The 5-digit number PQRST (all distinct digits) is such that T ≠ 0. P is thrice T. S is greater than Q by 4, while Q is greater than R by 3. How many such 5-digit numbers are possible?
Answer & explanation
Answer: (b) 4
P = 3T allows (T, P) = (1, 3), (2, 6) or (3, 9), and S = R + 7 allows (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9). Keeping all five digits different leaves 35291, 63072, 64182 and 94183 — four numbers.
- P = 3T with T ≠ 0 and P a single digit: (T, P) = (1, 3), (2, 6) or (3, 9).
- Q = R + 3 and S = Q + 4 = R + 7 ≤ 9, so R = 0, 1 or 2: (R, Q, S) = (0, 3, 7), (1, 4, 8) or (2, 5, 9).
- T = 1, P = 3: only (2, 5, 9) avoids repeats → 35291.
- T = 2, P = 6: (0, 3, 7) and (1, 4, 8) work → 63072 and 64182.
- T = 3, P = 9: only (1, 4, 8) works → 94183.
- Total: 4 numbers.
Remember · Tie the chained conditions to one variable (here R), list the few cases for each part, then strike out repeated digits.
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). ·