Certain 3-digit numbers have the following characteristics:
- 1.All the three digits are different.
- 2.The number is divisible by 7.
- 3.The number on reversing the digits is also divisible by 7.
How many such 3-digit numbers are there?
Answer & explanation
Answer: (b) 4
If a number and its reverse are both multiples of 7, so is their difference, 99 × (first digit − last digit). Since 99 is not a multiple of 7, the first and last digits must differ by 7 — giving just 168, 861, 259 and 952.
- Let the number be 100a + 10b + c; its reverse is 100c + 10b + a.
- Their difference is 99(a − c). Both are multiples of 7 and 99 is not, so a − c must be a multiple of 7.
- The digits are all different, so a − c = ±7: (a, c) = (1, 8), (8, 1), (2, 9), (9, 2) or (7, 0). The last fails: 7b0 is divisible by 7 only if b = 0 or 7, which repeats a digit.
- Find the middle digit in each case: 1b8 → 168 = 7 × 24; 8b1 → 861 = 7 × 123; 2b9 → 259 = 7 × 37; 9b2 → 952 = 7 × 136.
- Each case has exactly one suitable middle digit, so there are 4 such numbers.
- Check: 861 and 952 are the reverses of 168 and 259, and all four have three different digits.
Remember · A 3-digit number and its reverse differ by 99 × (first − last digit); this turns 'both divisible by 7' into an end-digit condition.
Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·