Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is
Answer & explanation
Answer: (d) 8
The units give 7 + 6 = 13 (carry 1), and the hundreds rise from 5 + 1 to 7, so the tens column must also carry: B + 4 = A + 10, i.e. A = B − 6. For 7A3 to be divisible by 3, 10 + A must be a multiple of 3, so A = 2 and B = 8.
- Units: 7 + 6 = 13 → write 3, carry 1.
- Hundreds: 5 + 1 + carry = 7, so the tens column must carry 1.
- Tens: B + 3 + 1 = A + 10 → A = B − 6; so B is 6, 7, 8 or 9 and A is 0, 1, 2 or 3.
- 7A3 divisible by 3 → 7 + A + 3 = 10 + A is a multiple of 3 → A = 2.
- B = 8. Check: 587 + 136 = 723, and 7 + 2 + 3 = 12.
Remember · In digit-sum puzzles, go column by column with carries, then use the divisibility rule to fix the digit.
Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·