🔢 Algebra Play
NCERT Ganita Prakash Part-II · Easy Notes for Exam Prep
Number Pyramids
Calendar Magic
Divisibility Tricks
Largest Product
6.1 Introduction to Algebra Play
In earlier classes, we used algebra to model real-life situations, solve equations, and find unknown values. In Class 8, we go a step further — we use algebra to understand, explain, and even create mathematical tricks and puzzles!
Using letter-numbers (variables like x, a, b, M, D) to prove why tricks always work, not just that they work.
- Algebra helps us model any number — not just specific ones.
- When a result is always the same regardless of the starting number, algebra can prove it.
- Topics covered: number tricks, pyramids, calendar grids, digit products, divisibility.
6.2 Think of a Number Tricks
🔹 Basic Trick — Always Gets 2
Follow these steps with any number you like:
- Think of a number → x
- Double it → 2x
- Add four → 2x + 4
- Divide by 2 → x + 2
- Subtract the original number → x + 2 − x = 2
To always get 3 instead of 2, change “Add four” to “Add six” (since 6÷2 = 3). To always get 5, add 10. The final answer = (the number you add) ÷ 2.
📅 Date-Guessing Trick
A more advanced trick — you can figure out someone’s chosen date (DD/MM) from just one number!
Let the Month = M and the Day = D. Here are the steps:
- Multiply month by 5 → 5M
- Add 6 → 5M + 6
- Multiply by 4 → 20M + 24
- Add 9 → 20M + 33
- Multiply by 5 → 100M + 165
- Add the day → 100M + 165 + D
To decode: Subtract 165 → you get 100M + D
Last 2 digits = Day (D), remaining digits = Month (M)
Mukta picks 26/01. After all steps she gets 291. Shubham subtracts 165 → 126. Last 2 digits = 26 (Day), remaining = 1 (Month). Date = 26th January ✓
🔹 Practice — Decode these answers
| Final Answer | Subtract 165 | Date |
|---|---|---|
| 1390 | 1225 | 25th December |
| 1269 | 1104 | 4th November |
| 394 | 229 | 29th February |
| 296 | 131 | 31st January |
6.3 Number Pyramids
Each number in the pyramid equals the sum of the two numbers directly below it.
▶ Example — 3-row pyramid
Row 2 (Middle): 10 | 13
Row 1 (Bottom): 1 | 9 | 4Check: 1+9=10 ✓ 9+4=13 ✓ 10+13=23 ✓
▶ Working Backwards (Given top and bottom)
If the top and two outer bottom values are given, use algebra to find the middle:
- Label the unknown middle bottom as c
- Middle-left = 12 + c, Middle-right = c + 8
- Sum of row 2 = Top: (12 + c) + (c + 8) = 60
- Solve: 20 + 2c = 60 → c = 20
▶ Algebraic Formula for Top of Pyramid
For a 3-row pyramid with bottom row a, b, c:
This means the middle value (b) is counted twice in the top!
Top = first + 2×middle + last. Example: bottom row = 4, 13, 8 → Top = 4 + 26 + 8 = 38
▶ Virahāṅka–Fibonacci Connection
The Virahāṅka-Fibonacci sequence is: 1, 2, 3, 5, 8, 13, 21, … (each number = sum of previous two).
If the first n Virahāṅka-Fibonacci numbers fill the bottom row of an n-row pyramid, every number in the pyramid is also a Virahāṅka-Fibonacci number!
6.4 Fun with Grids
📆 Calendar Magic
Pick any 2×2 block from a calendar (4 numbers). If someone tells you only the sum of those 4 numbers, you can figure out all 4!
Let the top-left number = a. Then:
▶ Finding the numbers from the sum
- You are told sum = 36
- Set up: 4a + 16 = 36
- Subtract 16: 4a = 20
- Divide by 4: a = 5
- The grid is: 5, 6, 12, 13 ✓
Top-left number = (Sum − 16) ÷ 4. For sum = 40 → a = (40−16)÷4 = 6. Grid = 6, 7, 13, 14.
🔷 Algebra Grids (Shapes = Numbers)
In an algebra grid, shapes represent numbers. The last column is the row sum. Solve like a system of equations!
Row 1: 🟦 + 🟦 + 🟦 = 27 → 🟦 = 9
Row 2: 🔴 + 🔴 + 🟦 = 19 → 2×🔴 + 9 = 19 → 🔴 = 5
- Start with the row that has only one shape (easiest to solve).
- Substitute found values into other rows to find remaining shapes.
- Always verify your answers by checking the row sums.
6.5 The Largest Product
Given three digits, arrange them in the form □□ × □ to get the largest possible product.
▶ Example with 2, 3, 5
All six arrangements:
| Arrangement | Product |
|---|---|
| 23 × 5 | 115 |
| 25 × 3 | 75 |
| 32 × 5 | 160 ✅ Largest |
| 35 × 2 | 70 |
| 52 × 3 | 156 |
| 53 × 2 | 106 |
▶ General Rule (Proved by Algebra)
Let the three digits be p < q < r. After comparing all six products using algebra:
Use the largest digit as the multiplier (units position). Arrange the other two digits in decreasing order to form the 2-digit multiplicand. Largest = qp × r
▶ Algebraic Proof Outline
qp × r = (10×q×r) + (p×r)
rp × q = (10×r×q) + (p×q)
First terms are EQUAL.
Second terms: p×r vs p×q
Since r > q → p×r > p×q
∴ qp × r is LARGER ✓
▶ Quick Practice
6.6 Decoding Divisibility Tricks
🔹 Trick 1 — Reverse & Subtract (Divisible by 9)
- Choose any 2-digit number with different digits, say ab
- Reverse it to get ba
- Find the difference: ba − ab
- The result is always divisible by 9, with no remainder
When you divide the difference by 9, the quotient = (larger digit − smaller digit). E.g., 74 − 47 = 27, 27÷9 = 3 = (7−4) ✓
🔹 Trick 2 — Reverse & Add (Divisible by 11)
Take any 2-digit number, reverse it, and add the two numbers. The result is always divisible by 11!
| Number | Reversed | Sum | Divisible by 11? |
|---|---|---|---|
| 31 | 13 | 44 | 44 ÷ 11 = 4 ✓ |
| 28 | 82 | 110 | 110 ÷ 11 = 10 ✓ |
| 12 | 21 | 33 | 33 ÷ 11 = 3 ✓ |
🔹 Trick 3 — 3-Digit Cyclic Sum (Divisible by 37)
Take a 3-digit number abc. Form two more numbers by cycling digits: bca and cab. Add all three.
= (100a+10b+c) + (100b+10c+a) + (100c+10a+b)
= 111a + 111b + 111c = 111(a + b + c)
Since 111 = 3 × 37, the sum is divisible by both 3 and 37.
🔹 Trick 4 — Repeating 3-digit number (abcabc)
Take a 3-digit number, repeat its digits to make a 6-digit number: abcabc.
And 1001 = 7 × 11 × 13
So dividing by 7, then 11, then 13 gives back the original abc!
Important Word Problems from the Chapter
🐴 Problem: Horses & Hens
A farm has horses and hens. Total heads = 55, total legs = 150. Find how many of each.
Legs equation: 4h + 2(55 − h) = 150
4h + 110 − 2h = 150
2h = 40
h = 20 horses, hens = 35
👩👧 Problem: Mother & Daughter Ages
Mother is 5 times daughter’s age. In 6 years, mother will be 3 times daughter’s age. Find daughter’s current age.
In 6 years: 5d + 6 = 3(d + 6)
5d + 6 = 3d + 18
2d = 12
d = 6 years (Daughter), Mother = 30 years
🐄 Problem: Gauri & Naina’s Cows
Naina has twice as many cows as Gauri. If Naina gives 3 to Gauri, they have equal numbers.
Equal after: 2g − 3 = g + 3
g = 6
Gauri = 6 cows, Naina = 12 cows
🌺 Problem: Karim & the Genie (Coins)
Karim goes around a magical tree 3 times. Each round doubles his coins but he pays 8 coins. After 3 rounds he has exactly 8 coins (to pay for the 3rd round).
After Round 1: 2x − 8
After Round 2: 2(2x−8) − 8 = 4x − 24
After Round 3: 2(4x−24) − 8 = 8x − 56He has 8 left (for genie): 8x − 56 = 8
8x = 64
x = 7 coins (Karim started with 7)
Important Exam Questions with Answers
Quick Revision Summary
Use variable x to track steps. The x cancels out, leaving a constant answer.
Final = 100M + 165 + D. Decode: subtract 165. Last 2 digits = Day, rest = Month.
Top = a + 2b + c for 3-row. Solve unknown cells using algebra equations.
2×2 grid sum = 4a + 16. Find a = (sum − 16) ÷ 4, then fill the grid.
Largest digit → multiplier. Other two in decreasing order → multiplicand.
Reverse-subtract → divisible by 9. Reverse-add → by 11. Cyclic sum → by 37. abcabc → by 7, 11, 13.
Whenever a question says “prove that … always happens” or “show that the result is always …”, the trick is to use a variable instead of a specific number and simplify. The answer will either cancel down to a fixed value or factor nicely (like 9k, 11k, etc.).
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