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)

• The equation PQR - PS = PPT can be rewritten as PQR - PPT = PS, which simplifies to 100P + 10Q + R - (100P + 10P + T) = 10P + S.

• Substituting Q = 3, the equation becomes 30 + R - 10P - T = 10P + S, or R - T - S = 20P - 30.

• Since P, Q, R, S, T are distinct non-zero digits, P cannot be 3 (as Q=3). Also, P must be at least 1.

• If P=1, R - T - S = -10. Since R, T, S are digits 1-9, R - T - S = -10 implies T + S - R = 10. Possible sets for {T, S, R} with T<6: (T=1, S=9, R=0 invalid), (T=2, S=8, R=0 invalid), (T=2, S=9, R=1 invalid P=1), (T=3 invalid Q=3), (T=4, S=6, R=0 invalid), (T=4, S=7, R=1 invalid), (T=5, S=5 invalid distinct), (T=5, S=6, R=1 invalid). Wait, let's re-evaluate P=1. R - T - S = -10. Max T+S is 9+8=17, min R is 1. 1-17=-16. Min T+S is 1+2=3, max R is 9. 9-3=6. So R-T-S ranges from -16 to 6. -10 is possible. T+S-R=10. T<6. T=1: S-R=9 -> S=9, R=0 (invalid). T=2: S-R=8 -> S=9, R=1 (P=1, invalid distinct). T=3: S-R=7 -> S=8, R=1 (P=1, invalid) or S=9, R=2 (valid? P=1, Q=3, T=3 invalid Q=T). T=4: S-R=6 -> S=7, R=1 (P=1 invalid) or S=8, R=2 (valid? P=1, Q=3, T=4, S=8, R=2. Distinct? 1,3,2,8,4. Yes. T<6. Valid.) or S=9, R=3 (Q=3 invalid). T=5: S-R=5 -> S=6, R=1 (P=1 invalid) or S=7, R=2 (valid? P=1, Q=3, T=5, S=7, R=2. Distinct? 1,3,2,7,5. Yes. T<6. Valid.) or S=8, R=3 (Q=3 invalid) or S=9, R=4 (valid? P=1, Q=3, T=5, S=9, R=4. Distinct? 1,3,4,9,5. Yes. T<6. Valid.). So for P=1, we have 3 solutions: (R,S) = (2,8), (2,7), (4,9).

• If P=2, R - T - S = 10. T+S-R = -10 -> R - T - S = 10. R = T + S + 10. Since R <= 9, T+S must be <= -1, impossible for positive digits.

• If P >= 2, 20P - 30 >= 10. R - T - S = 10. R = T + S + 10. Since T, S >= 1, R >= 12, which is not a digit. Thus P must be 1.

• The valid pairs (R, S) are (2, 8), (2, 7), and (4, 9). There are 3 possible values.

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.