Prelims · 8 Mental Ability · 8.1 Number system and simplification
8.1.5 Number puzzles: digits, sums and differences
Write a number in place-value form, turn the words into one or two equations, and most digit puzzles solve in two lines. (Counting digits and page numbers is in MAB-NS-14; general word problems are in MAB-NS-12.)
- APPSC Group 2 · Prelims
- Standard level
- 6 min read
- Free
Core facts
- A two-digit number with tens digit a and units digit b is 10a + b; reversed, it is 10b + a.
- Sum with its reverse = 11(a + b).
- Difference from its reverse = 9(a − b).
- A three-digit number abc = 100a + 10b + c. abc − cba = 99(a − c).
- Consecutive numbers: x, x + 1, x + 2. Consecutive even or odd: x, x + 2, x + 4.
- The middle of an odd count of consecutive numbers = sum ÷ count.
- Digits in AP: for three digits, the middle digit = digit sum ÷ 3.
- In cryptarithms, different letters are different digits and a leading letter is never 0.
Example 1. A two-digit number has digit sum 9. Reversing the digits increases it by 27. Find the number. Solution. 9(b − a) = 27, so b − a = 3; a + b = 9. a = 3, b = 6. Answer: 36
Example 2. A two-digit number exceeds its reverse by 63, and its digits add to 11. Find it. Solution. 9(a − b) = 63, so a − b = 7; a + b = 11. a = 9, b = 2. Answer: 92
Example 3. Three consecutive odd numbers add to 87. Find them. Solution. Middle = 87 ÷ 3 = 29. Answer: 27, 29, 31
Explained
Type 1: number as a multiple of its digit sum. Expand and simplify to a relation between digits.
Example 4. A two-digit number is 7 times the sum of its digits, and it exceeds its reverse by 36. Find it. (a) 42 (b) 63 (c) 84 (d) 73 Solution. 10a + b = 7a + 7b gives a = 2b. 9(a − b) = 36 gives a − b = 4. So b = 4, a = 8. Answer: (c) 84
Type 2: digit product with reversal.
Example 5. The product of the digits of a two-digit number is 18. Subtracting 27 reverses the digits. Find the number. Solution. 9(a − b) = 27, so a = b + 3. (b + 3)b = 18 gives b = 3, a = 6. Answer: 63
Type 3: three-digit numbers.
Example 6. The digits of a three-digit number are in AP and add to 12. Subtracting 396 reverses the number. Find it. Solution. Middle digit = 12 ÷ 3 = 4. 99(a − c) = 396, so a − c = 4; a + c = 8. a = 6, c = 2. Answer: 642
Type 4: sum of all numbers formed by permuting digits. With n distinct digits, each digit sits in each place (n − 1)! times. Sum = (digit sum) × (n − 1)! × 11…1 (n ones).
Example 7. Find the sum of all three-digit numbers formed with 2, 4 and 7, each used once. Solution. (2 + 4 + 7) × 2! × 111 = 13 × 2 × 111. Answer: 2,886
Type 5: operator puzzles. Find the rule from two cases, confirm it on the third, then apply.
Example 8. If 2 * 5 = 14, 3 * 4 = 21 and 5 * 2 = 35, find 6 * 3. (a) 45 (b) 54 (c) 63 (d) 36 Solution. Rule a² + ab: 4 + 10 = 14, 9 + 12 = 21, 25 + 10 = 35. So 36 + 18 = 54. Answer: (b) 54
Example 9. For whole numbers, a ◊ b = a − b if a > b, a + b if a < b, and ab if a = b. Find (7 ◊ 4) ◊ (3 ◊ 3). Solution. 7 ◊ 4 = 3; 3 ◊ 3 = 9; 3 ◊ 9 = 3 + 9 (since 3 < 9). Answer: 12
Speed tips.
- Use 11(a + b) and 9(a − b) directly instead of building full equations.
- In MCQs, test the options against the easiest condition first (the reversal difference), then the second.
- An operator rule must fit every example given, not just two.
Common traps
- ✗ "abc − cba = 9(a − c)." ✓ For three digits it is 99(a − c).
- ✗ "Three consecutive odd numbers: x, x + 1, x + 2." ✓ They step by 2: x, x + 2, x + 4.
- ✗ "Sum of permutations of 2, 4, 7 = 13 × 111." ✓ Each digit appears in each place 2! = 2 times: 13 × 2 × 111 = 2,886.
- ✗ "Three-digit numbers with digit sum 6 = C(8, 2) = 28." ✓ The hundreds digit cannot be 0: C(7, 2) = 21.
- ✗ "The rule fits the first two examples, so it is right." ✓ Check it on every example.
Quick revision
- Two-digit: 10a + b; reverse 10b + a.
- Sum with reverse 11(a + b); difference 9(a − b).
- Three-digit difference with reverse 99(a − c).
- Middle of consecutive numbers = sum ÷ count.
- Digits in AP: middle digit = sum ÷ 3.
- Permutation sum = (digit sum) × (n − 1)! × 11…1.
- Digit-sum counting: shift the leading digit by 1, then C(s + 2, 2).
- Cryptarithm: start from the units digit and the size of the result.
- Operator rule must fit every example.
Go deeper: Deeper (beyond what APPSC Group 2 usually asks)
Deeper Advanced
Example 10. ABC × 3 = CCC, where A, B, C are different digits. Find A + B + C. Solution. CCC = 111 × C, so ABC = 37 × C. The units digit of 37C must be C: 7C ends in C only for C = 0 or 5. C = 5 gives 185 (185 × 3 = 555). Answer: 14
Example 11. AB × B = CCB, where A, B, C are different non-zero digits and C is odd. Find AB. Solution. B × B must end in B, so B is 1, 5 or 6. B = 1 gives a two-digit product, too small. B = 5: 45 × 5 = 225, but C = 2 is even. B = 6: 56 × 6 = 336. Answer: 56
- Without the "C is odd" condition both 45 and 56 work; read every condition.
Example 12. How many three-digit numbers have digit sum 6? (a) 15 (b) 18 (c) 21 (d) 28 Solution. Write a = a′ + 1 (a ≥ 1). Then a′ + b + c = 5 in non-negative whole numbers. Ways = C(5 + 2, 2) = C(7, 2). Answer: (c) 21
Example 13. Which three-digit number equals 11 times the sum of its digits? Solution. 100a + 10b + c = 11(a + b + c) gives 89a = b + 10c. The right side is at most 99, so a = 1 and b + 10c = 89: c = 8, b = 9. Answer: 198
Example 14. 13 × 17 × 19 × 23 = 96,5X7. Find X. Solution. Fastest: compare remainders on 9. 13, 17, 19, 23 leave 4, 8, 1, 5, and 4 × 8 × 1 × 5 = 160 leaves 7. The digit sum of 965X7 is 27 + X, which leaves X. So X = 7. Check by multiplying: 221 × 19 = 4,199; 4,199 × 23 = 96,577. Answer: 7
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