Ch 6: Algebra Play Quick revision notes Class 8th Mathematics (Ganita Prakash-II)

Quick revision notes

Ch 6: Algebra Play Quick revision notes Class 8th Mathematics (Ganita Prakash-II)

Class 8 · Mathematics (Ganita Prakash) · Chapter 6 : Algebra Play · All Board · ENGLISH · 8 views

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Section 6.1

Algebra Play — What & Why

🎯 What is Algebra Play?

Using algebra to explain tricks, puzzles and patterns — and to invent new ones. Algebra shows us why a trick always works, regardless of the starting number.

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When we replace a specific number with a letter (like x) and simplify the expression, we can prove that the result is always the same — for any starting number.

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Section 6.2

'Think of a Number' Tricks

✨ Classic Trick — Always Gets 2

Think of any number

\(x\)

Double it

\(2x\)

Add four

\(2x + 4\)

Divide by 2

\(x + 2\)

Subtract the original number

Always = 2

\[ x + 2 - x = \boxed{2} \quad \text{(for any starting number } x\text{)} \]

📅 The Date Trick — Mukta & Shubham

Shubham asks Mukta to think of a date (day & month), follow steps, and tell him the final answer. He magically guesses the date!

Date trick comic — Shubham and Mukta, Republic Day 26/01, answer 291

Mukta thinks of 26/01 (Republic Day). After 7 steps → 291. Shubham decodes it instantly!

🔬 How the Trick Works — Algebra Proof

Let month = \(M\), day = \(D\). Follow the 7 steps:

\[ 5M \xrightarrow{+6} 5M+6 \xrightarrow{\times 4} 20M+24 \xrightarrow{+9} 20M+33 \xrightarrow{\times 5} 100M+165 \xrightarrow{+D} 100M + 165 + D \]

To decode: Subtract 165 from the final answer → get \(100M + D\).

Since \(D \leq 31\) (only 2 digits), the last 2 digits = day, the rest = month.

\[ 291 - 165 = 126 \quad \Rightarrow \quad M = 1,\; D = 26 \quad \Rightarrow \quad \text{26 January} \checkmark \]

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Key idea: The algebra always produces 100M + 165 + D. Subtract 165 to get 100M + D. Since days are ≤ 31 (2 digits), read off M and D separately.

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Section 6.3

Number Pyramids

📌 The Rule

Each cell = sum of the two cells directly below it. To fill backwards (top given, bottom missing) → use algebra (letter-numbers).

Standard 3-row pyramid: each cell = sum below

Number pyramid 23 at top, 10,13 middle, 1,9,4 bottom

Basic pyramid: 1+9=10, 9+4=13, 10+13=23

Step-by-step pyramid filling — working backwards from top

Working backwards: if top=10 and one bottom=1, use subtraction to find the missing cells

🔬 Using Algebra to Fill Pyramids

When middle values are unknown, label them as letters and set up equations.

Algebra pyramid: 60 at top, a and b in middle, 12 c 8 at bottom. Solution: c=20, a=32, b=28

Letters a, b, c → set up equations → solve for c=20, a=32, b=28

🔍 Worked Example

Top = 60, bottom row = 12, ?, 8. Let middle = \(a, b\) and unknown = \(c\).

\[ a + b = 60 \] \[ (12+c) + (c+8) = 60 \] \[ 20 + 2c = 60 \] \[ c = 20, \; a = 32, \; b = 28 \]

📐 General Formula for 3-row Pyramid

Bottom row: \(a, b, c\). The top is always:

\[ \text{Top} = a + 2b + c \]

The middle values are \(a+b\) and \(b+c\).

General pyramid formula: top = a+2b+c, middle = a+b and b+c, bottom = a, b, c

General 3-row pyramid — the middle element \(b\) is counted twice in the top!

Key Pattern

For any 3-row pyramid with bottom \(a, b, c\):

\[ \text{Top} = a + 2b + c \]

The middle element is counted twice because it contributes to both cells in the second row.

🗓️

Section 6.4

Fun with Grids — Calendar Magic & Algebra Grids

🗓️ Calendar Magic — 2×2 Grid Trick

Pick any 2×2 block from a calendar. If top-left number = \(a\), the other three cells are always:

August 2025 calendar with 2x2 grid 6,7,13,14 highlighted. Sum = 40

Any 2×2 block from the calendar — sum = 4a + 16 where a = top-left number

In any week-based calendar, consecutive numbers across rows differ by 7

\[ \text{Sum} = a + (a+1) + (a+7) + (a+8) = 4a + 16 \]

📐 Example — Find the 4 Numbers from the Sum

Sum = 36. Find the 2×2 grid.

\[ 4a + 16 = 36 \implies 4a = 20 \implies a = 5 \]

Grid: 5, 6, 12, 13 ✔

🔷 Algebra Grids — Shapes Represent Numbers

Each shape stands for an unknown number. Each row's last column = sum of the other values. Solve row by row, top to bottom.

Algebra grids with blue squares and red circles as unknowns. Row 1: three squares = 27, so square = 9. Row 2: two circles + one square = 19, so circle = 5

Row 1: ■+■+■=27 → ■=9. Row 2: ●+●+■=19 → 2●+9=19 → ●=5

Strategy for Algebra Grids

1
Find a row with only one type of shape → solve for that shape's value
2
Substitute known value into the next row → solve for the next unknown
3
Continue until all shapes are found → fill in the empty cells

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Section 6.5

The Largest Product

❓ The Problem

Given three digits, arrange them as: \(\boxed{\;} \; \boxed{\;} \times \boxed{\;}\) to get the largest possible product.

Example: digits 2, 3, 5 → which of 23×5, 25×3, 32×5, 35×2, 52×3, 53×2 is largest?

🏆 The Rule

For digits \(p < q < r\) (smallest to largest):

\[ \text{Largest product} = \underbrace{qp}_{\text{tens}=q,\;\text{units}=p} \times r \]

Put the largest digit as the multiplier. The other two go in the 2-digit number with the second-largest as the tens digit.

📐 Algebra Proof — Why qp × r wins

Compare \(qp \times r\) and \(rp \times q\):

\[ qp \times r = (10q \cdot r) + (p \cdot r) \qquad rp \times q = (10r \cdot q) + (p \cdot q) \]

First terms equal. Second: \(p \cdot r > p \cdot q\) since \(r > q\). So \(qp \times r\) wins! ✔

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Quick Rule: Digits 2, 3, 5 → Largest = 32 × 5 = 160 (not 53×2=106 or 52×3=156).

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Section 6.6

Decoding Divisibility Tricks

Mukta shows Shubham the reverse-digit trick: 47 reversed=74, difference=27, divide by 9 = 3 with no remainder

Mukta's trick: any 2-digit number → reverse → find difference → always divisible by 9!

🔬 Why the Difference is Always Divisible by 9

Let the 2-digit number be \(\overline{ab}\) (tens digit \(a\), units digit \(b\)).

\[ \overline{ab} = 10a + b \qquad \overline{ba} = 10b + a \] \[ \overline{ba} - \overline{ab} = (10b+a) - (10a+b) = 9b - 9a = 9(b-a) \]

Since the difference = \(9(b-a)\), it is always a multiple of 9. ✔

📐 Example: 47

Reversed = 74

\[ 74 - 47 = 27 = 9 \times 3 \]

Quotient = \(b - a = 7 - 4 = 3\) ✔

📐 Example: 82

Reversed = 28

\[ 82 - 28 = 54 = 9 \times 6 \]

Quotient = \(b - a\) (or \(a-b\) if \(a>b\)) ✔

🐎 Bonus: Horses and Hens Problem

A farm has 55 animals (horses + hens), 150 legs total.

\[ h + n = 55 \quad (\text{heads}) \] \[ 4h + 2n = 150 \quad (\text{legs}) \] \[ 2h = 150 - 2(55) = 40 \implies h = 20,\; n = 35 \]

20 horses and 35 hens

Boy counting horses and hens at a farm fence

55 heads, 150 legs — algebra solves it!

🧞 Karim and the Genie — Algebra Story

Each round: coins doubled, then pay 8 coins. After 3 rounds, Karim has only 8 coins left.

\[ \text{Start with } x \text{ coins} \] \[ \text{Round 1: } 2x - 8 \] \[ \text{Round 2: } 2(2x-8)-8 = 4x-24 \] \[ \text{Round 3: } 2(4x-24)-8 = 8x-56 = 8 \] \[ 8x = 64 \implies x = 7 \text{ coins} \]

Genie sitting on banyan tree with Karim walking below

Karim's 7 coins → after 3 rounds → exactly 8 left (the cost!)

⭐ Quick Summary

'Think of a Number' tricks always give the same result because the starting number \(x\) cancels out in the algebra.

Date trick: Steps produce \(100M + 165 + D\). Subtract 165 → get \(100M + D\). Read off month (M) and day (D) separately since days ≤ 31.

Number pyramids: each cell = sum of two below. For bottom row \(a, b, c\): top = \(a + 2b + c\) (middle element counted twice).

Calendar magic: any 2×2 block with top-left = \(a\) has sum \(4a + 16\). Given the sum, find \(a\) by solving \(4a + 16 = \text{sum}\).

Largest product from digits \(p < q < r\): arrange as \(qp \times r\) (largest digit as multiplier, others in decreasing order).

Divisibility trick: for any 2-digit number \(\overline{ab}\), the difference \(|\overline{ba} - \overline{ab}| = 9|b-a|\) — always divisible by 9.

Algebra is a powerful tool for proving why patterns and tricks work — not just for specific numbers, but for all numbers.