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UPSC Prelims 2025 CSAT, question 29

Topic: Number system

Q29

Let PQRPQR be a 3-digit number, PPTPPT be a 3-digit number and PSPS be a 2-digit number, where P,Q,R,S,TP,Q,R,S,T are distinct non-zero digits. Further, PQR−PS=PPTPQR-PS=PPT. If Q=3Q=3 and T<6T<6, then what is the number of possible values of (R,S)(R,S) ?

  1. (a) 2
  2. (b) 3Correct answer
  3. (c) 4
  4. (d) More than 4

Solution.

Answer: (b) 3

  1. Writing PQR−PS=PPTPQR - PS = PPT in place-value form gives (100P+10Q+R)−(10P+S)=100P+10P+T(100P + 10Q + R) - (10P + S) = 100P + 10P + T, which simplifies (with Q=3Q = 3) to 30+R−S=20P+T30 + R - S = 20P + T.
  2. Since R,S∈{1,…,9}R, S \in \{1, \dots, 9\}, we have R−S≤8R - S \le 8, so 20P+T≤3820P + T \le 38, which forces P=1P = 1 and S−R=10−TS - R = 10 - T; since T<6T < 6 and T∉{P,Q}={1,3}T \notin \{P, Q\} = \{1, 3\}, we have T∈{2,4,5}T \in \{2, 4, 5\}.
  3. Requiring P=1,Q=3,R,S,TP=1, Q=3, R, S, T to be distinct non-zero digits: T=2T = 2 (S−R=8S - R = 8) forces (R,S)=(1,9)(R, S) = (1, 9) which clashes with P=1P = 1; T=4T = 4 (S−R=6S - R = 6) gives (R,S)=(2,8)(R, S) = (2, 8) (since (1,7)(1, 7) and (3,9)(3, 9) clash with P=1,Q=3P=1, Q=3); T=5T = 5 (S−R=5S - R = 5) gives (R,S)=(2,7)(R, S) = (2, 7) and (4,9)(4, 9) (since (1,6)(1, 6) and (3,8)(3, 8) clash), yielding 33 valid pairs (R,S)(R, S).

Trap: Excluding the pairs where R=P=1R = P = 1 or R=Q=3R = Q = 3 reduces the count to 3 valid (R,S)(R, S) pairs: (2,8),(2,7),(2, 8), (2, 7), and (4,9)(4, 9).