๐ KEY TO ENJOY LEARNING MATHS
Class 8 Ganita Prakash · Chapter 6
Class 8 Ganita Prakash · Chapter 6
We Distribute, Yet Things Multiply
Page-wise textbook solutions (pages 136–158) · Every step · Figures · Common mistakes · Tips
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✔ Understood 0/48
๐️ Key ideas of this chapter
- Distributive property: \(a(b + c) = ab + ac\) — multiply every term inside by the term outside.
- Identity 1: \((a + m)(b + n) = ab + an + mb + mn\) — every term times every term.
- 1A: \((a+b)^2 = a^2 + 2ab + b^2\) · 1B: \((a-b)^2 = a^2 - 2ab + b^2\) · 1C: \((a+b)(a-b) = a^2 - b^2\)
- Only like terms (same letters, same powers) can be added.
- Fast multiplication: × 11, × 101, × 99, squares like 65², 99², 197² using identities.
๐ Textbook Page 136
๐ Page 136Think
For 23 × 27: by how much does the product grow if (1) 23 becomes 24 (2) 27 becomes 28 (3) both grow by 1?
1(1) 24 × 27 = 23 × 27 + 27 → grows by 27.
2(2) 23 × 28 = 23 × 27 + 23 → grows by 23.
3(3) 24 × 28 = 23 × 27 + 27 + 23 + 1 → grows by 51.
4In general: \(a(b+1) = ab + a\) and \((a+1)(b+1) = ab + a + b + 1\).
✅ Answer: By 27, by 23, and by 23 + 27 + 1 = 51.
๐ Textbook Page 138
๐ Page 138Try
Expand (a + 1)(b + 1) by first taking (b + 1) as one term.
1\((a+1)(b+1) = a(b+1) + 1(b+1)\)
2\(= ab + a + b + 1\)
✅ Answer: The same: ab + a + b + 1.
๐ Page 138Think
If one number grows by 1 and the other falls by 1, does the product always grow? Give 3 examples where it falls.
1\((a+1)(b-1) = ab + b - a - 1\) → change = b − a − 1.
2It falls when b − a − 1 < 0, i.e. when b ≤ a.
315 = 5 × 3 → 6 × 2 = 12 (falls by 3)
4100 = 10 × 10 → 11 × 9 = 99 (falls by 1)
528 = 7 × 4 → 8 × 3 = 24 (falls by 4)
✅ Answer: No. It falls whenever the number made smaller is not bigger than the other one.
๐ Page 138Check
Do these results hold for negative integers, e.g. a = −5, b = 8 or a = −4, b = −5?
1a = −5, b = 8: ab = −40; (a+1)(b+1) = −4 × 9 = −36; increase = a + b + 1 = 4 ✔
2a = −4, b = −5: ab = 20; (a+1)(b+1) = −3 × −4 = 12; a + b + 1 = −8 ✔ (12 − 20 = −8)
✅ Answer: Yes — integers follow the distributive law too.
๐ Textbook Page 139
๐ Page 139Identity 1
How does the product change if one number grows by m and the other by n?
1\((a+m)(b+n) = ab + an + mb + mn\)
✅ Answer: It grows by an + bm + mn.
๐ Textbook Page 140
๐ Page 140Practice
Use Identity 1 to find the change when (i) one number falls by 2 and the other grows by 3 (ii) one falls by 3 and the other by 4.
1(i) \((a-2)(b+3) = ab + 3a - 2b - 6\)
2(ii) \((a-3)(b-4) = ab - 4a - 3b + 12\)
3Check with a = 10, b = 7: (i) 8 × 10 = 80 = 70 + 30 − 14 − 6 ✔ (ii) 7 × 3 = 21 = 70 − 40 − 21 + 12 ✔
✅ Answer: (i) changes by 3a − 2b − 6 (ii) by −4a − 3b + 12.
๐ Page 140Practice
Expand (i) (a − u)(b + v) (ii) (a − u)(b − v).
1Multiply every term by every term and use sign rules.
✅ Answer: (i) ab + av − ub − uv (ii) ab − av − ub + uv
๐ Textbook Page 141
๐ Page 141Example
Expand \(\tfrac{3a}{2}(a - b + \tfrac{1}{5})\). Can any two terms be combined?
1\(\tfrac{3a}{2} \times a = \tfrac{3}{2}a^2\), \(\tfrac{3a}{2} \times b = \tfrac{3}{2}ab\), \(\tfrac{3a}{2} \times \tfrac{1}{5} = \tfrac{3}{10}a\)
2No two terms have the same letters and powers.
✅ Answer: \(\tfrac{3}{2}a^2 - \tfrac{3}{2}ab + \tfrac{3}{10}a\) — no like terms to combine.
๐ Page 141Example
Expand (a + b)(a + b) and (a + b)(a² + 2ab + b²).
1\((a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2\)
2\((a+b)(a^2+2ab+b^2) = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3\)
3Combine like terms: \(a^3 + 3a^2b + 3ab^2 + b^3\)
✅ Answer: \((a+b)^2 = a^2 + 2ab + b^2\); \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\)
๐ Textbook Page 142
๐ Page 142Figure it Out · Q1
In the multiplication grid, the middle of a 3 × 3 frame is pq. Write the other 8 numbers.
1Rows go p − 1, p, p + 1; columns go q − 1, q, q + 1.
| (p−1)(q−1) | (p−1)q | (p−1)(q+1) |
| p(q−1) | pq | p(q+1) |
| (p+1)(q−1) | (p+1)q | (p+1)(q+1) |
✅ Answer: Top row: (p−1)(q−1), (p−1)q, (p−1)(q+1); middle: p(q−1), pq, p(q+1); bottom: (p+1)(q−1), (p+1)q, (p+1)(q+1).
๐ Textbook Page 143
๐ Page 143Figure it Out · Q2
Expand: (i) (3 + u)(v − 3) (ii) \(\tfrac{2}{3}(15 + 6a)\) (iii) (10a + b)(10c + d) (iv) (3 − x)(x − 6) (v) (−5a + b)(c + d) (vi) (5 + z)(y + 9)
1(i) 3v − 9 + uv − 3u
2(ii) 10 + 4a
3(iii) 100ac + 10ad + 10bc + bd
4(iv) 3x − 18 − x² + 6x = −x² + 9x − 18
5(v) −5ac − 5ad + bc + bd
6(vi) 5y + 45 + yz + 9z
✅ Answer: See the steps.
⚠️ Common mistake: In (iv), (−x) × (−6) = +6x, not −6x.
๐ Page 143Figure it Out · Q3
Find 3 examples where a product stays the same when one number grows by 2 and the other falls by 4.
1\((a+2)(b-4) = ab - 4a + 2b - 8\)
2No change → −4a + 2b − 8 = 0 → b = 2a + 4
3a = 1, b = 6: 6 and 3 × 2 = 6 ✔; a = 2, b = 8: 16 = 4 × 4 ✔; a = 3, b = 10: 30 = 5 × 6 ✔
✅ Answer: 1 × 6, 2 × 8, 3 × 10 (any a with b = 2a + 4).
๐ Page 143Figure it Out · Q4
Expand (i) (a + ab − 3b²)(4 + b) (ii) (4y + 7)(y + 11z − 3).
1(i) 4a + 4ab − 12b² + ab + ab² − 3b³ = 4a + 5ab + ab² − 12b² − 3b³
2(ii) 4y² + 44yz − 12y + 7y + 77z − 21 = 4y² + 44yz − 5y + 77z − 21
✅ Answer: (i) 4a + 5ab + ab² − 12b² − 3b³ (ii) 4y² + 44yz − 5y + 77z − 21
๐ Page 143Figure it Out · Q5
Expand (i) (a − b)(a + b) (ii) (a − b)(a² + ab + b²) (iii) (a − b)(a³ + a²b + ab² + b³). What is the next identity?
1(i) a² − b²
2(ii) a³ + a²b + ab² − a²b − ab² − b³ = a³ − b³
3(iii) a⁴ − b⁴ (middle terms cancel in pairs)
4Next: (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵
✅ Answer: a² − b², a³ − b³, a⁴ − b⁴; next a⁵ − b⁵.
๐ Textbook Page 144
๐ Page 144Math Talk
Rule for × 11 in one line. Find (i) 94 × 11 (ii) 495 × 11 (iii) 3279 × 11 (iv) 4791256 × 11.
1Write the last digit; then add each pair of neighbouring digits from the right, carrying when needed; finally write the first digit (+ carry).
2(i) 9 | 9+4 | 4 → 1034
3(ii) 4 | 4+9 | 9+5 | 5 → 5445 (with carries)
4(iii) 36069
5(iv) 52703816
✅ Answer: (i) 1034 (ii) 5445 (iii) 36069 (iv) 52703816
๐ Page 144Math Talk
Rule for × 101, 1001 … Find 3874 × 101, 89 × 101, 949 × 101, 265831 × 1001, 1111 × 1001, 9734 × 99, 23478 × 999.
1× 101: add the number to itself shifted 2 places (dcba00 + dcba). × 1001: shift 3 places, and so on.
23874 × 101 = 387400 + 3874 = 391274
389 × 101 = 8989; 949 × 101 = 95849
4265831 × 1001 = 265831000 + 265831 = 266096831; 1111 × 1001 = 1112111
59734 × 99 = 973400 − 9734 = 963666
623478 × 999 = 23478000 − 23478 = 23454522
✅ Answer: 391274, 8989, 95849, 266096831, 1112111, 963666, 23454522
๐ก Tip: × 99 = × 100 − the number; × 999 = × 1000 − the number.
๐ Textbook Page 145
๐ Page 145Area model
Find 65² using a square of side 60 + 5.
165² = 60² + 2 × 60 × 5 + 5² = 3600 + 600 + 25
✅ Answer: 4225
๐ Page 145Math Talk
Is (a + b)² always greater than a² + b²?
1\((a+b)^2 - (a^2+b^2) = 2ab\)
2Greater when ab > 0 (same signs); equal if a or b is 0; smaller when a and b have opposite signs: (3 + (−2))² = 1 < 9 + 4.
✅ Answer: No — only when a and b are both positive or both negative.
๐ Page 145Practice
Use Identity 1A to find 104² and 37².
1104² = (100 + 4)² = 10000 + 800 + 16 = 10816
237² = (30 + 7)² = 900 + 420 + 49 = 1369
✅ Answer: 10816 and 1369
๐ Textbook Page 146
๐ Page 146Practice
Write (i) (m + 3)² (ii) (6 + p)² (iii) (6x + 5)² (iv) (3j + 2k)².
1(i) m² + 6m + 9
2(ii) 36 + 12p + p²
3(iii) 36x² + 60x + 25
4(iv) 9j² + 12jk + 4k²
✅ Answer: See the steps.
๐ Textbook Page 147
๐ Page 147Area model
Use 60² and 5² to find 55². Then write (a − b)².
1Take the 60 × 60 square, remove two strips 60 × 5, and add back the 5 × 5 square removed twice.
255² = 3600 − 300 − 300 + 25 = 3025
3\((a-b)^2 = a^2 - 2ab + b^2\)
✅ Answer: 3025; (a − b)² = a² − 2ab + b²
๐ Page 147Practice
Find 99² and 58² using (a − b)².
199² = (100 − 1)² = 10000 − 200 + 1 = 9801
258² = (60 − 2)² = 3600 − 240 + 4 = 3364
✅ Answer: 9801 and 3364
๐ Page 147Practice
Expand (i) (b − 6)² (ii) (−2a + 3)² (iii) \((7y - \tfrac{3}{4z})^2\)
1(i) b² − 12b + 36
2(ii) 4a² − 12a + 9
3(iii) \(49y^2 - 2 \times 7y \times \tfrac{3}{4z} + \tfrac{9}{16z^2} = 49y^2 - \tfrac{21y}{2z} + \tfrac{9}{16z^2}\)
✅ Answer: See the steps.
๐ Textbook Page 148
๐ Page 148Pattern 1
2(2² + 1²) = 3² + 1², 2(3² + 1²) = 4² + 2² … Explain the pattern.
1\((a+b)^2 + (a-b)^2 = 2a^2 + 2b^2\)
2Example: 2(5² + 6²) = 11² + 1² = 122
✅ Answer: \(2(a^2+b^2) = (a+b)^2 + (a-b)^2\)
๐ Page 148Pattern 2
9 × 9 − 1 × 1 = 10 × 8, 8 × 8 − 6 × 6 = 14 × 2 … Write it in algebra. Use it for 98 × 102 and 45 × 55.
1\(a^2 - b^2 = (a+b)(a-b)\)
298 × 102 = (100 − 2)(100 + 2) = 10000 − 4 = 9996
345 × 55 = (50 − 5)(50 + 5) = 2500 − 25 = 2475
✅ Answer: 9996 and 2475
๐ Textbook Page 149
๐ Page 149Think
Why is \(a^2 = (a+b)(a-b) + b^2\) true? Use it for 31² and 197².
1\((a+b)(a-b) + b^2 = a^2 - b^2 + b^2 = a^2\)
231² = 32 × 30 + 1 = 961
3197² = 200 × 194 + 9 = 38809
✅ Answer: 961 and 38809 (Sridharacharya's method)
๐ Page 149Figure it Out · Q1
Which is greater: (a − b)² or (b − a)²?
1b − a = −(a − b), and a negative number squared equals the positive number squared.
✅ Answer: They are equal.
๐ Page 149Figure it Out · Q2
Express 100 as a difference of two squares.
1a² − b² = (a + b)(a − b) = 100 with a + b and a − b both even (or both odd).
250 × 2 → a = 26, b = 24; 10 × 10 → a = 10, b = 0.
✅ Answer: 100 = 26² − 24² (also 10² − 0²)
๐ Page 149Figure it Out · Q3
Find 406², 72², 145², 1097² and 124².
1406² = (400 + 6)² = 160000 + 4800 + 36 = 164836
272² = (70 + 2)² = 4900 + 280 + 4 = 5184
3145² = (140 + 5)² = 19600 + 1400 + 25 = 21025
41097² = (1100 − 3)² = 1210000 − 6600 + 9 = 1203409
5124² = (120 + 4)² = 14400 + 960 + 16 = 15376
✅ Answer: 164836, 5184, 21025, 1203409, 15376
๐ Page 149Figure it Out · Q4
Do Patterns 1 and 2 work for negative integers and fractions?
1They are identities proved with the distributive law, which holds for all numbers.
2Example: a = −3, b = ½: 2(9 + ¼) = 18.5 and (−2.5)² + (−3.5)² = 6.25 + 12.25 = 18.5 ✔
✅ Answer: Yes, for all numbers.
๐ Textbook Page 150
๐ Page 150Mind the Mistake
Check each of the 12 simplifications, explain any mistake and correct it.
1Expand again carefully; only like terms can be added; every term inside a bracket is multiplied.
| # | Given | What went wrong | Correct |
|---|---|---|---|
| 1 | −3p(−5p + 2q) = −3p + 5p − 2q | Multiplied only the first term / dropped p | 15p² − 6pq |
| 2 | 2(x − 1) + 3(x + 4) = 2x − 1 + 3x + 4 | Did not multiply the second terms | 5x + 10 |
| 3 | y + 2(y + 2) = (y + 2)² | Treated y + 2(…) as (y + 2)(…) | 3y + 4 |
| 4 | (5m + 6n)² = 25m² + 36n² | Forgot the middle term 2ab | 25m² + 60mn + 36n² |
| 5 | (−q + 2)² = q² − 4q + 4 | ✔ Correct | q² − 4q + 4 |
| 6 | 3a(2b × 3c) = 6ab × 9ac | Distributed over multiplication | 18abc |
| 7 | ½(10s − 6) + 3 = 5s | ✔ Correct | 5s |
| 8 | 5w² + 6w = 11w² | Added unlike terms | 5w² + 6w (cannot simplify) |
| 9 | 2a³ + 3a³ + 6a²b + 6ab² = 5a³ + 12a²b² | Added unlike terms a²b and ab² | 5a³ + 6a²b + 6ab² |
| 10 | (x + 2)(x + 5) = x² + 7x + 10 | ✔ Correct | x² + 7x + 10 |
| 11 | (a + 2)(b + 4) = ab + 8 | Missed the cross terms | ab + 4a + 2b + 8 |
| 12 | ab² + a²b + a²b² = ab(a + b + ab) | ✔ Correct | ab(b + a + ab) |
✅ Answer: Cards 5, 7, 10 and 12 are correct; the others are fixed in the table.
๐ Page 150Pattern
Circle pattern: draw the next figure. How many circles in Step 10, Step 15 and Step k?
1Steps 1, 2, 3 have 3, 8, 15 circles.
2All four methods give \(k^2 + 2k\): e.g. k rows of (k + 2), or \((k+1)^2 - 1\).
3Step 4: 24; Step 10: 100 + 20 = 120; Step 15: 225 + 30 = 255
✅ Answer: Step k: k² + 2k; Step 10: 120; Step 15: 255
๐ Textbook Page 152
๐ Page 152Pattern
Square-tile frames: how many tiles in each figure, Step 4, Step 10 and Step n?
1Step n is an (n + 2) × (n + 2) square with an n × n hole.
2Tiles = (n + 2)² − n² = 4n + 4 (also 4 sides of n + 1 tiles: 4(n + 1)).
3Steps 1, 2, 3: 8, 12, 16; Step 4: 20; Step 10: 44
✅ Answer: 4n + 4; Step 4: 20; Step 10: 44
๐ Textbook Page 153
๐ Page 153Check
Shaded square inside four m × n rectangles: show Tadang's (m + n)² − 4mn equals Yusuf's (n − m)².
1\((m+n)^2 - 4mn = m^2 + 2mn + n^2 - 4mn = m^2 - 2mn + n^2 = (n-m)^2\)
✅ Answer: They are equal.
๐ Page 153Check
Fig. 1 (three x × y rectangles): show x² − xy, x(x + 2y) − 3xy and x(x − y) are equal; find the area for x = 8, y = 3.
1x(x + 2y) − 3xy = x² + 2xy − 3xy = x² − xy
2x(x − y) = x² − xy
3x = 8, y = 3: 64 − 24 = 40
✅ Answer: All equal x² − xy; area = 40 square units.
๐ Textbook Page 154
๐ Page 154Math Talk
Write the area of the dashed region (big rectangle s × p with an L-shaped strip of width r). Find it for p = 6, r = 3.5, s = 9.
1The dashed region is a rectangle of sides (s − r) and (p − r).
2Area = (s − r)(p − r) = ps − pr − rs + r²
3(9 − 3.5)(6 − 3.5) = 5.5 × 2.5 = 13.75
✅ Answer: (p − r)(s − r) = 13.75 square units
๐ Page 154Figure it Out · Q1
Compute: (i) 46² by (a + b)² (ii) 397 × 403 by (a + b)(a − b) (iii) 91² by (a − b)² (iv) 43 × 45 by (a + b)(a − b)
1(i) (40 + 6)² = 1600 + 480 + 36 = 2116
2(ii) (400 − 3)(400 + 3) = 160000 − 9 = 159991
3(iii) (100 − 9)² = 10000 − 1800 + 81 = 8281
4(iv) (44 − 1)(44 + 1) = 1936 − 1 = 1935
✅ Answer: 2116, 159991, 8281, 1935
๐ Page 154Figure it Out · Q2
Find: (i) (p − 1)(p + 11) (ii) (3a − 9b)(3a + 9b) (iii) −(2y + 5)(3y + 4) (iv) (6x + 5y)² (v) (2x − ½)² (vi) (7p)(3r)(p + 2)
1(i) p² + 11p − p − 11 = p² + 10p − 11
2(ii) 9a² − 81b²
3(iii) −(6y² + 8y + 15y + 20) = −6y² − 23y − 20
4(iv) 36x² + 60xy + 25y²
5(v) 4x² − 2x + ¼
6(vi) 21pr(p + 2) = 21p²r + 42pr
✅ Answer: See the steps.
⚠️ Common mistake: In (iii), the minus sign changes EVERY term.
๐ Textbook Page 155
๐ Page 155Figure it Out · Q3
Choose the right expressions: (i) two more than a square number (ii) the sum of the squares of two consecutive numbers.
1(i) square number s², two more → s² + 2
2(ii) m² + (m + 1)² (or m² + (m − 1)²)
✅ Answer: (i) s² + 2 (ii) m² + (m + 1)² and m² + (m − 1)²
๐ Page 155Figure it Out · Q4
In any 2 × 2 square of a calendar, compare the diagonal products. Why?
1Label a, a + 1 (top) and a + 7, a + 8 (bottom).
2(a + 1)(a + 7) − a(a + 8) = a² + 8a + 7 − a² − 8a = 7
3Example: 4 × 12 = 48, 5 × 11 = 55, difference 7.
✅ Answer: The diagonal products always differ by 7.
๐ Page 155Figure it Out · Q5
True or false? (i) (k + 1)(k + 2) − (k + 3) is always 2. (ii) (2q + 1)(2q − 3) is a multiple of 4. (iii) Even squares are multiples of 4; odd squares are 1 more than a multiple of 8. (iv) (6n + 2)² − (4n + 3)² is 5 less than a square.
1(i) = k² + 2k − 1 → k = 1 gives 2, k = 2 gives 7 → false
2(ii) = 4q² − 4q − 3, which is odd → false
3(iii) (2n)² = 4n² ✔; (2n + 1)² = 4n(n + 1) + 1 and n(n + 1) is even → 8m + 1 ✔ → true
4(iv) = 20n² − 5 → 5 less than 20n², which is not a square for n = 1, 2 … → false
✅ Answer: (i) F (ii) F (iii) T (iv) F
๐ Page 155Figure it Out · Q6
One number leaves remainder 3 and another leaves 5 on dividing by 7. Find the remainders of their sum, difference and product.
1Numbers 7a + 3 and 7b + 5.
2Sum = 7(a + b) + 8 → remainder 1
3Difference (7b + 5) − (7a + 3) = 7(b − a) + 2 → remainder 2 (the other way round it is 5)
4Product = 49ab + 35a + 21b + 15 = 7(…) + 15 → remainder 1
✅ Answer: Sum 1, difference 2 (or 5), product 1
๐ Page 155Figure it Out · Q7
Square the middle of three consecutive numbers and subtract the product of the other two. What happens?
1\(n^2 - (n-1)(n+1) = n^2 - (n^2 - 1) = 1\)
2Example: 7² − 6 × 8 = 49 − 48 = 1
✅ Answer: Always 1.
๐ Textbook Page 156
๐ Page 156Figure it Out · Q8
Add two numbers, multiply by half their sum. Show the result is half the square of the sum.
1\((a+b) \times \tfrac{1}{2}(a+b) = \tfrac{1}{2}(a+b)^2\)
✅ Answer: \(\tfrac{1}{2}(a+b)^2\)
๐ Page 156Figure it Out · Q9
Which is larger? (i) 14 × 26 or 16 × 24 (ii) 25 × 75 or 26 × 74
1(i) 14 × 26 = (20 − 6)(20 + 6) = 400 − 36; 16 × 24 = 400 − 16 → 16 × 24 is larger
2(ii) 25 × 75 = 50² − 25²; 26 × 74 = 50² − 24² → 26 × 74 is larger
✅ Answer: (i) 16 × 24 (ii) 26 × 74
๐ก Tip: Numbers closer together give a bigger product (same sum).
๐ Page 156Figure it Out · Q10
A park has two g × g green squares; the rest is a path w ft wide. Find the area to be tiled.
1From the plan, the gap between the two squares looks like 2w (a path on each side).
2Length = w + g + 2w + g + w = 2g + 4w; breadth = w + g + w = g + 2w
3Tiled area = (2g + 4w)(g + 2w) − 2g² = 2g² + 8gw + 8w² − 2g² = 8gw + 8w²
4(If the middle gap is only w: (2g + 3w)(g + 2w) − 2g² = 7gw + 6w².)
✅ Answer: 8w(g + w) square feet
๐ Page 156Figure it Out · Q11
For each pattern draw the next figure, find the units in Step 10 and in Step y.
1(a) Step 1: 9, Step 2: 16, Step 3: 25 → Step y: (y + 2)² (a block y × (y + 2) plus two arms of y + 2)
2(a) Step 10: 12² = 144
3(b) Step 1: 2² + 1 = 5, Step 2: 3² + 2 = 11, Step 3: 4² + 3 = 19 → Step y: (y + 1)² + y
4(b) Step 10: 121 + 10 = 131
✅ Answer: (a) (y + 2)², Step 10: 144 (b) (y + 1)² + y, Step 10: 131
๐ Textbook Page 158
๐ Page 158Puzzle: Coin Conjoin
Turn a triangle of 10 coins upside down by moving the fewest coins. What about 15 coins? Is there a rule?
13 coins: 1 move; 6 coins: 2 moves; 10 coins: 3 moves (move the three corner coins).
215 coins: 5 moves; 21 coins: 7 moves.
3Rule: minimum moves = (number of coins) ÷ 3, ignoring the remainder.
✅ Answer: 10 coins: 3 moves; 15 coins: 5 moves; in general ⌊n ÷ 3⌋.
๐ Where do we use this?
- Mental maths: 98 × 102, 45 × 55, 99² in seconds.
- Finding areas of paths, borders and frames.
๐ Link to higher classes
- Class 9: (a + b + c)², (a + b)³, factorisation. Class 10: quadratic equations. Class 11: binomial theorem.
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