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Class 8 Ganita Prakash · Chapter 3
Class 8 Ganita Prakash · Chapter 3
A Story of Numbers
Page-wise textbook solutions (pages 48–81) · Every step · Figures · Common mistakes · Tips
✔ Understood 0/44
๐️ Key ideas of this chapter
- To count, we need a fixed standard sequence and a one-to-one mapping with the objects.
- Big ideas in order: counting in groups → landmark numbers (Roman) → a base (Egyptian, base 10) → place value (Mesopotamian, Mayan, Chinese) → zero as a digit and a number (India).
- Landmark numbers of a base-n system: \(1, n, n^2, n^3, \dots\)
- The Hindu (Indian) number system: base 10, place value, 10 digits including 0 — used all over the world.
๐ Textbook Page 48
๐ Page 48Think
Reema's questions: Since when have humans been counting? What were they counting and why?
1Even in the Stone Age (tens of thousands of years ago), people counted.
2They counted food, animals in their herd, goods for trade and offerings in rituals.
3They also counted days to predict the new moon, full moon and seasons.
✅ Answer: Humans have counted since the Stone Age — for food, animals, trade and calendars.
๐ Textbook Page 51
๐ Page 51Math Talk
Without number names or written numbers: (Q1) how do we check that all cows came back? (Q2) do we have fewer cows than our neighbour? (Q3) how many more do we need?
1Q1: Keep one stick for each cow in the morning. In the evening, match each returning cow with one stick. A stick left over means a cow is missing.
2Q2: Put our sticks and the neighbour's sticks side by side and pair them one with one.
3If the neighbour has sticks left over, we have fewer cows.
4Q3: The number of the neighbour's left-over sticks = the number of cows we need.
| Cows | ๐ ๐ ๐ ๐ ๐ |
| Our sticks | | | | | | |
| Neighbour's sticks | | | | | | | | |
Pair one of our sticks with one of the neighbour's. The 2 sticks left over are how many more cows we need.
✅ Answer: Use one-to-one matching with sticks: leftovers tell us who has more and by how many.
๐ก Tip: This idea — pairing one object with one object — is called a one-to-one mapping. It is the heart of counting.
๐ Textbook Page 53
๐ Page 53Math Talk
How many numbers can you represent using the sounds of the letters of your language?
1English has 26 letters → numbers 1 to 26 only.
2Hindi (Devanagari) has about 46–52 letters (vowels + consonants) → about 50 numbers.
3Any language has a fixed number of letters, so it stops somewhere.
✅ Answer: Only as many numbers as there are letters — a limited list.
๐ Page 53Think
In Table 1 (I, II, III, IV, V …, XX), how could you extend the method to bigger numbers?
1Keep using X for 10: XXX = 30.
2Make new symbols for bigger groups: L = 50, C = 100, D = 500, M = 1000.
3Example: 67 = 50 + 10 + 5 + 1 + 1 = LXVII.
✅ Answer: Introduce new symbols for bigger landmark numbers (L, C, D, M, …).
๐ Textbook Page 54
๐ Page 54Figure it Out · Q1
Using only sticks (no number names or digits), give a method to add, subtract, multiply and divide two collections of sticks.
1Add: put both collections together into one heap.
2Subtract: pair each stick of the smaller heap with a stick of the bigger heap and remove the pairs. What is left is the difference.
3Multiply A × B: for every stick in B, place a full copy of heap A. All the copies together are the product.
4Divide A ÷ B: keep taking away groups the size of B from A. Put one stick aside for each group taken. Those sticks = quotient; leftover sticks of A = remainder.
✅ Answer: Adding = joining; subtracting = pairing and removing; multiplying = repeated copies; dividing = repeated taking away.
๐ Page 54Figure it Out · Q2
Method 2 uses a to z for 1 to 26. Extend it with strings of letters (like ‘aa’ for 27) to represent all numbers.
1One way: after z (26) use two letters: aa = 27, ab = 28, …, az = 52, ba = 53, …, zz = 702.
2After zz use three letters: aaa = 703, and so on.
3Another simple way: aa = 27, bb = 28, …, zz = 52, aaa = 53, … (but this needs very long strings).
✅ Answer: Use longer and longer letter-strings in a fixed order — there are many correct ways.
๐ก Tip: The first way (aa, ab, ac …) is like a base-26 counting system — just like a car odometer.
๐ Page 54Figure it Out · Q3
Try making your own number system.
1Choose symbols, e.g. ● = 1, ▲ = 5, ■ = 25.
2Rule: group as many ■ as possible, then ▲, then ●.
3Example: 38 = 25 + 5 + 5 + 1 + 1 + 1 = ■ ▲ ▲ ● ● ●
✅ Answer: Any clear system with symbols and grouping rules is fine.
๐ Textbook Page 56
๐ Page 56Think
Gumulgal numbers: 1 urapon, 2 ukasar, 3 ukasar-urapon, 4 ukasar-ukasar, 5 ukasar-ukasar-urapon, 6 ukasar-ukasar-ukasar. How are the names formed?
1They count in 2s.
23 = 2 + 1, 4 = 2 + 2, 5 = 2 + 2 + 1, 6 = 2 + 2 + 2.
3Any number bigger than 6 was called ‘ras’ (many).
✅ Answer: Each name is made by joining ‘ukasar’ (2) as many times as possible and ‘urapon’ (1) if needed.
๐ Textbook Page 57
๐ Page 57Activity
Look at boxes with different numbers of dots. Up to what group size can you see the number at a glance, without counting?
1Most people see 1, 2, 3 and 4 dots instantly.
2With 5 or more dots we start counting one by one.
✅ Answer: Usually up to 4. That is why tally marks are often grouped in 5s.
๐ Textbook Page 58
๐ Page 58Math Talk
What are the difficulties of counting only in groups of one size? How would you write 1345 if you count only in 5s?
1\(1345 \div 5 = 269\)
2So 1345 needs 269 copies of the ‘5’ symbol!
3Writing and reading such a long numeral is slow and error-prone.
✅ Answer: 1345 = 269 fives — far too long. One group size is not enough for big numbers.
๐ Textbook Page 59
๐ Page 59Figure it Out · Q1
Write in Roman numerals: (i) 1222 (ii) 2999 (iii) 302 (iv) 715
1(i) 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1 = MCCXXII
2(ii) 2999 = 1000 + 1000 + 900 + 90 + 9 = MM + CM + XC + IX = MMCMXCIX
3(Using only adding: MMDCCCCLXXXXVIIII — also allowed in old times.)
4(iii) 302 = 100 + 100 + 100 + 1 + 1 = CCCII
5(iv) 715 = 500 + 100 + 100 + 10 + 5 = DCCXV
✅ Answer: (i) MCCXXII (ii) MMCMXCIX (iii) CCCII (iv) DCCXV
⚠️ Common mistake: Writing 2999 as IMMM (3000 − 1). Only I, X, C go before the next two bigger symbols (IV, IX, XL, XC, CD, CM).
๐ Page 59Worked
Add without converting: (a) CCXXXII + CCCCXIII
1Collect symbols: C: 2 + 4 = 6, X: 3 + 1 = 4, I: 2 + 3 = 5.
25 C = D, so 6 C = D + C.
35 I = V.
4Sum = D C XXXX V = DCXLV
✅ Answer: DCXLV (= 645)
๐ Textbook Page 60
๐ Page 60Practice
Add without converting: (b) LXXXVII + LXXVIII
1Collect: L: 1 + 1 = 2, X: 3 + 2 = 5, V: 1 + 1 = 2, I: 2 + 3 = 5.
25 I = V → now V: 3. Two V = X → X: 6 and one V left.
35 X = L → L: 3, X: 1. Two L = C → C: 1, L: 1.
4Sum = C L X V = CLXV
✅ Answer: CLXV (= 165)
๐ก Tip: Group from the smallest symbol upwards, just like carrying in normal addition.
๐ Page 60Try This
Multiply landmark numbers without converting: V × L, L × D, V × D, VII × IX. Also try CCXXXI × MDCCCLII.
1V × L: 5 fifties = 250 = CCL
2L × D: 50 five-hundreds = 25 000 = 25 M's (MMMMMMMMMMMMMMMMMMMMMMMMM)
3V × D: 5 five-hundreds = 2500 = MMD
4VII × IX = 63 = LXIII
5CCXXXI × MDCCCLII = 231 × 1852 = 4,27,812 → 427 M's followed by DCCCXII — very hard to do in Roman numerals!
✅ Answer: CCL, 25 M's, MMD, LXIII. The big product shows why Roman numerals are bad for multiplication.
๐ Page 60Figure it Out · Q1
Some people of a Pacific island use different sequences of number names to count different objects. Why might they do this?
1Different objects are counted differently: coconuts may be counted in pairs, fish in bundles, canoes one by one.
2The name of the number also tells what is being counted, so trade is clear.
✅ Answer: Because different things were grouped and traded in different ways.
๐ Page 60Figure it Out · Q2
Extend the Gumulgal system beyond 6 (counting in 2s) and evaluate:
(i) (uk-uk-uk-uk-ur) + (uk-uk-uk-ur)
(ii) (uk-uk-uk-uk-ur) − (uk-uk-uk)
(iii) (uk-uk-uk-uk-ur) × (uk-uk)
(iv) (uk×8) ÷ (uk-uk)
uk = ukasar (2), ur = urapon (1)
(i) (uk-uk-uk-uk-ur) + (uk-uk-uk-ur)
(ii) (uk-uk-uk-uk-ur) − (uk-uk-uk)
(iii) (uk-uk-uk-uk-ur) × (uk-uk)
(iv) (uk×8) ÷ (uk-uk)
uk = ukasar (2), ur = urapon (1)
1(i) Join them: uk×4 + uk×3 and ur + ur. Two ur make one uk → uk×8. (9 + 7 = 16)
2(ii) Remove 3 uk from 4 uk → uk-ur. (9 − 6 = 3)
3(iii) uk-uk means ‘double twice’. (uk×4 + ur) doubled = uk×8 + uk = uk×9; doubled again = uk×18. (9 × 4 = 36)
4(iv) Make groups of uk-uk from uk×8 → 4 groups = uk-uk. (16 ÷ 4 = 4)
✅ Answer: (i) ukasar ×8 (ii) ukasar-urapon (iii) ukasar ×18 (iv) ukasar-ukasar
๐ Textbook Page 61
๐ Page 61Figure it Out · Q3
Which features make the Hindu number system more efficient than the Roman system?
1Only 10 symbols (0–9) are needed for every number — Roman needs new symbols for bigger numbers.
2Place value: the position tells the value (5 in 50 vs 500).
3Zero as a digit and as a number — no gaps or confusion.
4Easy written methods for +, −, ×, ÷.
✅ Answer: Base 10, place value, zero, and easy calculation.
๐ Page 61Figure it Out · Q4
Refine the number system you made earlier using these ideas.
1Use landmark numbers that grow by the same factor, e.g. ● = 1, ▲ = 5, ■ = 25, ◆ = 125.
2Then you never need more than 4 of any one symbol.
✅ Answer: Own answer — use a base!
๐ Textbook Page 62
๐ Page 62Figure it Out · Q1
Write in Egyptian numerals: 10458, 1023, 2660, 784, 1111, 70707.
1Break each number into place values, then draw that many symbols of each kind.
210458 = 1 (10 000) + 4 (100) + 5 (10) + 8 (1)
31023 = 1 (1000) + 2 (10) + 3 (1)
42660 = 2 (1000) + 6 (100) + 6 (10)
5784 = 7 (100) + 8 (10) + 4 (1)
61111 = one of each: 1000, 100, 10, 1
770707 = 7 (10 000) + 7 (100) + 7 (1)
✅ Answer: See the drawings.
๐ก Tip: A zero digit simply means: draw nothing for that symbol.
๐ Page 62Figure it Out · Q2
What numbers do these numerals stand for? (i) 2 coils, 7 heel bones, 6 strokes (ii) 4 lotus flowers, 3 coils, 2 heel bones, 2 strokes
1(i) 2 × 100 + 7 × 10 + 6 × 1 = 276
2(ii) 4 × 1000 + 3 × 100 + 2 × 10 + 2 × 1 = 4322
✅ Answer: (i) 276 (ii) 4322
๐ Textbook Page 63
๐ Page 63Worked
Write 143 in our base-5 system.
1Largest landmark not more than 143 is 125.
2143 = 125 + 5 + 5 + 5 + 1 + 1 + 1
3= 1 circle, 3 squares, 3 triangles
✅ Answer: ○ □□□ △△△
๐ Page 63Figure it Out · Q1
Write in the base-5 system: 15, 50, 137, 293, 651.
115 = 5 + 5 + 5 → 3 squares
250 = 25 + 25 → 2 hexagons
3137 = 125 + 5 + 5 + 1 + 1 → 1 circle, 2 squares, 2 triangles
4293 = 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1 → 2 circles, 1 hexagon, 3 squares, 3 triangles
5651 = 625 + 25 + 1 → 1 wave, 1 hexagon, 1 triangle
✅ Answer: See the drawings.
⚠️ Common mistake: Using 5 or more of one shape. 5 of any shape must be swapped for 1 of the next shape.
๐ Page 63Figure it Out · Q2
Is there a number that cannot be written in our base-5 system? Why or why not?
1Every counting number can be written: keep taking the biggest landmark that fits.
2But there is no symbol for zero (nothing to draw).
✅ Answer: All counting numbers can be written; only 0 cannot (no symbol for zero).
๐ Page 63Figure it Out · Q3
Find the landmark numbers of a base-7 system. What are the landmark numbers of a base-n system?
1Base 7: \(7^0 = 1,\ 7^1 = 7,\ 7^2 = 49,\ 7^3 = 343,\ 7^4 = 2401, \dots\)
2Base n: \(1, n, n^2, n^3, \dots\)
✅ Answer: 1, 7, 49, 343, 2401, …; in general \(n^0, n^1, n^2, n^3, \dots\)
๐ Textbook Page 65
๐ Page 65Figure it Out · Q1
Add the Egyptian numerals: (i) [9 lotus, 6 coils, 8 strokes] + [5 coils, 7 strokes] (ii) [1 lotus, 8 heel bones] + [4 heel bones, 6 strokes]
1(i) First number = 9608, second = 507.
2Strokes: 8 + 7 = 15 → 1 heel bone + 5 strokes.
3Heel bones: 0 + 0 + 1 = 1.
4Coils: 6 + 5 = 11 → 1 lotus + 1 coil.
5Lotus: 9 + 1 = 10 → 1 finger (10 000).
6Sum = 1 finger, 1 coil, 1 heel bone, 5 strokes = 10115
7(ii) 1080 + 46: heel bones 8 + 4 = 12 → 1 coil + 2 heel bones; strokes 6.
8Sum = 1 lotus, 1 coil, 2 heel bones, 6 strokes = 1126
✅ Answer: (i) 10115 (ii) 1126
๐ก Tip: Exactly like carrying in our addition: every 10 of one symbol becomes 1 of the next.
๐ Page 65Figure it Out · Q2
Add in base-5: [○ ⬡ ⬡ □ △ △] + [○ ○ ○ ⬡ □ □ △ △]
1First = 125 + 25 + 25 + 5 + 1 + 1 = 182. Second = 375 + 25 + 10 + 2 = 412.
2Triangles: 2 + 2 = 4
3Squares: 1 + 2 = 3
4Hexagons: 2 + 1 = 3
5Circles: 1 + 3 = 4
6No shape reaches 5, so no regrouping. Sum = ○○○○ ⬡⬡⬡ □□□ △△△△ = 594
✅ Answer: 4 circles, 3 hexagons, 3 squares, 4 triangles (= 594)
๐ Textbook Page 66
๐ Page 66Think
What is any landmark number multiplied by 10 (heel bone)? Find: 10 × 10, 100 × 10, 1000 × 10, 10 000 × 10.
1\(10 \times 10 = 10^2 = 100\) (coil)
2\(10^2 \times 10 = 10^3\) (lotus)
3\(10^3 \times 10 = 10^4\) (finger)
4\(10^4 \times 10 = 10^5\) (tadpole)
✅ Answer: Always the NEXT landmark number.
๐ Page 66Think
Multiply each by 100 (coil): 10 × 100, 100 × 100, 1000 × 100, 10 000 × 100.
1\(10 \times 10^2 = 10^3\)
2\(10^2 \times 10^2 = 10^4\)
3\(10^3 \times 10^2 = 10^5\)
4\(10^4 \times 10^2 = 10^6\)
✅ Answer: It jumps 2 landmarks up each time.
๐ Textbook Page 67
๐ Page 67Think
Find: (i) \(10 \times 10^5\) (ii) \(10^2 \times 10^3\) (iii) \(10^3 \times 10^3\) (iv) \(10^4 \times 10^6\)
1Add the powers: \(10^a \times 10^b = 10^{a+b}\).
2(i) \(10^6\) (ii) \(10^5\) (iii) \(10^6\) (iv) \(10^{10}\)
3Note: Egyptians had no symbol for \(10^{10}\) — their symbols stopped at \(10^7\).
✅ Answer: (i) \(10^6\) (ii) \(10^5\) (iii) \(10^6\) (iv) \(10^{10}\) — the product of two landmark numbers is always a landmark number.
๐ Page 67Math Talk
Does this property hold in our base-5 system? In any system with a base?
1Base 5: \(5^a \times 5^b = 5^{a+b}\), another power of 5. Example: 5 × 25 = 125 (□ × ⬡ = ○).
2Base n: \(n^a \times n^b = n^{a+b}\) always.
✅ Answer: Yes, for every number system with a base.
๐ Textbook Page 68
๐ Page 68Practice
Find: (i) (5 coils, 2 heel bones, 2 strokes) × heel bone (ii) (1 lotus, 1 heel bone) × heel bone. What is a simple rule for multiplying by 10?
1(i) 522 × 10: each symbol becomes the next one → 5 lotus, 2 coils, 2 heel bones = 5220
2(ii) 1010 × 10: lotus → finger, heel bone → coil → 1 finger, 1 coil = 10100
3Rule: replace every symbol by the next bigger symbol.
✅ Answer: (i) 5220 (ii) 10100. Rule: move every symbol up one landmark.
๐ Textbook Page 69
๐ Page 69Abacus
On the abacus, 2907 and 43 are placed on the two sides. How do we find the sum? What if a line gets 10 or more?
1Bring the counters on each line together.
2Ones: 7 + 3 = 10 → remove them and put 1 counter on the tens line.
3Tens: 0 + 4 + 1 = 5 → one counter above the tens line (worth 5).
4Hundreds: 9; Thousands: 2.
5Sum = 2950.
✅ Answer: 2950
๐ Page 69Figure it Out · Q1
Can a number's Egyptian numeral have one symbol 10 or more times? Why not?
110 of any symbol can be swapped for 1 of the next symbol.
2So in the shortest form, each symbol appears at most 9 times.
✅ Answer: No (up to 10⁷). Only the biggest symbol, the sun (10⁷), might need to repeat 10 or more times for numbers of 10 crore or more — because there is no bigger symbol.
๐ Textbook Page 70
๐ Page 70Figure it Out · Q2
Create your own base-4 system and write 1 to 16.
1Landmarks: 1, 4, 16. Symbols: ● = 1, ▲ = 4, ■ = 16.
21 ●, 2 ●●, 3 ●●●, 4 ▲, 5 ▲●, 6 ▲●●, 7 ▲●●●, 8 ▲▲
39 ▲▲●, 10 ▲▲●●, 11 ▲▲●●●, 12 ▲▲▲, 13 ▲▲▲●, 14 ▲▲▲●●, 15 ▲▲▲●●●, 16 ■
✅ Answer: See the list — never more than 3 of one symbol.
๐ Page 70Figure it Out · Q3
Give a simple rule to multiply a number by 5 in our base-5 system.
1Every landmark × 5 = the next landmark (△→□, □→⬡, ⬡→○, ○→wave).
2So replace every shape by the next bigger shape.
✅ Answer: Change each shape to the next shape.
๐ Textbook Page 72
๐ Page 72Worked
Write 640 and 7530 in the compact Mesopotamian way.
1640 = 10 × 60 + 40 → [10] [40]
27530 = 2 × 3600 + 5 × 60 + 30 → [2] [5] [30]
✅ Answer: See the drawing.
๐ Textbook Page 73
๐ Page 73Figure it Out · Q1
Write in the Mesopotamian system: (i) 63 (ii) 132 (iii) 200 (iv) 60 (v) 3605
1(i) 63 = 1 × 60 + 3 → [1] [3]
2(ii) 132 = 2 × 60 + 12 → [2] [12]
3(iii) 200 = 3 × 60 + 20 → [3] [20]
4(iv) 60 = 1 × 60 + 0 → [1] [blank]
5(v) 3605 = 1 × 3600 + 0 × 60 + 5 → [1] [blank] [5]
✅ Answer: See the drawings.
⚠️ Common mistake: For 60 or 3605, forgetting the blank place. Without it, 60 looks the same as 1!
๐ Page 73Think
Look at the representation of 60. What will 3600 look like?
160 = one nail in the 60s place, blank in the 1s place.
23600 = one nail in the 3600s place, blanks in the 60s and 1s places.
3Since blanks are hard to see, 1, 60 and 3600 all look like a single nail!
✅ Answer: Just one nail — exactly like 1 and 60. This confusion is why a zero symbol is needed.
๐ Textbook Page 76
๐ Page 76Practice
Write in the Mayan system (landmarks 1, 20, 360, 7200 …; dot = 1, bar = 5, shell = 0): (i) 77 (ii) 100 (iii) 361 (iv) 721
1(i) 77 = 3 × 20 + 17 → top: 3 dots; bottom: 17 = 3 bars + 2 dots
2(ii) 100 = 5 × 20 + 0 → top: 1 bar; bottom: shell
3(iii) 361 = 1 × 360 + 0 × 20 + 1 → 1 dot / shell / 1 dot
4(iv) 721 = 2 × 360 + 0 × 20 + 1 → 2 dots / shell / 1 dot
✅ Answer: See the drawings (read from top to bottom: biggest landmark first).
⚠️ Common mistake: Using 400 as the third landmark. In the Mayan system it is 18 × 20 = 360.
๐ Textbook Page 77
๐ Page 77Worked
Read the rod numeral heng-2, zong-6, heng-3, zong-4.
1Positions from the right: 1 (zong), 10 (heng), 100 (zong), 1000 (heng).
2\(2 \times 1000 + 6 \times 100 + 3 \times 10 + 4 = 2634\)
✅ Answer: 2634
๐ Textbook Page 80
๐ Page 80Figure it Out · Q1
Why did the Chinese switch between zong and heng? If only zong symbols were used, how would 41 look? Could it be read another way?
1Switching makes it easy to see where one place ends and the next begins.
2Only zong: 41 = |||| | (4 upright rods, then 1 upright rod).
3With no clear space it looks like ||||| — that could be read as 5, or 32 (||| ||), 23 (|| |||), 14, 311, …
✅ Answer: Alternating avoids confusion between neighbouring places.
๐ Page 80Figure it Out · Q2
Make a base-2 place value system with ‘ukasar’ and ‘urapon’ as the two digits. Compare with the Gumulgal system.
1Let urapon = 0 and ukasar = 1. Landmarks: 1, 2, 4, 8, 16 …
21 = uk; 2 = uk ur; 3 = uk uk; 4 = uk ur ur; 5 = uk ur uk; 6 = uk uk ur; 7 = uk uk uk; 8 = uk ur ur ur
3In Gumulgal: 8 = uk-uk-uk-uk (adding 2s). In base 2: 8 = uk ur ur ur (place value).
4The base-2 system can write every number; Gumulgal stopped at 6 and grows very long.
| Number | Base-2 (uk = 1, ur = 0) | Digits | Gumulgal |
|---|---|---|---|
| 1 | uk | 1 | urapon |
| 2 | uk ur | 10 | ukasar |
| 3 | uk uk | 11 | uk-ur |
| 4 | uk ur ur | 100 | uk-uk |
| 5 | uk ur uk | 101 | uk-uk-ur |
| 6 | uk uk ur | 110 | uk-uk-uk |
| 7 | uk uk uk | 111 | ras |
| 8 | uk ur ur ur | 1000 | ras |
✅ Answer: Base-2 place value is shorter and never-ending; Gumulgal is just repeated adding of 2s.
๐ Page 80Figure it Out · Q3
Where do Hindu numerals and 0 play an important role in daily life and jobs? How would life be different without them?
1Shopping bills, bank accounts, phone numbers, PINs, dates and time.
2Jobs: shopkeepers, accountants, engineers, doctors (doses), scientists, programmers (computers use 0 and 1!).
3Without them: slow calculation, no easy written sums, no computers, no modern science.
✅ Answer: They are everywhere — modern life depends on them.
๐ Page 80Figure it Out · Q4
If we used base 8 or base 5 instead of base 10, how would we write numbers? Write 25 in base 8, base 5 and base 2.
1Base 8 uses digits 0–7 and places 1, 8, 64, …
225 = 3 × 8 + 1 → 31 (base 8)
3Base 5: places 1, 5, 25 → 25 = 1 × 25 + 0 × 5 + 0 → 100 (base 5)
4Base 2: places 1, 2, 4, 8, 16 → 25 = 16 + 8 + 1 = 1·16 + 1·8 + 0·4 + 0·2 + 1·1 → 11001 (base 2)
✅ Answer: \(25 = 31_8 = 100_5 = 11001_2\)
๐ก Tip: Divide by the base again and again; the remainders (read upwards) are the digits.
๐ Where do we see these ideas today?
- Base 60 lives on in clocks: 60 seconds = 1 minute, 60 minutes = 1 hour.
- Computers work in base 2 (0 and 1). Roman numerals still appear on clocks, book chapters and film titles.
๐ Link to higher classes
- Expanded form and powers of 10 (Class 8 Ch 2), binary numbers in computer science, number systems in Class 9.
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