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) ?
- (a) 2
- (b) 3 ✓ UPSC's answer
- (c) 4
- (d) More than 4
Why the answer is (b)
• In place-value form: PQR = 100P + 10Q + R, PS = 10P + S and PPT = 110P + T.
• PQR − PS = PPT gives 90P + 10Q + R − S = 110P + T, i.e. 10Q + R − S − T = 20P. With Q = 3: 30 + R − S − T = 20P.
• R − S − T is at most 9 − 1 − 2 = 6, so 20P ≤ 36 and P = 1. Then S + T = R + 10.
• With P = 1, Q = 3, T < 6 and all digits distinct and non-zero, T can be 2, 4 or 5:
• T = 2: S = R + 8 needs R = 1, which is already P — no solution.
• T = 4: S = R + 6 gives R = 2, S = 8 (R = 3 is already Q).
• T = 5: S = R + 5 gives R = 2, S = 7 and R = 4, S = 9 (R = 1 and 3 are taken).
• Checks: 132 − 18 = 114, 132 − 17 = 115, 134 − 19 = 115. So (R, S) has 3 possible values, option (b).
Why the other options are wrong
- (a) 2
- There are 3 valid pairs, not 2.
- (c) 4
- There are 3 valid pairs, not 4.
- (d) More than 4
- There are exactly 3 valid pairs, which is not more than 4.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2025, held on 25 May 2025. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.