Class 8 Maths Chapter 6 Algebra Play Visual Infographic | NCF 2023 Ganita Prakash
Chapter 6: Algebra Play
Decoding Number Tricks, Birthday Predictions, Pascal Pyramid Sums, Calendar Invariants & Divisibility Proofs
๐ช 1. 'Think of a Number'
Variable EliminationTricks where algebraic operations cancel the unknown starting number to produce a constant result:
- Think of a number: x
- Double it: 2x
- Add 4: 2x + 4
- Divide by 2: x + 2
- Subtract original x: (x + 2) − x = 2 (Always!)
To make the final answer k, simply add 2k in step 3!
๐ 2. Birthday Date Decoder
Positional Value MagicGiven Month (M) and Day (D), follow these steps:
1. Month × 5 → 5M
2. Add 6 → 5M + 6
3. Multiply by 4 → 20M + 24
4. Add 9 → 20M + 33
5. Multiply by 5 → 100M + 165
6. Add Day → 100M + 165 + D
The Secret: Subtract 165 from final answer → 100M + D gives Month & Day directly!
๐ 3. Digit Reversal Invariants
Divisibility ProofsLet a 2-digit number be ab = 10a + b and reversed ba = 10b + a:
- Difference: (10b + a) − (10a + b) = 9(b − a) → Always divisible by 9!
- Sum: (10a + b) + (10b + a) = 11(a + b) → Always divisible by 11!
Cyclic 3-Digit Sum: abc + bca + cab = 111(a + b + c) = 37 × 3(a + b + c) → Divisible by 37 and 3!
๐งฑ 4. Additive Number Pyramids
Pascal Binomial WeightsEach brick is the sum of the two directly underneath:
• 3-Row Top: a + 2b + c
• 4-Row Top: a + 3b + 3c + d
• Virahฤแน
ka Sequence: If base has Fibonacci terms, every brick above is also in the sequence!
๐ฏ 5. The Largest Product Rule
Optimization ProofTo arrange 3 digits (p < q < r) in [ _ _ × _ ] to form the maximum product:
Structure: qp × r
(Largest digit = Multiplier, remaining in descending order)
- Digits 2, 3, 5: 32 × 5 = 160 (beats 52 × 3 = 156).
- Digits 1, 3, 7: 31 × 7 = 217.
- Digits 3, 5, 9: 53 × 9 = 477.
✨ 6. The 7-11-13 Repeating Trick
1001 FactorizationWrite any 3-digit number twice to make a 6-digit number abcabc:
abcabc = 1000 × abc + abc
= abc × 1001
Since 1001 = 7 × 11 × 13:
Dividing abcabc successively by 7, 11, and 13 returns the exact original 3-digit number abc!
๐ Master Algebraic Puzzles & Problem Models Matrix
Algebraic setups and exact solutions for word problems, logic grids, and financial scenarios:
| Puzzle Archetype | Algebraic Model / Equation | Working & Method | Final Solution |
|---|---|---|---|
| Calendar 2 × 2 Grid | Sum = a + (a+1) + (a+7) + (a+8) = 4a + 16 | If Sum = 36 → 4a = 20 → a = 5 | Dates: 5, 6, 12, 13 |
| Horses & Hens (Heads/Legs) | Heads: h + c = 55 Legs: 4h + 2c = 150 |
If all 55 were hens = 110 legs. Extra 40 legs / 2 = 20 horses | 20 Horses, 35 Hens |
| Mother & Daughter Ages | Now: M = 5D In 6 years: (5D + 6) = 3(D + 6) |
5D + 6 = 3D + 18 → 2D = 12 → D = 6 | Daughter is 6 years old |
| Gauri & Naina's Cows | N = 2G N − 3 = G + 3 |
2G − 3 = G + 3 → G = 6 → N = 12 | Gauri: 6, Naina: 12 cows |
| Dosa Cart Profit | Cost = 5000 + 10x Revenue = P × x |
Target Profit = ₹2000 → Total Rev = ₹7000 Case 1: 100 dosas → Price = 7000/100 Case 2: At ₹50 → 50x − 10x − 5000 = 2000 → 40x = 7000 |
(i) Price = ₹70 (ii) 175 Dosas |
| Odd Fraction Sequences | Sum of first n odds = n2 1/3, (1+3)/(5+7), (1+3+5)/(7+9+11)... |
Numerator = n2; Denominator = (2n)2 − n2 = 3n2 Fraction = n2 / (3n2) |
Always 1/3 (Constant!) |
๐ง 7. Karim and the Genie
Walking around the banyan tree doubles coins; genie takes 8 coins per round:
- Start: C coins
- Round 1: 2C − 8
- Round 2: 2(2C − 8) − 8 = 4C − 24
- Round 3: 2(4C − 24) − 8 = 8C − 56 = 0 → 8C = 56 → C = 7 coins!
• Karim gains money if tax < C coins.
• For n rounds, zero remaining requires start: C = Tax × (1 − 1/2n).
๐บ 8. The Three Shrines & Magical Ponds
Flowers double in each pond; equal flowers F placed in each shrine:
Let starting flowers = x, placed per shrine = F.
• After Pond 1 & Shrine 1: 2x − F
• After Pond 2 & Shrine 2: 2(2x − F) − F = 4x − 3F
• After Pond 3 & Shrine 3: 2(4x − 3F) − F = 8x − 7F = 0
→ 8x = 7F → x/F = 7/8
Smallest Solution: Start with 7 flowers and place 8 flowers at each shrine!
⚠️ Common Pitfalls & Algebraic Traps
Multiplication Placement Trap
To maximize a 2-digit × 1-digit product, students mistakenly put the largest digit in the tens place (52 × 3 = 156). Putting the largest digit as the single multiplier yields the true maximum (32 × 5 = 160)!
Digit vs Value Confusion
A two-digit number with digits a and b is represented as 10a + b, not a × b. Reversing digits gives 10b + a. Forgetting place values corrupts algebra proofs!
Pyramid Row Weighting
In a 3-row number pyramid, the top is NOT a + b + c. Middle elements contribute to multiple paths, giving top value a + 2b + c following binomial coefficients!
