🔢 Class 8 Mathematics — Playing with Numbers Interactive Visual Revision Notes Aligned with the Class 8 Mathematics syllabus. Independently created study material. Not affiliated with or endorsed by NCERT.

📋 What is Inside?

📝 Section 1 — Classroom Notes

🌎 Real-World Connection: You are at a shop and the bill is ₹437. You want to quickly check if it is divisible by 9 without a calculator. Just add the digits: 4 + 3 + 7 = 14. Not divisible by 9. What if the bill was ₹432? 4 + 3 + 2 = 9. Yes! Divisibility rules like this help cashiers, accountants, and puzzle-solvers every day. Numbers have hidden patterns — this chapter reveals them!

🔢 Topic 1: General Form of a Number

Any 2-digit number = 10a + b

🔢 Topic 2: Reversing a 2-Digit Number

Number + Reverse = 11(a + b)

🔢 Topic 3: General Form of 3-Digit Numbers

Any 3-digit number = 100a + 10b + c

🔢 Topic 4: Divisibility by 2

🔢 Topic 5: Divisibility by 3

🔢 Topic 6: Divisibility by 5

🔢 Topic 7: Divisibility by 9

🔢 Topic 8: Divisibility by 10

🔢 Topic 9: Puzzles with Numbers

🔢 Topic 10: Fun Number Patterns

1 + 3 + 5 + ... + (2n−1) = n²

🎨 Section 2 — Easy Illustrations

Picture 1: General Form of 47

General Form of a 2-Digit Number 4 7 Tens (a) Ones (b) ↓ 47 = 4 × 10 + 7 × 1 General Form: 10a + b

Break any 2-digit number into tens and ones. Multiply and add!

Picture 2: Reversing a Number

Reverse a Number and Add Original 53 → Swap digits Reversed 35 53 + 35 = 88 (10a + b) + (10b + a) = 11(a + b) Sum is always a multiple of 11!

53 has a=5, b=3. So 11×(5+3) = 11×8 = 88. It always works!

Picture 3: Divisibility Rules Chart

Divisibility Rules at a Glance ÷ 2 Last digit is even 0, 2, 4, 6, 8 ÷ 3 Digit sum ÷ by 3 531: 5+3+1=9 ✓ ÷ 5 Ends in 0 or 5 735 ✓ 490 ✓ ÷ 9 Digit sum ÷ by 9 729: 7+2+9=18 ✓ ÷ 10 Ends in 0 only 570 ✓ 345 ✗ Remember: Divisible by 9 → always divisible by 3 too!

Five simple rules. Check the last digit or the digit sum. That is all!

Picture 4: Missing Digit Puzzle

Puzzle: Find the Missing Digit 3 A 5 + 2 B 1 6 1 6 Ones place: 5 + 1 = 6 ✓ No carry. Tens place: A + B = 1, so maybe A=0, B=1. Hundreds: 3 + 2 = 5. But answer says 6! So carry = 1.

Use place values and carry rules to find missing digits in addition puzzles.

Picture 5: Odd Number Pattern

Sum of Odd Numbers = Perfect Square! 1 1 = 1² = 1 1 + 3 1+3 = 4 = 2² = 4 1 + 3 + 5 1+3+5 = 9 = 3² = 9 1 + 3 + 5 + 7 = 16 = 4² = 16 Sum of first n odd numbers = n²

Add odd numbers one by one. The sum is always a perfect square!

✏️ Section 3 — Worked Examples

Example 1: Write 374 in General Form and Verify

Question: Write 374 in general form. Then check it.

1 The number 374 has 3 digits.
2 Hundreds digit = 3, Tens = 7, Ones = 4.
3 General form = 100a + 10b + c.
4 Put the values: 100×3 + 10×7 + 4.
5 Calculate: 300 + 70 + 4.
6 Total = 374. Matches the original!
374 = 100×3 + 10×7 + 4 ✓

a = 3, b = 7, c = 4

Example 2: Check if 2538 is Divisible by 3 and 9

Question: Is 2538 divisible by 3? By 9?

1 Find the sum of digits: 2 + 5 + 3 + 8.
2 Digit sum = 18.
3 Is 18 divisible by 3? 18 ÷ 3 = 6. Yes!
4 So 2538 is divisible by 3.
5 Is 18 divisible by 9? 18 ÷ 9 = 2. Yes!
6 So 2538 is divisible by 9 too.
2538 is divisible by both 3 and 9 ✓

Digit sum = 18. Divisible by 3 and 9.

Example 3: Find the Missing Digit

Question: Find A: 7A3 + 2_8 = 1021. The blank = 3.

So the sum is: 7A3 + 238 = 1021. Find A.

1 Ones place: 3 + 8 = 11. Write 1, carry 1.
2 Tens place: A + 3 + 1 (carry) = must give 2 in tens.
3 If A + 4 = 12 (write 2, carry 1), then A = 8.
4 Hundreds: 7 + 2 + 1 (carry) = 10. Write 0, carry 1.
5 Thousands: 0 + 0 + 1 (carry) = 1. Correct!
6 Check: 783 + 238 = 1021. It matches!
A = 8, so the number is 783

783 + 238 = 1021 ✓

📊 Section 4 — Cheat Sheet

TopicWhat It MeansKey Rule
General Form (2-digit)Break into tens and ones10a + b
General Form (3-digit)Hundreds, tens, ones100a + 10b + c
Reverse and AddSwap digits and addSum = 11(a+b)
Divisible by 2Last digit is even0, 2, 4, 6, 8
Divisible by 3Digit sum ÷ by 3Add all digits
Divisible by 5Last digit 0 or 5Only 0 or 5
Divisible by 9Digit sum ÷ by 9Stricter than 3
Divisible by 10Last digit is 0Only 0
Number PuzzlesFind missing digitsUse place value
Odd Number SumAdd first n odd numbersSum = n²
PalindromeSame forwards and backwards121, 1331
÷9 → ÷3Divisible by 9 means by 3 tooAlways true

📄 Section 5 — Take-Home Handout

📝 A. Fill in the Blanks

✓ B. True or False

✏️ C. Solve These

1. Write 862 in general form and verify.

Show your work here

2. Check if 4572 is divisible by 3 and by 9.

Show your work here

🔗 D. Match the Columns

Column AColumn B
1. Divisible by 2a. Last digit is 0
2. Divisible by 5b. Digit sum ÷ by 9
3. Divisible by 9c. Last digit is even
4. Divisible by 10d. Ends in 0 or 5

🎨 E. Find the Missing Digit

In the addition below, find the value of P:

4P6 + 327 = 813

Show your work here

✅ Detailed Solutions — Section 5 Handout

A. Fill in the Blanks — Answers

1. The general form of 58 is 10 × 5 + 8. (58 = 5 tens + 8 ones.)
2. A number is divisible by 2 if its last digit is even (0, 2, 4, 6 or 8).
3. The sum of digits of 729 is 18. (7 + 2 + 9 = 18.)
4. A number is divisible by 10 if it ends in 0.
5. The sum of the first 5 odd numbers is 25. (1 + 3 + 5 + 7 + 9 = 25.)

B. True or False — Answers

1. 435 is divisible by 5. — True. (Last digit is 5, so divisible by 5. ✓)
2. 123 is divisible by 9. — False. (1 + 2 + 3 = 6. 6 is not divisible by 9.)
3. If a number is divisible by 9, it is also divisible by 3. — True. (9 = 3 × 3, so every multiple of 9 is also a multiple of 3. ✓)
4. The reverse of 74 is 37. — False. (The reverse of 74 is 47, not 37.)
5. 1 + 3 + 5 + 7 + 9 = 25. — True. (= 25 = 5². ✓)

C1. Write 862 in general form and verify

1 Hundreds digit = 8, Tens digit = 6, Units digit = 2.
2 General form: 8 × 100 + 6 × 10 + 2.
3 Verify: 800 + 60 + 2 = 862. ✓
862 = 8 × 100 + 6 × 10 + 2

C2. Check if 4572 is divisible by 3 and by 9

1 Find the digit sum: 4 + 5 + 7 + 2 = 18.
2 Is 18 divisible by 3? 18 ÷ 3 = 6. Yes. So 4572 is divisible by 3.
3 Is 18 divisible by 9? 18 ÷ 9 = 2. Yes. So 4572 is divisible by 9.
4572 is divisible by both 3 and 9.

D. Match the Columns — Answers

1. Divisible by 2 → c. Last digit is even
2. Divisible by 5 → d. Ends in 0 or 5
3. Divisible by 9 → b. Digit sum ÷ by 9
4. Divisible by 10 → a. Last digit is 0

E. Find the Missing Digit: 4P6 + 327 = 813

1 We know 4P6 + 327 = 813.
2 So 4P6 = 813 − 327.
3 813 − 327 = 486.
4 The number is 4P6 = 486, so P = 8.
5 Check: 486 + 327 = 813. ✓
P = 8

🃏 Section 6 — Revision Flashcards

👉

🔢

What is the general form of a 2-digit number?

If tens digit = a and ones digit = b, then the number = 10a + b.

Example: 63 = 10×6 + 3

🔄

What happens when you reverse a 2-digit number and add?

The sum is always a multiple of 11.

Formula: 11(a + b)

✅

How do you check if a number is divisible by 3?

Add all the digits. If the sum is divisible by 3, the number is too.

Example: 531 → 5+3+1=9 → Yes!

💡

What is the divisibility rule for 9?

Add all the digits. If the sum is divisible by 9, the number is too.

If ÷9 works, ÷3 also works!

🔎

How do you check divisibility by 2?

Look at the last digit. If it is even (0, 2, 4, 6, 8), it is divisible by 2.

📈

What is the sum of first n odd numbers?

n²

1+3+5+7 = 4² = 16

🔧

How do you find a missing digit in an addition?

Work column by column from the ones place. Use carry rules and place values.

🎯

What is a palindrome number?

A number that reads the same forwards and backwards.

Examples: 121, 1331, 14641

🧠 Section 7 — Mind Map

Playing with Numbers General Form (2-digit) Number = 10a + b Example: 47 = 10×4 + 7 General Form (3-digit) 100a + 10b + c Reverse and Add Sum = 11(a + b) Always multiple of 11 ÷ 2, 5, 10 ÷2: last digit even ÷5: ends 0 or 5 ÷10: ends 0 ÷ 3 and ÷ 9 Add all digits ÷9 → also ÷3 Number Puzzles Find missing digits Use place value + carry Number Patterns Palindromes: 121, 1331 Sum of n odd = n² General Form Helps prove divisibility and solve puzzles

Hover on any branch to focus it. All key ideas connected together.