Playing with Numbers — Class 8 Mathematics

1. Numbers in General Form

Any two-digit number 'ab' (where a is tens digit, b is ones digit) can be written as: 10a + b. A three-digit number 'abc' = 100a + 10b + c. This is the GENERAL FORM or expanded notation.

Examples

• 57 = 10×5 + 7
• 328 = 100×3 + 10×2 + 8 = 300 + 20 + 8
• 100a + 10b + c, where a,b,c are digits (a ≠ 0, 0≤b,c≤9)

Reversing Digits

A two-digit number = 10a + b. Its reverse = 10b + a. Sum: (10a+b)+(10b+a) = 11a+11b = 11(a+b). 'The sum of a two-digit number and its reverse is ALWAYS divisible by 11!' Difference: (10a+b)−(10b+a) = 9a−9b = 9(a−b). 'The difference is ALWAYS divisible by 9!'

Three-Digit Number Games

Take any three-digit number abc. Reverse it: cba. Subtract the smaller from larger: |(100a+10b+c)−(100c+10b+a)| = 99|a−c|. The middle digit ALWAYS cancels out. 'The result is always a multiple of 99. Such patterns reveal the hidden STRUCTURE of numbers.'

2. Divisibility Tests — Complete Reference

Divisibility by 11 — Detailed

For 2728: Odd positions (1st, 3rd): 2+2=4. Even positions (2nd, 4th): 7+8=15. Difference = |15−4| = 11 → divisible by 11 ✓. Check: 2728÷11 = 248.

Divisibility by 7 (Beyond Syllabus but Useful)

Double the last digit, subtract from the rest. Repeat until you get a small number. If it's divisible by 7, the original is too. Example 343: 34−(2×3) = 28 → 28÷7=4 ✓.

3. Cryptarithmetic — Letters for Digits

In these puzzles, digits are REPLACED by letters. Each letter stands for a UNIQUE digit (0-9). The same letter always represents the same digit. Decode using divisibility rules and carry-over logic.

Example 1 — Simple Addition

AB + BA = 77. Find A and B. (10A+B)+(10B+A) = 11A+11B = 11(A+B) = 77. A+B = 7. Possible pairs: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). 'Unless additional constraints are given, multiple solutions exist. Common additional constraint: A > B.'

Example 2 — with Carry

1A × A = 9A. Find A. 1A = 10+A. (10+A)×A = 9A → 10A+A² = 90+A → 10A−A+A² = 90 → 9A+A² = 90 → A(A+9) = 90. A=6: 6×15=90 ✓. Check: 16×6 = 96 ✓.

Example 3 — Two-Digit Multiplication

AB × 3 = CAB. 3(10A+B) = 100C+10A+B. 30A+3B = 100C+10A+B. 20A+2B = 100C. 10A+B = 50C. Since A,B,C are single digits and C≠0: If C=1: 10A+B=50 → A=5, B=0. Check: 50×3=150 ✓ (C=1,A=5,B=0).

4. Why Divisibility Tests Work — The Proof

Divisibility by 3: A three-digit number abc = 100a+10b+c = 99a+a+9b+b+c = 99a+9b+(a+b+c) = 9(11a+b)+(a+b+c). Since 9(11a+b) is divisible by 3, the whole number is divisible by 3 IF AND ONLY IF (a+b+c) is divisible by 3. 'This is why we only need to check the SUM OF DIGITS. The rest of the number — 99a+9b — is ALREADY a multiple of 3.'

Divisibility by 9: Same proof. 9(11a+b) is always divisible by 9. So the number is divisible by 9 iff (a+b+c) is divisible by 9.

5. Number Patterns — Fun Explorations

• Palindromic numbers: 121, 1331, 12321. The reverse equals the original.
• Kaprekar's Constant (6174) : For any 4-digit number (not all digits same), sort descending and ascending, subtract. Repeat. You ALWAYS reach 6174 within 7 steps.
• Cubes ending in the same digit: 4³=64 (ends in 4). 5³=125 (ends in 5). 6³=216 (ends in 6). 9³=729 (ends in 9). 'Numbers ending in 4, 5, 6, 9, or 0 have cubes ending in the SAME digit.'

6. Common Mistakes

1. Divisibility by 3 and 9 confusion: For 3, the sum must be divisible by 3. For 9, the sum must be divisible by 9. 'A number divisible by 9 is ALWAYS divisible by 3. But NOT vice versa.'
2. Using digits that repeat in cryptarithmetic: Each letter = unique digit.
3. Divisibility by 4: Check the last TWO digits, NOT just the last one. 124 is divisible by 4 (24÷4=6). 122 is not (22÷4=5.5).

7. AP Exam Focus

Key Exam Tips

• For divisibility questions: STATE the test. SHOW the calculation. WRITE the conclusion.
• For 'find digits A, B, C' questions: use the general form (10A+B etc.). Set up equations. Use divisibility where helpful.
• Divisibility by 11 is the MOST frequently tested in 4-5 mark questions. Practice the alternating sum method until it's automatic.

Quick Self-Test

1. Is 392 divisible by 8? (Answer: Last 3 digits: 392. 392÷8=49. Yes.)
2. Is 143 divisible by 11? (Answer: (1+3)−4=0. Yes. 143÷11=13.)
3. Write 75 in general form. (Answer: 10×7+5.)
4. A two-digit number and its reverse sum to 33. Find the number(s). (Answer: 11(A+B)=33 → A+B=3. Possible: 12, 21.)
5. Is 519 divisible by 3? (Answer: 5+1+9=15. 15÷3=5. Yes.)

Cryptarithmetic — More Challenging Puzzles

Example — AB × AB = 4B1, find A and B. (10A+B)² = 400+10B+1 = 401+10B. The square of a two-digit number ending in B gives a result ending in 1. Only numbers ending in 1 or 9 have squares ending in 1. Try 21²=441. 21: A=2,B=1. LHS: (21)²=441. RHS: 401+10(1)=411 ≠ 441. Try 29²=841. 841: A=2,B=9. RHS: 401+10(9)=401+90=491 ≠ 841. Try other possibilities through systematic checking.

Example — Find the digit A: 31A + 1A3 = 477. (310+A)+(100+10A+3) = 477 → 413+11A = 477 → 11A = 64 → A is not a single digit. Try: 31A means 3 hundreds, 1 ten, A ones = 310+A. 1A3 = 100+10A+3. Sum = 413+11A = 477. 11A = 64. Not a digit. This problem has no solution — a valid conclusion.

Deeper Divisibility — Divisible by Composite Numbers

To check divisibility by 12: check divisibility by 3 AND 4 (since 12=3×4, and 3,4 are coprime). For 14: check 2 AND 7. For 15: check 3 AND 5. For 18: check 2 AND 9. 'A number is divisible by a composite number if it is divisible by EACH of its COPRIME factors. Do NOT use 6 = 2×3 to check 2 and 6 separately — 2 and 6 are not coprime.'

Divisibility by 7, 13, and Other Primes

By 7: Double the last digit, subtract from the rest. Repeat. 343: 34−(2×3)=28. 28÷7=4 ✓. By 13: Multiply last digit by 4, ADD to the rest. 299: 29+(4×9)=65. 65÷13=5 ✓. By 17: Multiply last digit by 5, subtract from the rest. By 19: Multiply last digit by 2, add to the rest.

Number Patterns — Fascinating Regularities

• 111111111² = 12345678987654321 (palindromic pyramid — works up to nine 1s).
• 37×3 = 111. 37×6 = 222. 37×9 = 333... 37×27 = 999. '37 multiplied by multiples of 3 gives repeated digits.'
• 12345679×9 = 111111111 (the digit 8 is missing — that's deliberate!).
• 1089 trick: Take any 3-digit number where first and last digits differ by ≥2. Reverse it. Subtract smaller from larger. Reverse the result. Add. You always get 1089.

Magic of 9

Cast out nines — a quick check for arithmetic errors. Add the digits of a number. If the sum is ≥9, add again. The single-digit result is the 'digital root.' For correct addition/multiplication, the digital roots must match. 'Casting out nines catches ~90% of arithmetic errors. It was used by accountants before calculators.'

AP Exam Tips for Divisibility

• 'Is 27648 divisible by 4?' → Check last 2 digits: 48÷4=12. YES.
• 'Is 531 divisible by 6?' → Check 2: last digit is odd → NO. Don't need to check 3.
• For divisibility by 11, the DIFFERENCE must be 0 or a multiple of 11 (11, 22, 33...).