Chapter 5 of 14

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

Number Play

"'Number Play' is about divisibility reasoning — not just the tests but WHY they work. Learn to add and subtract remainders directly, solve 'always/sometimes/never true' claims with algebra, master the digital root and its link to division by 9, combine coprime tests for divisors like 18, 36 and 44, and crack cryptarithms by reasoning from the units column. 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…

1Express consecutive and consecutive-even numbers algebraically and use them to prove statements about sums
2Classify statements as always, sometimes or never true, justifying with algebra and with matching examples and non-examples
3Add and subtract remainders directly, without computing the original large numbers
4Solve remainder problems by shifting to a common multiple — the 'each remainder is one less than its divisor' trick
5State and explain the divisibility tests for 2, 3, 4, 5, 6, 8, 9, 10 and 11, including why each works
6Define the digital root and use the fact that it equals the remainder mod 9 (with 9 standing for 0)
7Solve cryptarithms by reasoning from the units column and the carry
8Combine two coprime divisibility tests to handle composite divisors like 18, 36, 44 and 15

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Why this chapter matters

This chapter turns divisibility from a set of tricks into something you can explain. Instead of memorising 'add the digits for 9', you find out why it works — every power of 10 leaves remainder 1 on division by 9 — and that single insight then generates the tests for 3 and 11 as well. It also teaches the habit that matters most in mathematics: telling apart 'always true', 'sometimes true' and 'never true', and knowing that one counter-example destroys a claim while a hundred examples never prove one.

Before you start — revise these

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

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Factors, multiples, HCF and LCM

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Prime factorisation

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Basic algebraic expressions and forming simple equations

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The idea of a remainder in division

Number Play — Class 8 Mathematics (Ganita Prakash)

"Learning maths is not just about knowing some shortcuts and following procedures but about understanding 'why' something works." — Ganita Prakash, Grade 8

What the book actually covers (2026-27) This chapter is built on three ideas, and every exam question comes from them:

Remainder reasoning — writing numbers as divisor × quotient + remainder, adding and subtracting remainders on their own, and deciding whether a claim is always, sometimes or never true using algebra rather than examples.

Divisibility tests and why they work — the book is emphatic that the reason matters more than the rule. It derives the tests for 3, 9 and 11 from what powers of 10 leave behind, and introduces digital roots.

Cryptarithms — letter puzzles solved by reasoning from the units column. Fibonacci, Pascal's triangle, magic squares, Vedic multiplication shortcuts and the tests for 7 and 13 are not in this chapter. They have been kept at the end of this page under Appendix — beyond the current syllabus, because they are genuinely enjoyable, but do not spend exam-preparation time on them.

1. About the Chapter

'Number Play' teaches you to reason about divisibility and remainders. It covers:

Consecutive numbers in algebra — and using them to prove statements about sums
Remainder arithmetic — finding the remainder of 4779 + 661 without adding the numbers
Divisibility tests for 2, 3, 4, 5, 6, 8, 9, 10 and 11 — and the reason each one works
Digital roots and their link to division by 9
Cryptarithms — puzzles where letters stand for digits

Why This Matters

The chapter's real subject is justification. You are constantly asked whether a claim is always, sometimes or never true — and a hundred supporting examples never prove "always", while a single counter-example destroys it.

1A. Remainder Reasoning

Writing a number by its remainder

Any number divided by d can be written as N = dq + r, where r is the remainder and 0 ≤ r < d. This one line does most of the work in this chapter.

Remainders add and subtract on their own

You do not need the large numbers at all.

661 leaves remainder 3 on division by 7, and 4779 leaves remainder 5. Sum: 3 + 5 = 8, which is more than 7 — so regroup: one more 7, remainder 1. Difference: 5 − 3 = 2.

Check: 5440 = 7 × 777 + 1 ✓ and 4118 = 7 × 588 + 2 ✓

Two shortcuts worth memorising

Situation	Method
Same remainder r for every divisor	The number is LCM + r. Remainder 2 for both 3 and 4 → 12n + 2
Every remainder is one less than its divisor	The number is LCM − 1. Remainders 2, 3, 4 for divisors 3, 4, 5 → 60 − 1 = 59

Both work by shifting the number so that all the awkward remainders vanish at once.

Always, sometimes or never true?

Claim type	What is needed
Always true	Algebra. 6x + 9y = 3(2x + 3y), so it is always a multiple of 3
Sometimes true	An example and a non-example. 18 + 9 = 27 ✓ but 12 + 9 = 21 ✗
Never true	Algebra showing a fixed leftover. 8(7b−3) − 4(11b+1) = 12b − 28 = 12(b−2) − 4, always 4 short

1B. Why the Divisibility Tests Work

This is the part the book cares about most, and it is regularly examined as a "justify" question.

Why adding digits tests for 9 (and 3)

Every power of 10 is one more than a multiple of 9: 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, …

So expanding a number splits it into a multiple-of-9 part plus the digits themselves:

$3547 = 3 (999 + 1) + 5 (99 + 1) + 4 (9 + 1) + 7 = multiples of 9 (2997 + 495 + 36) + digit sum (3 + 5 + 4 + 7)$

The first bracket contributes nothing to the remainder, so the number and its digit sum leave the same remainder. Since 3 also divides 9, 99 and 999, the identical argument gives the test for 3.

Why the test for 11 alternates

Here the powers of 10 flip sign: 10 = 11 − 1 leaves −1, 100 = 99 + 1 leaves +1, 1000 = 1001 − 1 leaves −1. So each digit contributes with alternating sign, starting from + at the units place.

857076 → 6 − 7 + 0 − 7 + 5 − 8 = −11, a multiple of 11, so divisible ✓

The trap: a negative result is fine. −11 means divisible; −3 means remainder −3 + 11 = 8.

Why 4 and 8 look only at the end

100 is a multiple of 4, so everything above the last two digits is already divisible by 4. Similarly 1000 is a multiple of 8, so only the last three digits matter.

Composite divisors — split into COPRIME factors

This is the highest-yield idea in the chapter.

Divisor	Correct split	Wrong split
6	2 × 3	—
12	4 × 3	~~2 × 6~~ (18 passes both, isn't a multiple of 12)
15	3 × 5	—
18	2 × 9	~~3 × 6~~
36	4 × 9	~~6 × 6~~
44	4 × 11	—

The factors must share no common factor, otherwise passing both tests proves nothing.

1C. Digital Roots

Add the digits, then add again, until a single digit remains. That digit is the digital root.

3547 → 19 → 10 → 1

The key property: the digital root is the remainder on division by 9 — with one relabelling, that a multiple of 9 has digital root 9, not 0. (The process always ends on a digit from 1 to 9, so it can never produce 0.)

Consequences

Digital root	Remainder ÷ 9	Remainder ÷ 3
1, 4, 7	1, 4, 7	1
2, 5, 8	2, 5, 8	2
3, 6	3, 6	0
9	0	0

Digital roots cycle

The digital roots of the multiples of k repeat with cycle length 9 ÷ HCF(k, 9):

multiples of 3 → 3, 6, 9 repeating (length 3)
multiples of 6 → 6, 3, 9 repeating (length 3)
multiples of 4 → 4, 8, 3, 7, 2, 6, 1, 5, 9 repeating (length 9 — every root appears, since 4 shares no factor with 9)

Note: parity and digital root have no relation, because parity is about division by 2 while the digital root is about division by 9, and 2 and 9 share no factor.

1D. Cryptarithms

Letters stand for digits. Each letter is one digit, different letters are different digits, and no number starts with 0.

How to start

Always begin at the units column — it is the only column with no carry coming into it.

Useful openings:

A multiple of 5 ends in 0 or 5
If the product's units digit equals a letter already in the multiplicand, only a couple of values fit
Repdigit answers factorise: 111 = 3 × 37, so EF × E = GGG forces EF = 37
Bound the size: if a 2-digit number × 4 stays 2-digit, the number is at most 24

Worked example

QR + QR + QR = PRR The answer's units digit is R, and it comes from 3 × R. So 3R ends in R, meaning R = 0 or 5. R = 0 would force a multiple of 100, so R = 5. Then 3 × Q5 = P55, and Q = 8 gives 85 × 3 = 255 ✓ Q = 8, R = 5, P = 2

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

Not in Ganita Prakash — included as an extra.

Method: Take the last digit, double it, and subtract from the rest. Repeat. If the final result is divisible by 7, so is the original.

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

Not in Ganita Prakash — included as an extra.

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 ✓

Appendix — beyond the current syllabus

Everything from here to the end of the page is enrichment, not examinable content for this chapter. Fibonacci numbers, triangular and pentagonal numbers, Pascal's triangle, Vedic-style multiplication shortcuts, magic squares and arithmetic/geometric sequences do not appear in Ganita Prakash Chapter 5. They are kept because they are genuinely enjoyable and several are picked up in later years — but for exam preparation, work from sections 1A to 1D above and from the chapter's own Figure it Out exercises.

The examinable content of this chapter is: remainder reasoning, always/sometimes/never-true justification, the divisibility tests for 2-6, 8, 9, 10 and 11 with their reasons, digital roots, and cryptarithms.

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.

Consecutive numbers

n, n+1, n+2, … (sum of four = 4n + 6)

Consecutive EVEN numbers step by 2: 2n, 2n+2, 2n+4.

Remainder form

N = dq + r, with 0 ≤ r < d

Remainders add and subtract on their own; regroup if the result reaches d.

Divisibility by 3 and 9

test the digit sum

Works because every power of 10 leaves remainder 1 on division by 9 (and by 3).

Divisibility by 11

alternating digit sum from the units digit

0 or any multiple of 11 (including negative) means divisible.

Divisibility by 4 and 8

last two digits for 4; last three for 8

Because 100 is a multiple of 4 and 1000 is a multiple of 8.

Composite divisors

split into coprime factors

18 = 2×9, 36 = 4×9, 44 = 4×11, 15 = 3×5, 6 = 2×3. Both tests must pass.

Digital root

repeated digit sum = remainder mod 9

A multiple of 9 has digital root 9, not 0.

Product of k consecutive integers

always a multiple of k!

2 for two, 6 for three, 24 for four, 120 for five.

⚠️

Common mistakes & fixes

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

WATCH OUT

✗ Proving 'always true' by giving examples

✓ Use algebra. To prove 6x + 9y is a multiple of 3, factorise it as 3(2x + 3y). One counter-example, though, IS enough to disprove a claim.

WATCH OUT

✗ Treating a negative alternating sum as 'not divisible by 11'

✓ Check whether the value is a multiple of 11 including negatives. For a true remainder, add 11 until the result is between 0 and 10.

WATCH OUT

✗ Saying the digital root of a multiple of 9 is 0

✓ A remainder of 0 shows up as a digital root of 9. Every other remainder equals the digital root directly.

WATCH OUT

✗ Splitting a composite divisor into factors that share a common factor

✓ Split into COPRIME factors: 12 = 4 × 3, 36 = 4 × 9, 44 = 4 × 11. Then both tests together are enough.

WATCH OUT

✗ Writing consecutive even numbers as n, n+1, n+2

✓ Use 2n, 2n+2, 2n+4 — or centre them on the middle value, e.g. 5p−4, 5p−2, 5p, 5p+2, 5p+4.

WATCH OUT

✗ Assuming a shuffled multiple of 9 might stop being a multiple of 9

✓ Any shuffle of a multiple of 9 (or 3) is still one. This does NOT extend to 2, 4, 5, 8 or 11, whose tests depend on position.

WATCH OUT

✗ Starting a cryptarithm from the leftmost column

✓ Always begin at the units digit, and note that the units of the product often pins one letter immediately.

NCERT exercises (with solutions)

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

Figure it Out — Multiples, Remainders and Consecutive Numbers

Questions

Figure it Out — Divisibility by 9

Questions

In-text — The Divisibility Shortcut for 11

Questions

In-text — The Divisibility Table

Questions

In-text — Digital Roots

Questions

Figure it Out — Digital Roots

Questions

In-text — Cryptarithms

Questions

Figure it Out — Divisibility, Conjectures and Cryptarithms

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Number Play?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~10 marks in Punjab (PSEB) exams

5-minute revision

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

•Divisibility tests: 2 (last digit even), 3 (digit sum), 4 (last two digits), 5 (last digit 0 or 5), 6 (by 2 and 3), 8 (last three digits), 9 (digit sum), 10 (ends in 0), 11 (alternating digit sum).
•The 3 and 9 tests work because every power of 10 leaves remainder 1 on division by 9; the 11 test works because powers of 10 alternate between remainder +1 and −1.
•For a composite divisor, split into coprime factors and apply both tests: 18 = 2×9, 36 = 4×9, 44 = 4×11, 15 = 3×5.
•Remainders can be added and subtracted independently of the original numbers; regroup if the total reaches the divisor.
•If every remainder is one less than its divisor, the number is one less than the LCM of the divisors.
•If all remainders are equal to r, the number is LCM + r.
•Digital root = remainder on division by 9, with a multiple of 9 giving 9 rather than 0.
•Digital roots of multiples of k cycle with length 9 ÷ HCF(k, 9): length 3 for k = 3 and 6, length 9 for k = 4.
•Any rearrangement of a multiple of 9 (or 3) is still a multiple of 9 (or 3), since the digit sum is unchanged.
•The product of k consecutive integers is always a multiple of k factorial: 2, 6, 24, 120 for k = 2, 3, 4, 5.
•In cryptarithms each letter is one digit, different letters are different digits, and no number starts with 0. Start at the units column.

Punjab (PSEB) 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

Where this shows up in the real world

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

Bank account numbers

Bank account numbers, Aadhaar and ISBN codes use check digits computed from exactly this kind of modular arithmetic to catch typing errors

Barcodes and UPI IDs verify themselves with a weighted di…

Barcodes and UPI IDs verify themselves with a weighted digit-sum check before a transaction is accepted

Casting out nines was the standard way accountants and cl…

Casting out nines was the standard way accountants and clerks checked long multiplication before calculators

Cryptography and computer hashing rest on remainder arith…

Cryptography and computer hashing rest on remainder arithmetic — the same modular ideas at much larger scale

Scheduling problems (when do three events on 3-

Scheduling problems (when do three events on 3-, 4- and 5-day cycles coincide?) are LCM-and-remainder problems in disguise

Exam strategy

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

Four question types recur. (1) Divisibility tests and tables — never divide, and for composite divisors state the coprime split first, since that alone often carries a mark. (2) Find the digits a and b — always convert the condition into a digit-sum equation and a last-digits condition, then tabulate every even (or allowed) value rather than guessing. (3) Always/sometimes/never true — give algebra for 'always', and give BOTH an example and a non-example for 'sometimes'; a single counter-example is enough for 'never'. (4) Cryptarithms — start at the units column, note whether the units digit forces a value (multiples of 5 end in 0 or 5; a repdigit answer like GGG factorises as 111 = 3 × 37), and always verify the finished sum. Show the reasoning line before the answer; most of the marks sit there.

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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Because every power of 10 leaves remainder 1 when divided by 9 — 10 = 9+1, 100 = 99+1, 1000 = 999+1. So each digit contributes just itself to the remainder, and the number leaves the same remainder as its digit sum. The same argument works for 3, since 3 also divides 9.

Because the powers of 10 alternate: 10 = 11 − 1 leaves −1, 100 = 99 + 1 leaves +1, 1000 = 1001 − 1 leaves −1, and so on. So each digit contributes with alternating sign, starting from + at the units place.

No. 2 and 6 share a common factor, so passing both is not enough — 18 is divisible by 2 and by 6 but not by 12. You must split into coprime factors: 12 = 4 × 3. The same care is needed for 36 (= 4 × 9, not 6 × 6) and 8.

9. The process 4 + 5 = 9 stops there, and a digital root is always a digit from 1 to 9. It corresponds to remainder 0 on division by 9 — that relabelling is the only quirk of digital roots.

Not usually. The test for 4 looks at the last two digits, so moving digits around changes the answer — 124 is a multiple of 4 but 142 is not. Only the tests that depend purely on the digit SUM (3 and 9) survive shuffling.

At the units column, because no carry comes into it. Look for a digit the units column forces: a multiple of 5 must end in 0 or 5; if the answer's units digit equals one of the operand's, only a few values fit; and repdigit answers like GGG factorise usefully (111 = 3 × 37, 444 = 4 × 111).

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