - One ? Mark

    Context: This page shows the addition of two Egyptian numerals and asks for the final sum.

    The ? Mark Question: The page shows a calculation: 15 n and 15 | are grouped. Since 10 n gives 9, the sum is partially shown as 9 and nnnnn and then more. The final step has a ? mark.

    • Answer: Let's complete the calculation:

      1. We have 15 n (each n = 10) and 15 | (each | = 1).

      2. Group 10 of the n symbols: 10 n = 1 9 (100). This leaves 5 n.

      3. Group 10 of the | symbols: 10 | = 1 n (10). This leaves 5 |.

      4. Now add the new symbols to what we have: We now have 1 9, 5 n + 1 n = 6 n, and 5 |.

      5. The final sum is: 9 + 6 n + 5 |, which is 100 + 60 + 5 = 165.


    Page 65 (Bottom) - Figure it Out (2 Questions)

    1. Add the following Egyptian numerals:
    (i) and (ii) [Specific numerals are not provided in your query, so the general method is...]

    • Method:

      1. Count all the symbols of each type (9n|).

      2. For every 10 |, convert them to 1 n.

      3. For every 10 n, convert them to 1 9.

      4. For every 10 9, convert them to 1 .

      5. Write down the final count of each symbol after all conversions.

    2. Add the following numerals that are in the base-5 system...: ○○○□△△ + ○○○○□□△△

    • Method: First, convert each symbol group to its value in the base-5 system (○=125, □=25, △=5).

      • First number: ○○○□△△ = 125+125+125 + 25 + 5+5 = 375 + 25 + 10 = 410

      • Second number: ○○○○□□△△ = 125*4 + 25*2 + 5*2 = 500 + 50 + 10 = 560

      • Sum: 410 + 560 = 970

    • Now represent 970 in base-5:

      • 970 ÷ 625 = 1, remainder 345. (○=625)

      • 345 ÷ 125 = 2, remainder 95. (□=125? Wait, let's use consistent symbols. If ○=125, what is the symbol for 625? Let's assume  = 625)

      • 95 ÷ 25 = 3, remainder 20. (□=25)

      • 20 ÷ 5 = 4, remainder 0. (△=5)

      • So 970 = (1 × 625) + (2 × 125) + (3 × 25) + (4 × 5) + (0 × 1)

      • Final Answer:  ○○ □□□ △△△△


    Page 66 (Bottom) - Three ? Marks

    Context: This page is about multiplying numbers in the Egyptian system.

    The Three ? Marks are part of these product problems:

    1. (i) ∩ × ∩ -> 10 × 10 = 100. 100 is represented by 9. So the product is 9.

    2. (ii) 9 × ∩ -> 100 × 10 = 1000. 1000 is represented by . So the product is .

    3. (iii) 999 × ∩ -> This is 300 × 10 = 3000. 3000 = 3 × 1000, so it is represented by ∩∩∩.

    The key insight is: In a base-10 system like the Egyptian one, multiplying any landmark number by 10 gives the next highest landmark number. This is what makes multiplication by the base easy.


    Page 67 (Bottom) - Three ? Marks

    Context: This page continues the multiplication exercise.

    The Three ? Marks are part of these problems:

    1. (i) ∩ × 9 -> 10 × 100 = 1000. The product is .

    2. (ii) 9 × 9 -> 100 × 100 = 10,000. 10,000 is the next landmark (8). The product is 8.

    3. (iii) 999 × 999 -> This is 300 × 300 = 90,000. 90,000 = 9 × 10,000, so it is represented by nine 8 symbols: 8 8 8 8 8 8 8 8 8.

    The key conclusion is: The product of any two landmark numbers in a base system is another landmark number. This is a huge advantage over systems like the Roman numerals.


    Page 68 (Bottom) - Two ? Marks

    Context: This page shows how an abacus based on the decimal system was used in Europe.

    The Two ? Marks are in the abacus diagram:
    The abacus has lines for 1000, 100, 10, and 1. The problem is to add 2907 + 43.

    • ? Mark 1 (How to represent 2907):

      • 2907 = 2000 + 900 + 0 + 7.

      • On the abacus:

        • 1000s line: 2 counters (for 2000)

        • 100s line: 9 counters (for 900). But wait, the abacus also uses a counter above the line worth 5. So 900 is better shown as 1 counter above (500) and 4 counters below (4x100=400). Total: 500+400=900.

        • 10s line: 0 counters (for 0)

        • 1s line: 7 counters (for 7). This would be 1 counter above (5) and 2 counters below (2x1=2). Total: 5+2=7.

    • ? Mark 2 (How to represent 43):

      • 43 = 0 + 0 + 40 + 3.

      • On the abacus:

        • 1000s line: 0

        • 100s line: 0

        • 10s line: 4 counters (for 40)

        • 1s line: 3 counters (for 3)

    The hint explains the final step: To add, combine the counters on each line.

    • 1s line: 7 (from 2907) + 3 (from 43) = 10. This is exactly 10 ones, which is equal to 1 ten. So, remove all 10 counters from the 1s line and add 1 counter to the 10s line.

    • 10s line: 0 (from 2907) + 4 (from 43) + 1 (from the 1s line) = 5.

    • 100s line: 9 (from 2907) + 0 (from 43) = 9.

    • 1000s line: 2 (from 2907) + 0 (from 43) = 2.

    • Final Sum: The abacus now shows 2 on the 1000s line, 9 on the 100s line, 5 on the 10s line, and 0 on the 1s line. This represents 2950. (2907 + 43 = 2950).

      Page 69-70 - Figure it Out

      Context: This section discusses the shortcomings of the Egyptian system and introduces the final, crucial idea: the place value system.

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

      • Answer: No, there cannot be. This is the fundamental rule and advantage of the Egyptian system. If you have 10 of any symbol, you must group them into the next higher landmark number.

        • 10 | (1s) become 1 n (10).

        • 10 n (10s) become 1 9 (100).

        • 10 9 (100s) become 1  (1000).

        • This rule continues for all higher powers of 10. This grouping principle is what makes the system efficient and prevents representations from becoming impossibly long for large numbers.

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

      • Answer: Let's create a simple base-4 (quaternary) system.

        • Landmark Numbers: 4⁰ = 1, 4¹ = 4, 4² = 16, 4³ = 64...

        • Symbols for 0-3: Since it's base-4, we need symbols for 0, 1, 2, and 3. Let's use: 0 = ร˜, 1 = |, 2 = ‖, 3 = ¶.

        • Representing Numbers (Positional System): We'll use a place value system where the rightmost digit is 1s, the next is 4s, the next is 16s, etc.

      NumberCalculation (Base-4)Representation
      11|
      22
      33
      41×4 + 0×1|ร˜
      51×4 + 1×1||
      61×4 + 2×1|‖
      71×4 + 3×1
      82×4 + 0×1‖ร˜
      92×4 + 1×1‖|
      102×4 + 2×1‖‖
      112×4 + 3×1‖¶
      123×4 + 0×1¶ร˜
      133×4 + 1×1¶|
      143×4 + 2×1¶‖
      153×4 + 3×1¶¶
      161×16 + 0×4 + 0×1|ร˜ร˜

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

      • Answer: In a base-5 system, multiplying a number by 5 (which is the base, written as 10 in base-5) has the same effect as multiplying by 10 in base-10: it shifts every digit one place to the left and adds a zero at the end.

        • Simple Rule: To multiply any number by 5 in base-5, append a zero to the right-hand end of the number.

        • Example 1: 3 (base-5) is 3 (decimal). 3 × 5 = 15.

          • 15 in base-5 is (3×5) + (0×1) = 30.

          • 3 -> 30 (Appending a zero works).

        • Example 2: 12 (base-5) is (1×5 + 2×1) = 7 (decimal). 7 × 5 = 35.

          • 35 in base-5 is (1×25) + (2×5) + (0×1) = 120.

          • 12 -> 120 (Appending a zero works).


      Page 71 (Bottom) - One ? Mark

      Context: This page introduces the Mesopotamian (Babylonian) base-60 (sexagesimal) system.

      The ? Mark Question: Note that we have actually used Indian numerals in creating these symbols.

      • Answer: This is a note from the authors. They are using familiar Hindu-Arabic numerals (1, 2, 3, ...) to stand in for the actual, more complex cuneiform symbols used by the Mesopotamians. They did this to make the concept easier for you to understand and remember. They are not claiming the Mesopotamians used Indian numerals. The real Mesopotamian symbols were wedges pressed into clay tablets (e.g.,  for 1,  for 10).


      Page 72 (Bottom) - Two ? Marks

      Context: This page shows how to represent numbers in the Mesopotamian system.

      The First ? Mark: ? Can we represent this more compactly? (Referring to the number 640).

      • Answer: Yes. Instead of drawing ten separate symbols for 60 and four for 10, we can use the principle of place value. We can write the symbol for 10 (how many 60s we have) followed by the symbol for 40 (the remainder). The position of the '10' symbol tells us it represents 10×60, not 10×1. The compact representation is 10 40 (in Mesopotamian symbols).

      The Second ? Mark: ? Example: Let us try another number — 7530.

      • Answer: Let's represent 7530.

        1. 7530 ÷ 3600 = 2, remainder 330. (2 groups of 3600)

        2. 330 ÷ 60 = 5, remainder 30. (5 groups of 60)

        3. The remainder is 30. (30 groups of 1)

        • So, 7530 = (2 × 3600) + (5 × 60) + (30 × 1)

        • Its compact representation is 2 5 30 (where the 2 is in the 3600s place, the 5 in the 60s place, and the 30 in the 1s place).


      Page 73 (Bottom) - Figure it Out & One ? Mark

      FIGURE IT OUT:

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

      • Method: Break each number down into groups of 3600, 60, and 1.

        • (i) 63: 63 = (1 × 60) + (3 × 1) -> 1, 3 -> 1 3

        • (ii) 132: 132 = (2 × 60) + (12 × 1) -> 2, 12 -> 2 12

        • (iii) 200: 200 = (3 × 60) + (20 × 1) -> 180 + 20 -> 3, 20 -> 3 20

        • (iv) 60: 60 = (1 × 60) + (0 × 1) -> 1, 0 -> 1 (This creates ambiguity! This is the big problem the system had before zero).

        • (v) 3605: 3605 = (1 × 3600) + (0 × 60) + (5 × 1) -> 1, 0, 5 -> 1 5 (Extremely ambiguous without a zero placeholder).

      ONE ? MARK:

      The ? Mark Question: ? Can we represent this more compactly by dropping the symbols for the different powers of 60 altogether?

      • Answer: Yes, and this is the brilliant final step. We can drop the special symbols for 1, 60, 3600, etc. The position of a numeral now tells us what power of 60 it multiplies.

        • The rightmost position is for 60⁰ (1s).

        • The next position to the left is for 60¹ (60s).

        • The next is for 60² (3600s).

        • And so on.

      • This creates a true place-value system. The representation for 640 becomes just 10 40, where the '10' is understood to be in the 60s place. This is incredibly compact and efficient.

        Page 76 (Bottom) - One ? Mark

        Context: This page discusses the Mayan number system. It shows a Mayan numeral and breaks down its value based on its vertical position.

        The ? Mark Question: How is this to be read? (Referring to a specific Mayan numeral).

        • Answer: Mayan numerals are read from the bottom up.

          1. The bottommost group of symbols represents the number of 1s (20⁰).

          2. The group above it represents the number of 20s (20¹).

          3. The group above that represents the number of 360s (20 × 18). This is the puzzling part—instead of 400 (20²), they used 360.

          4. The next group up would represent the number of 7200s (360 × 20), and so on.

        The page provides the calculation for the specific numeral:

        • Bottom: 3 dots (3) - 3 ones

        • Middle: 2 dots and 2 bars (2 + 10 = 12) - 12 twenties (12 × 20 = 240)

        • Top: 4 dots (4) - 4 three-hundred-sixties (4 × 360 = 1440)

        • Total: 1440 + 240 + 3 = 1683

        The calculation in your image ((4)×360 + (11)×20 + (3)×1 = 1660) seems to have a typo. Based on the standard Mayan system and the description of the symbols (2 dots + 2 bars = 12, not 11), the total should be 1683.


        Page 78 (Bottom) - One ? Mark

        Context: This page introduces the Chinese rod numeral system. It shows a numeral and explains that the orientation of the digits (Zong vs. Heng) alternates to indicate place value and prevent ambiguity.

        The ? Mark Question: How is this to be read? (Referring to a specific Chinese rod numeral).

        • Answer: Chinese rod numerals use alternating digit orientations to clearly mark place value.

          • The rightmost digit is in Zong (vertical) orientation. This is the 1s place (10⁰).

          • The digit to its left is in Heng (horizontal) orientation. This is the 10s place (10¹).

          • The next digit to the left is back to Zong orientation. This is the 100s place (10²).

          • The next digit is Heng again. This is the 1000s place (10³).

          • This pattern continues alternating for every higher place value.

        The page provides the calculation for the specific numeral (from top to bottom in the image):

        • Top digit (Heng): ไบŒ (2) - This is in the 1000s place: 2 × 1000 = 2000

        • Next digit (Zong): ๐ญ (6) - This is in the 100s place: 6 × 100 = 600

        • Next digit (Heng): ไธ‰ (3) - This is in the 10s place: 3 × 10 = 30

        • Bottom digit (Zong): ๐ฃ (4) - This is in the 1s place: 4 × 1 = 4

        • Total: 2000 + 600 + 30 + 4 = 2634


        Page 80 (Bottom) - Figure it Out

        Context: This is the final "Figure it Out" section, asking reflective questions about the evolution of number systems.

        1. 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?

        • Answer:

          • Reason for Alternation: They alternated symbols to clearly distinguish each place value and prevent ambiguity. Without alternation, a number like ไธ…ไธ€ (21) could be misread as ไธ‰ (3) if the spacing was unclear. The change in orientation acts as a built-in separator.

          • Representing 41 with only Zong: 41 is 4 tens and 1 one. Using only Zong symbols, it would be written as ๐ฃ (4) followed by | (1): ๐ฃ|

          • Ambiguity: This numeral could easily be misinterpreted. Without clear spacing, ๐ฃ| might look like a single digit. It could be misread as ๐ค (5) or ๐ฅ (6), or simply as 41 if the spacing was large enough. The alternating system removes this uncertainty entirely.

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

        • Answer:

          • Base-2 (Binary) System:

            • Let urapon (1) be the digit 1.

            • Let ukasar (2) be the digit 0. (We need a second symbol, and 0 is perfect).

            • This is a standard binary system where the place values are powers of 2 (1, 2, 4, 8, 16...).

            • Examples:

              • 1 (urapon) = 1

              • 10 (urapon ukasar) = (1×2) + (0×1) = 2

              • 11 (urapon urapon) = (1×2) + (1×1) = 3

              • 100 (urapon ukasar ukasar) = (1×4) + (0×2) + (0×1) = 4

          • Comparison with Gumulgal System:

            • Similarity: Both systems are built upon the idea of grouping by twos.

            • Difference: The Gumulgal system is additive and unary. The number is the sum of its parts, and the length of the number name grows linearly with the number itself (e.g., 6 is "ukasar-ukasar-ukasar").

            • The base-2 system is positional. It uses place value, so the length of the numeral only grows logarithmically with the number. It is infinitely more efficient for representing large numbers (e.g., 6 is just 110).

        3. 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?

        • Answer:

          • Daily Life & Professions: Hindu numerals and 0 are the foundation of almost every modern field.

            • Daily Life: Telling time, managing money (banking, shopping), using phones/computers, measuring ingredients, reading addresses.

            • Professions:

              • Science & Engineering: All calculations, physics, chemistry, construction.

              • Computing: The entire digital world is built on binary (a base-2 system using 0 and 1).

              • Finance & Economics: Accounting, stock markets, economic modeling.

              • Navigation & Surveying: GPS, maps, architecture.

          • Life Without This System: Our world would be unrecognizably primitive.

            • Science and Technology would be severely hampered. Complex calculations for astronomy, physics, and engineering would be nearly impossible without efficient algorithms.

            • Computers would not exist. The binary system is a direct descendant of the Hindu place-value system.

            • Global trade and commerce would be inefficient and limited, relying on clumsy tools like abacuses or complex mental math only experts could perform.

            • General literacy would be lower, as mathematics would remain a specialized skill for experts rather than a universal tool.

        4. 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?

        • Answer: The symbols (0,1,2,3,4,5,6,7,8,9) would be the same, but we would only use a subset of them, and the place values would change.

          • Base-8 (Octal): Digits used: 0, 1, 2, 3, 4, 5, 6, 7.

            • Place values: ..., 512, 64, 8, 1.

            • 25 in Base-8: 25 ÷ 8 = 3, remainder 1. So, 25₁₀ = 31₈ (3×8 + 1×1).

          • Base-5: Digits used: 0, 1, 2, 3, 4.

            • Place values: ..., 125, 25, 5, 1.

            • 25 in Base-5: 25 ÷ 5 = 5, remainder 0. 5 ÷ 5 = 1, remainder 0. So, 25₁₀ = 100₅ (1×25 + 0×5 + 0×1).

          • Base-2 (Binary): Digits used: 0, 1.

            • Place values: ..., 32, 16, 8, 4, 2, 1.

            • 25 in Base-2: 25 - 16 = 9. 9 - 8 = 1. 1 - 1 = 0.

            • So, it uses one 16 (1), one 8 (1), zero 4 (0), zero 2 (0), and one 1 (1).

            • 25₁₀ = 11001₂

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