Chapter 5 of 14

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CBSE Class 8 Mathematics★ 8-10 marks · CBSE Class 8 Maths (Ganita Prakash Chapter 5)

Number Play

'Number Play' is a discovery chapter — exploring patterns in numbers, divisibility rules, mental-math tricks, magic squares, sequences, and the playfulness behind serious mathematics. Master Vedic-style fast computation, divisibility for 2-13, generating sequences, and the curious patterns that delight mathematicians. CBSE Class 8 Maths (Ganita Prakash) Chapter 5.

⏱ 22 min readEasy difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Apply divisibility rules for 2 through 13
2Recognise famous patterns (triangular, square, Fibonacci, Pascal)
3Use Vedic mental-math shortcuts
4Construct and verify magic squares
5Connect playful tricks to algebraic explanations

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Why this chapter matters

Trains pattern recognition — the universal skill in mathematics. Master divisibility rules (used everywhere), Vedic mental-math shortcuts (save exam time), famous patterns (Fibonacci, triangular, magic squares).

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Class 7 Number Patterns

Number Play — Class 8 Mathematics (Ganita Prakash)

"Numbers are not just for serious work — they are also for play. Patterns, puzzles, and tricks are the gateway to seeing what mathematics really IS."

1. About the Chapter

'Number Play' is the most playful chapter in Ganita Prakash. It introduces you to:

Divisibility rules (how to tell at a glance if a number is divisible by 2-13)
Patterns in numbers (Fibonacci, triangular numbers, Pascal's triangle)
Mental-math shortcuts (Vedic tricks)
Magic squares
Number puzzles (riddles solved by algebra)

Why This Matters

Most of mathematics is PATTERN RECOGNITION. This chapter trains your eye to see patterns — a skill that benefits every later chapter.

2. Divisibility Rules (Master ALL)

Divisibility by 2

A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.

Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3.

Example: 4827 → 4+8+2+7 = 21; 21 ÷ 3 = 7 ✓

Divisibility by 4

A number is divisible by 4 if the last two digits form a number divisible by 4.

Example: 12,316 → last two digits 16 → 16 ÷ 4 = 4 ✓

Divisibility by 5

A number is divisible by 5 if its last digit is 0 or 5.

Divisibility by 6

A number is divisible by 6 if it is divisible by both 2 AND 3.

Divisibility by 7

Method (Indian Vedic): Take the last digit, double it, and subtract from the rest. Repeat. If final result is divisible by 7, original is too.

Example: 343 → 34 − (2×3) = 34 − 6 = 28 → 28 ÷ 7 = 4 ✓

Divisibility by 8

A number is divisible by 8 if its last three digits form a number divisible by 8.

Divisibility by 9

A number is divisible by 9 if the sum of its digits is divisible by 9.

Example: 729 → 7+2+9 = 18 → 18 ÷ 9 = 2 ✓

Divisibility by 10

A number is divisible by 10 if its last digit is 0.

Divisibility by 11

Alternating sum of digits (from right) must be divisible by 11.

Example: 121 → 1 − 2 + 1 = 0; 0 ÷ 11 = 0 ✓
Example: 9482 → 2 − 8 + 4 − 9 = −11 → divisible by 11 ✓

Divisibility by 12

Divisible by both 3 AND 4.

Divisibility by 13

Method: Add 4 times the last digit to the rest. Repeat.

Example: 845 → 84 + (4×5) = 84 + 20 = 104. Repeat: 10 + (4×4) = 26. 26 ÷ 13 = 2 ✓

3. Famous Number Patterns

Triangular Numbers

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Pattern: T(n) = n(n+1)/2

T(1) = 1
T(2) = 3
T(3) = 6 (drawn as triangle of 3-row dots)
T(10) = 55

Square Numbers

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...

(See Chapter 1: Squares)

Pentagonal Numbers

1, 5, 12, 22, 35, ... Pattern: P(n) = n(3n−1)/2

Fibonacci Sequence

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...

Rule: Each term = sum of previous two.

F₁ = 1, F₂ = 1, F₃ = 2, F₄ = 3, F₅ = 5...

Appears in nature: sunflower spirals, pinecones, rabbit population growth.

Pascal's Triangle

            1
          1   1
        1   2   1
      1   3   3   1
    1   4   6   4   1
  1   5  10  10   5   1
1   6  15  20  15   6   1

Each entry = sum of two above. Appears in binomial coefficients (Class 11 onwards).

4. Mental-Math Shortcuts (Vedic-Inspired)

Squaring numbers ending in 5

Rule: ab5² → write 25 at the end; before it, write a(a+1) where a is the tens digit.

25² = 2 × 3 | 25 = 625
35² = 3 × 4 | 25 = 1225
65² = 6 × 7 | 25 = 4225
95² = 9 × 10 | 25 = 9025

Multiplying by 11 (two-digit)

Rule: ab × 11 = a (a+b) b. If (a+b) ≥ 10, carry.

23 × 11 = 2 | 2+3 | 3 = 253
47 × 11 = 4 | 4+7 | 7 = 4 | 11 | 7 = (4+1) | 1 | 7 = 517

Multiplying by 5, 25, 125

×5 = ×10 ÷ 2
×25 = ×100 ÷ 4
×125 = ×1000 ÷ 8

Example: 87 × 25 = 8700 ÷ 4 = 2175 ✓

Subtracting from 100, 1000, 10000

Rule (subtract from 1000): All digits subtract from 9, except last digit subtracts from 10.

1000 − 467 = (9−4)(9−6)(10−7) = 533

Multiplication of close numbers

Rule: ab × ac = a(a+b+c) | bc — works when first digit same, last digits sum to 10.

67 × 63 = 6×7 | 7×3 = 42 | 21 = 4221

Actually for 23 × 27: first digit 2, last digits 3+7=10. So 2(3) | 3×7 = 6 | 21 = 621.

5. Magic Squares

A magic square is an n×n grid filled with distinct numbers such that every row, column, and main diagonal sums to the same value (the magic constant).

Classic 3×3 (Lo Shu Square / Indian Vedic)

8  1  6
3  5  7
4  9  2

Magic constant = 15 (every row, column, diagonal sums to 15).

Construction Method (Odd n)

Place 1 in the middle of the top row
Move up-and-right one cell at a time
If you go off the grid, wrap around
If the cell is occupied, move down one instead

4×4 (Even)

More complex. The Lo Shu / Indian Vedic 4×4 magic squares were studied by ancient Indian mathematicians.

Magic Constant Formula

For an n×n square filled with 1 to n²: Magic constant = n(n² + 1) / 2

For n = 3: 3(9+1)/2 = 15 ✓
For n = 4: 4(16+1)/2 = 34
For n = 5: 5(25+1)/2 = 65

6. Number Puzzles and Tricks

Classic Puzzle: Cross-Number Verification

Find a 4-digit number where:

Sum of digits = 18
Reverse of the number = 4 times the original
Divisible by 11

This is solved by setting up equations — connects to algebra.

The Sum of Consecutive Integers

1 + 2 + 3 + ... + n = n(n+1)/2
1 + 2 + ... + 100 = 100 × 101 / 2 = 5050

This is the Gauss formula — the young Gauss is said to have computed this in seconds.

Sum of Consecutive Odd Numbers = Square

1 + 3 + 5 + ... + (2n−1) = n² (From Chapter 1)

Sum of Consecutive Squares

1² + 2² + 3² + ... + n² = n(n+1)(2n+1)/6

Sum of Consecutive Cubes

1³ + 2³ + ... + n³ = [n(n+1)/2]²

7. Sequences and Series

Arithmetic Sequence (AP)

A sequence with common DIFFERENCE: a, a+d, a+2d, ...

3, 7, 11, 15, ... (d = 4)

n-th term: aₙ = a + (n−1)d

Geometric Sequence (GP)

A sequence with common RATIO: a, ar, ar², ar³, ...

2, 6, 18, 54, ... (r = 3)

n-th term: aₙ = ar^(n−1)

Other Famous Sequences

Fibonacci (each term = sum of previous two)
Triangular, Square, Cube numbers
Prime numbers (2, 3, 5, 7, 11, 13, ...)

8. Number Tricks

"Think of a number"

Think of any number
Double it
Add 10
Divide by 2
Subtract original number
Result: 5

Why does this always give 5? Let x = number. ((2x + 10) / 2) − x = (x + 5) − x = 5 ✓

This is the magic of algebra explaining tricks.

Divisibility by 9 trick

Choose any number, e.g., 7283
Add digits: 7+2+8+3 = 20
20 is not divisible by 9, so 7283 is not.
Try 729: 7+2+9 = 18; 18 ÷ 9 = 2 ✓ — divisible.

9. Worked Examples

Example 1: Divisibility

Is 13,572 divisible by 6?

Divisible by 2? Last digit 2 → yes
Divisible by 3? Sum: 1+3+5+7+2 = 18; 18 ÷ 3 = 6 → yes
Therefore divisible by 6 ✓

Example 2: Number from Pattern

Find the 10th triangular number.

T(10) = 10 × 11 / 2 = 55 ✓

Example 3: Vedic Squaring

Compute 75².

7 × 8 | 25 = 56 | 25 = 5625 ✓

Example 4: Multiplying by 11

Find 35 × 11.

3 | 3+5 | 5 = 3 | 8 | 5 = 385 ✓

Example 5: Subtracting from 10000

Find 10000 − 6789.

(9−6)(9−7)(9−8)(10−9) = 3211 ✓

Example 6: Fibonacci

What is F₁₀?

1, 1, 2, 3, 5, 8, 13, 21, 34, 55
F₁₀ = 55

Example 7: Magic Square Verification

Verify this is a magic square:

2 7 6
9 5 1
4 3 8

Rows: 2+7+6 = 15, 9+5+1 = 15, 4+3+8 = 15 ✓
Columns: 2+9+4 = 15, 7+5+3 = 15, 6+1+8 = 15 ✓
Diagonals: 2+5+8 = 15, 6+5+4 = 15 ✓
This IS a magic square with constant 15.

10. Common Mistakes

Confusing divisibility tests
- Divisible by 9 needs SUM of digits divisible by 9 (NOT by 3 alone)
- Divisible by 4 needs LAST TWO digits (NOT one digit)
Vedic squaring wrong format
- 65² = 6×7 | 25 = 4225 (not 6725)
11-multiplication carry
- 47 × 11 = 4|11|7 = 517 (don't forget the carry)
Fibonacci confusion
- Each term is sum of TWO previous, not 'multiply by 2'
Magic square wrong constant
- For 3×3 with 1-9: constant is 15 (not 9 or 10)

11. Tips for Mastery

For Divisibility

Memorise the rules in tabular form
Practise QUICK identification on 4-5 digit numbers
These rules are tested EVERY year

For Mental Math

Practise daily — 5 problems per type
After 2 weeks, you'll do them automatically

For Patterns

Always look for PATTERNS in any sequence
Try fitting formulas: n², n³, n(n+1)/2

For Puzzles

Use ALGEBRA to solve tricks (let x = unknown)
This connects 'play' to 'serious math'

12. Historical Notes

Indian Vedic Mathematics

'Vedic Mathematics' (Bharati Krishna Tirthaji, 1965) — modern compilation of ancient Sanskrit shortcuts
16 'sutras' (formulas) for fast mental computation
Roots in Sulba Sutras and later Indian math traditions

Magic Squares in India

Earliest known 4×4 magic square in India — by Khajuraho temples (~1000 CE)
Narayana Pandit (14th century CE) wrote 'Ganita Kaumudi' with magic-square theory
Ramanujan discovered new methods for constructing magic squares

Gauss and the Schoolboy

Carl Gauss (1777-1855), age 9, summed 1 to 100 in seconds using the pairing trick:
- 1+100, 2+99, ..., 50+51 — each pair sums to 101
- 50 pairs × 101 = 5050 ✓

13. Conclusion

Mathematics is full of patterns, puzzles, and shortcuts — not just rules to memorise. 'Number Play' is the chapter that reveals this playful side.

The divisibility rules will help you in every later math chapter. The patterns (triangular, Fibonacci, Pascal) will appear again and again in higher mathematics. The Vedic shortcuts will save you HOURS in exams.

Most importantly, this chapter teaches you that mathematics is delightful — once you see the patterns, you can't stop seeing them. Number play is the gateway to mathematical thinking.

Practise the tricks, master the divisibility rules, and let yourself be amazed by the elegant patterns hiding in plain sight. Indian mathematics has always celebrated this playful spirit — from Lilavati's poetic puzzles to Ramanujan's astonishing identities. Now it's your turn to play.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Sum of 1 to n

1 + 2 + ... + n = n(n+1)/2

Gauss formula; triangular numbers

Sum of squares

1² + 2² + ... + n² = n(n+1)(2n+1)/6

Sum of cubes

1³ + 2³ + ... + n³ = [n(n+1)/2]²

Nicomachus's theorem

Magic constant (n×n square 1 to n²)

M = n(n²+1)/2

AP n-th term

aₙ = a + (n−1)d

GP n-th term

aₙ = ar^(n−1)

Triangular number

Tₙ = n(n+1)/2

Pentagonal number

Pₙ = n(3n−1)/2

Fibonacci

Fₙ = Fₙ₋₁ + Fₙ₋₂; F₁=F₂=1

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Wrong divisibility test

✓ For 9: sum of digits divisible by 9 (NOT 3). For 4: last TWO digits (not 1). For 8: last THREE digits.

WATCH OUT

✗ Vedic-multiply mistakes

✓ 65² = 6×7 | 25 = 4225. Not 6 × 25 = 150 or 6725.

WATCH OUT

✗ Fibonacci misunderstanding

✓ Each term = SUM of previous TWO. Not 'double previous' (that's GP with ratio 2).

WATCH OUT

✗ Sum formula off-by-one

✓ 1 + 2 + ... + n = n(n+1)/2. So 1 to 10 is 10×11/2 = 55, not 50 or 60.

WATCH OUT

✗ Magic square wrong constant

✓ For 3×3 with 1-9: constant is 15 (use M = n(n²+1)/2).

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Try These - Divisibility

Apply divisibility tests

Questions

Try These - Patterns

Identify, extend sequences

Questions

Try These - Mental Math

Mixed problems on patterns

Questions

Ex 5.2

Exercise 5.2

Magic squares and number tricks

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Divisibility

Is 13,572 divisible by 6?

▶ Show solution

✦ Answer: Yes. Test: (a) Last digit 2 → divisible by 2. (b) Sum of digits 1+3+5+7+2 = 18; 18 ÷ 3 = 6 → divisible by 3. Since divisible by both 2 and 3, divisible by 6.

Q2EASY· Pattern

What is the 10th Fibonacci number?

▶ Show solution

✦ Answer: F₁=1, F₂=1, F₃=2, F₄=3, F₅=5, F₆=8, F₇=13, F₈=21, F₉=34, F₁₀=55. So F₁₀ = 55.

Q3MEDIUM· Mental Math

Using Vedic method, compute 85².

▶ Show solution

Step 1 — Identify the pattern for numbers ending in 5.
  For ab5²: write a × (a+1) followed by 25.

Step 2 — Apply for 85².
  a = 8
  a × (a+1) = 8 × 9 = 72
  Append 25: 7225

Step 3 — Verify.
  85 × 85: 85 × 85 = 85 × 80 + 85 × 5 = 6800 + 425 = 7225 ✓

✦ Answer: 85² = 7225.

Q4MEDIUM· Magic square

Find the magic constant for a 5×5 magic square containing numbers 1 to 25.

▶ Show solution

Step 1 — Apply magic constant formula.
  M = n(n² + 1)/2
  where n = order of the square = 5

Step 2 — Compute.
  M = 5(5² + 1)/2
     = 5(25 + 1)/2
     = 5 × 26 / 2
     = 130 / 2
     = 65

Step 3 — Verify intuition.
  Sum of 1 to 25 = 25 × 26 / 2 = 325
  Each row should sum to 325/5 = 65 ✓

✦ Answer: Magic constant for 5×5 = 65. Every row, column, diagonal sums to 65.

Q5HARD· Application

(a) Find the sum 1 + 2 + 3 + ... + 200. (b) Find the sum of squares 1² + 2² + 3² + ... + 20². (c) Find the sum of cubes 1³ + 2³ + 3³ + ... + 10³.

▶ Show solution

Part (a) — Sum 1 to 200.
  Formula: n(n+1)/2
  = 200 × 201 / 2
  = 100 × 201
  = 20,100

Part (b) — Sum of squares 1² to 20².
  Formula: n(n+1)(2n+1)/6
  = 20 × 21 × 41 / 6
  = 17,220 / 6
  Wait, let me recompute: 20 × 21 = 420; 420 × 41 = 17,220
  17,220 / 6 = 2870

Part (c) — Sum of cubes 1³ to 10³.
  Formula: [n(n+1)/2]²
  = [10 × 11 / 2]²
  = 55²
  = 3,025

✦ Answer: (a) 20,100. (b) 2,870. (c) 3,025.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Divisibility by 2: last digit 0,2,4,6,8
•Divisibility by 3: sum of digits divisible by 3
•Divisibility by 4: last two digits divisible by 4
•Divisibility by 5: last digit 0 or 5
•Divisibility by 6: divisible by 2 AND 3
•Divisibility by 9: sum of digits divisible by 9
•Divisibility by 10: last digit 0
•Divisibility by 11: alternating sum divisible by 11
•Triangular numbers: 1, 3, 6, 10, 15, ... formula n(n+1)/2
•Fibonacci: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
•Magic constant (n×n with 1-n²): n(n²+1)/2
•3×3 magic constant = 15; 4×4 = 34; 5×5 = 65
•Vedic X5² rule: a × (a+1) | 25
•Vedic ×11 (two digit): a | a+b | b (carry if needed)
•Gauss sum: 1 + 2 + ... + n = n(n+1)/2
•Sum of squares: n(n+1)(2n+1)/6
•Sum of cubes: [n(n+1)/2]²

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-10 marks per chapter

Question type	Marks each	Typical count	What it tests
MCQ / Very Short	1	3	Divisibility tests; pattern identification
Short Answer	2-3	2	Vedic shortcuts; magic squares; sum formulas
Long Answer	5	1	Multi-step pattern problems; constructing magic squares

Prep strategy

Memorise divisibility tests for 2-13
Practise Vedic squaring with all 9 'X5²' values
Memorise sum formulas (1+2+...+n, sum of squares, sum of cubes)
Construct a 3×3 magic square from scratch
Practice Fibonacci sequence up to F₁₅

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Mental arithmetic competitions

Vedic methods used in international mental-math contests. India often performs well due to this tradition.

Bank check digit

Banks use divisibility rules to verify account numbers. The last digit is often a 'check digit' computed using divisibility math.

Music theory

Octave relationships are powers of 2; musical intervals use ratios. Sequences and patterns underlie musical structure.

Cryptography

Modern encryption uses divisibility properties of huge numbers. Vedic-inspired fast computation is used in some cryptographic implementations.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Memorise all divisibility tests (2-13)
Use Vedic shortcuts to save time on big calculations
For 'find magic constant' questions, use formula M = n(n²+1)/2
Show pattern recognition for full marks
Connect to algebra (e.g., explain 'think of a number' tricks)

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Wilson's theorem and prime divisibility
Pascal's triangle and binomial coefficients
Fibonacci-related identities (Cassini, etc.)
Squaring and multiplying with Trachtenberg method
Indian Vedic sutras — all 16 with examples
Ramanujan's work on magic squares and theta functions

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 8 School ExamVery High

Class 8 Maths Olympiad (IMO)Very High

NTSE Mental AbilityVery High

NMTCVery High

Aryabhata Maths CompetitionHigh

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Every digit's place value (1, 10, 100, 1000, ...) is 1 + (a multiple of 9). For example, 10 = 1 + 9, 100 = 1 + 99, etc. So a number like 729 = 7(100) + 2(10) + 9 = 7(1+99) + 2(1+9) + 9 = (7+2+9) + (7×99 + 2×9). The second part is divisible by 9. So 729 is divisible by 9 iff (7+2+9) = 18 is — which it is. This works for any number — proof by place-value.

Yes, surprisingly often. Sunflower seed spirals (21, 34, 55 or 34, 55, 89 — all Fibonacci), pinecone scales, nautilus shells, branching of trees, arrangement of leaves on stems — all show Fibonacci numbers. The reason: in growth processes, each new structure depends on the previous two, mimicking the Fibonacci rule. Beautiful coincidence between math and biology!

Partly. The 16 'sutras' were compiled in modern times (by Bharati Krishna Tirthaji, 1965) drawing on Sanskrit math traditions. The TECHNIQUES are genuine — derived from Indian mathematical thinking that goes back to Aryabhata (5th c. CE) and the Sulba Sutras (~800 BCE). But the 'Vedic' framing as a single ancient text is a modern reconstruction. The methods themselves are mathematically valid and useful.

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