Playing with Numbers
1. Generalised Form of Numbers
Any two-digit number can be written as 10a + b where a is the tens digit and b is the units digit.
Two-digit numbers: 37 = 10 × 3 + 7 (a = 3, b = 7) 89 = 10 × 8 + 9 (a = 8, b = 9)
Three-digit numbers: 100a + 10b + c where a is the hundreds digit, b is the tens digit, c is the units digit. 347 = 100 × 3 + 10 × 4 + 7
General rule: A number with digits d₁d₂d₃...dₙ = d₁ × 10ⁿ⁻¹ + d₂ × 10ⁿ⁻² + ... + dₙ × 10⁰
'Writing numbers in generalised form helps us PROVE divisibility rules and solve cryptarithmetic puzzles.'
2. Reversing Digits
Two-Digit Reversal
Original number: 10a + b Reversed number: 10b + a Sum: 10a + b + 10b + a = 11a + 11b = 11(a + b) — ALWAYS divisible by 11. Difference: (10a + b) — (10b + a) = 9a — 9b = 9(a — b) — ALWAYS divisible by 9.
Worked Example: The sum of a two-digit number and its reverse is 121. The digits differ by 3. Find the number.
Let the number be 10a + b. Sum = 11(a + b) = 121 → a + b = 11 Difference = a — b = 3 (or b — a = 3) Adding: 2a = 14 → a = 7, b = 4 Number = 74 (or 47).
Three-Digit Reversal
Original: 100a + 10b + c. Reversed: 100c + 10b + a. Difference: 99(a — c) — ALWAYS divisible by 99.
3. Divisibility Tests
| Divisor | Test | Example |
|---|---|---|
| 2 | Last digit is EVEN (0, 2, 4, 6, 8) | 3478 ✓ (ends in 8) |
| 3 | Sum of digits divisible by 3 | 471 → 4+7+1 = 12 ✓ |
| 4 | Last TWO digits divisible by 4 | 732 → 32 ÷ 4 = 8 ✓ |
| 5 | Last digit is 0 or 5 | 1275 ✓ |
| 6 | Divisible by BOTH 2 and 3 | 738 ✓ (ends in 8, sum = 18) |
| 7 | Double last digit, subtract from rest, check | 203 → 20 — 6 = 14 ✓ |
| 8 | Last THREE digits divisible by 8 | 5128 → 128 ÷ 8 = 16 ✓ |
| 9 | Sum of digits divisible by 9 | 486 → 4+8+6 = 18 ✓ |
| 10 | Last digit is 0 | 4500 ✓ |
| 11 | (Sum of odd position digits) — (sum of even position digits) = 0 or ±11 | 121 → (1+1) — 2 = 0 ✓ |
Proving Divisibility by 9: A number 100a + 10b + c = 99a + 9b + (a + b + c) 99a and 9b are divisible by 9. So the number is divisible by 9 IF (a + b + c) is divisible by 9.
4. Cryptarithmetic Puzzles
Letters represent UNIQUE digits. Each letter stands for the SAME digit throughout.
Worked Example: Solve AB + BA = 121 (A and B are digits, AB is a two-digit number).
10A + B + 10B + A = 11(A + B) = 121 A + B = 11 Pairs: (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2) The puzzle has MULTIPLE solutions unless additional conditions are given.
Worked Example (ICSE Style): Find A and B if 3AB + AB3 = 689.
3AB = 300 + 10A + B AB3 = 100A + 10B + 3 Sum = 300 + 10A + B + 100A + 10B + 3 = 303 + 110A + 11B = 689 110A + 11B = 386 11(10A + B) = 386 10A + B = 386/11 = 35.09... (not integer — no solution in integers)
5. Finding Missing Digits
Worked Example: 45 × □ = 315
315 ÷ 45 = 7. So the missing digit is 7.
Worked Example: 5□3 is divisible by 9. Find the missing digit.
5 + □ + 3 = 8 + □ must be divisible by 9. 8 + □ = 9 or 18. So □ = 1 (or 10 — not a digit). Answer: □ = 1. Number = 513.
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| 'Divisibility by 7 is same as by 3' | Divisibility by 7 has a DIFFERENT test: double last digit, subtract from rest |
| 'A number divisible by 2 and 4 is divisible by 8' | FALSE! 12 is divisible by 2 and 4 but NOT by 8 |
| 'Confusing digits and numbers in puzzles' | AB = 10A + B, NOT A × B. Always expand to generalised form |
| 'Forgetting that A and B must be DIFFERENT' | In cryptarithmetic, different letters represent DIFFERENT digits |
ICSE Exam Focus (4–6 marks)
- 2-mark questions: Divisibility tests — checking given numbers
- 3-mark questions: Generalised form — finding sums and differences
- 4-mark questions: Cryptarithmetic puzzles
- 6-mark questions: Word problems combining generalised form with equations
Self-Test
Q1. Check if 12345 is divisible by 11. A1. Odd positions: 1 + 3 + 5 = 9. Even positions: 2 + 4 = 6. Difference = 3. 3 ≠ 0 and 3 ≠ ±11. NOT divisible by 11.
Q2. Write 347 in generalised form. A2. 347 = 100 × 3 + 10 × 4 + 7.
Q3. Find the missing digit: 72□4 is divisible by 8. A3. Last three digits: 2□4. Check possibilities: 204 ÷ 8 = 25.5, 214 ÷ 8 = 26.75, 224 ÷ 8 = 28, 234 ÷ 8 = 29.25, 244 ÷ 8 = 30.5. □ = 2 gives 224 ÷ 8 = 28. Answer: □ = 2.
Q4. If AB × 3 = 1AB, find A and B. A4. AB = 10A + B. 3(10A + B) = 100 + 10A + B. 30A + 3B = 100 + 10A + B. 20A + 2B = 100. 10A + B = 50. So AB = 50. A = 5, B = 0. Check: 50 × 3 = 150 ✓.
Q5. The difference between a two-digit number and its reverse is 27. The sum of the digits is 9. Find the number. A5. 9(a — b) = 27 → a — b = 3. a + b = 9. Adding: 2a = 12 → a = 6, b = 3. Number = 63 or 36.
Q6. Prove that any three-digit number and its reverse have a difference divisible by 99. A6. Original = 100a + 10b + c. Reverse = 100c + 10b + a. Difference = 99a — 99c = 99(a — c). Since 99 is a factor, the difference is divisible by 99.
