Let PQR be a 3-digit number, PPT be a 3-digit number and PS be a 2-digit number, where P, Q, R, S, T are distinct non-zero digits. Further, PQR − PS = PPT. If Q = 3 and T < 6, then what is the number of possible values of (R, S)?
Answer & explanation
Answer: (b) 3
Writing the numbers in place-value form gives 30 + R − S − T = 20P, which forces P = 1 and S + T = R + 10. With distinct digits, Q = 3 and T below 6, only (R, S) = (2, 8), (2, 7) and (4, 9) work.
- PQR − PS = PPT means (100P + 10Q + R) − (10P + S) = 100P + 10P + T.
- Simplify: 90P + 10Q + R − S = 110P + T, so 10Q + R − S − T = 20P. With Q = 3: 30 + R − S − T = 20P.
- The left side is at most 30 + 9 − 1 − 1 = 37, so 20P = 20 and P = 1. Then S + T = R + 10.
- P = 1 and Q = 3 are used, so R, S, T are different digits from 2, 4, 5, 6, 7, 8, 9, with T = 2, 4 or 5.
- T = 2: S = R + 8 needs R = 1 (not allowed). T = 4: S = R + 6 gives R = 2, S = 8. T = 5: S = R + 5 gives (R, S) = (2, 7) or (4, 9).
- So (R, S) has 3 possible values.
- Check: 132 − 18 = 114, 132 − 17 = 115, 134 − 19 = 115.
Remember · For letter-digit equations, expand in place values, simplify, and use digit limits to fix the leading digit first.
Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·