Consider the following multiplication problem:
(PQ) × 3 = RQQ, where P, Q and R are different digits and R ≠ 0.
What is the value of (P + R) ÷ Q?
Answer & explanation
Answer: (b) 2
Three times Q must end in Q, so Q is 0 or 5, and only Q = 5 works. Then 85 × 3 = 255 is the only fit, so P = 8, R = 2 and (P + R) ÷ Q = 10 ÷ 5 = 2.
- The units digit of 3 × Q must be Q, so Q = 0 or 5.
- Q = 0: P0 × 3 = R00 needs 30P = 100R, i.e. 3P = 10R — no digit P works with R ≠ 0.
- Q = 5: (10P + 5) × 3 = 100R + 55 gives 30P = 100R + 40, i.e. 3P = 10R + 4, so P = 8 and R = 2.
- 85 × 3 = 255, and (P + R) ÷ Q = (8 + 2) ÷ 5 = 2.
Remember · In digit puzzles, start with the units place — digits whose multiples end in themselves narrow things quickly.
Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·